content
stringlengths 228
999k
| pred_label
stringclasses 1
value | pred_score
float64 0.5
1
|
|---|---|---|
Android Build an Interactive Story App The Model-View-Presenter Pattern Creating a Data Model
Jorge Ponce
Jorge Ponce
2,003 Points
Next, we need getter and setter methods for our member variable. Let's start with the getter method. Add a method to get
i dont now try to make the best method and everthing error error help with this
Spaceship.java
public class Spaceship {
public String shipType;
public String getshipType() {
return shipType;
}
}
1 Answer
andren
andren
28,469 Points
The only issue with your code is the name of the getter method. The name of the getter has to be written in camelCase, meaning that every word other than the first one is capitalized. So instead of getshipType the name should be getShipType.
If you fix that name like this:
public class Spaceship {
public String shipType;
public String getShipType() {
return shipType;
}
}
Then you'll be able to pass task 3.
Jorge Ponce
Jorge Ponce
2,003 Points
Thanks i dont see the letter small thanks thanks.
|
__label__pos
| 0.985993 |
OML Search
Geometry Worksheets - Surface of Sphere
Objective: I know how to calculate the surface area of a sphere.
The surface area of a sphere is given by the formula:
V = 4πr2
If you are given the diameter, remember to first divide the diameter by 2 to get the radius before using the formula.
Read this lesson on the surface area of spheres if you need additional help.
Fill in all the gaps, then press "Check" to check your answers. Use the "Hint" button to get a free letter if an answer is giving you trouble. You can also click on the "[?]" button to get a clue. Note that you will lose points if you ask for hints or clues!
Find the surface areas of the following spheres.
Take π = 3.14
Give your answer to the nearest tenths.
(Remember to include the relevant unit in your answer)
diameter = 4 cm
surface area =
radius = 10 ft
surface area =
diameter = 36 m
surface area =
radius = 5 in
surface area =
diameter = 43 mm
surface area =
radius = 21 yd
surface area =
radius = 11 in
surface area =
diameter = 14 yd
surface area =
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We hope that the free math worksheets have been helpful. We encourage parents and teachers to select the topics according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles.
OML Search
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
[?] Subscribe To This Site
XML RSS
follow us in feedly
Add to My Yahoo!
Add to My MSN
Subscribe with Bloglines
|
__label__pos
| 0.830695 |
Kevin Kevin - 11 months ago 43
Node.js Question
Something weird about express session store
If I store an object in session like this:
user.name = "Kelvin"; // user is an object pass by "mongoose" findOne's callback.
req.session.user = user;
console.log(req.session.user.name); // Kelvin
and after that, I access "user" in other routes of express:
app.get("/somepath", function(req, resp) {
console.log(req.session.user.name); // undefined
});
I want to know why req.session.user.name is undefined besides the function I set it?
Answer Source
After looking into mongoose source code I can say that this is because how mongoose models and session works. When you call req.session.user = user then req.session.user points to the object but actually the data needs to be stored somewhere (like memory or Redis).
In order to do that Express calls JSON.stringify(sess) and string is stored in memory. Here's where mongoose enters. Models are constructed in such a way, that when you stringify them only attributes predefined in Schema are included. So when you set req.session.user = user Express stringifies user (you loose name attribute) and saves the data. When you call req.session.user in another route Express parses stringified data and you obtain object without name attribute.
Now how to fix this? There are several ways.
• Add name attribute to the Schema;
• Define new literal object for user:
var newuser = { id: user.id, name : "Kelvin", pwd: user.pwd, etc. }; req.session.user = newuser;
• Store name in another field: req.session.name = "Kelvin"; (the best solution imho)
By the way: You shouldn't hold the user object in session. What if some other user like administrator makes changes to the user object? You won't see them at all. I advice holding only the id of the user in session and make custom middleware to load user from DB and store it in request object.
|
__label__pos
| 0.952555 |
PHP Arrays Tutorial
by Yogesh Khanna 20-Jan-16
0 1143
An array is a data structure that saves one or more same type of values in a single value. For instance, if you like to save 100 numbers, then instead of showing 100 variables, it's simple to define an array of 100 lengths.
There are three different types of arrays and each array value is accessed using an ID, which is known as array index.
Numeric array - An array with a numeric index. It stores and accesses the values in linear form.
Associative array- An array with strings as index. This stores element values in the association with key values rather than in a strict linear index order.
Multidimensional array - An array containing one or more arrays and values is accessed using numerous indices.
PHP Built - In array functions are given in the function reference PHP Array Functions
<?php
$school= array("library","classroom", "sportsroom","playground");
echo "In our school" . $school[0] . ", " . $school[1] . ", " . $school[12]. " and " . $school[3] . ".";
?>
In our school library, classroom, sportsroom and playground.
Please see above example showing only one variable ($school) but stored multiple values i.e. library, classroom, sportsroom, playground. This is the benefit of PHP Array.
PHP Arrays Tutorial
Indexed Arrays : Indexed arrays can store numbers, strings but their index will be represented by numbers only and index only auto starts from zero value.
<?php
$numbers = array(1, 2, 3, 4, 5, 6, 7);
foreach($numbers as $value)
{
echo "Value is $value";
}
?>
Associative Arrays: Associative arrays use named keys that you assign to them, its similar as indexed arrays in functionality but in terms of index it's different.
<?php
$age= array("john" => 45, "raj" => 52, "peter" => 37);
echo "Age of john is ". $age['john'] . "<br />";
echo "Age of raj is ". $age['raj'] . "<br />";
echo "Age of peter is ". $age['peter'] . "<br />";
?>
Age of John is 45
Age of Raj is 52
Age of Peter is 37
Multidimensional Arrays: In multidimensional array, you can create array under array means like sub array.
It's very interesting to use.
<?php
$marks = array(
"john" => array(
"english" => 50,
"maths" => 30,
"science" => 40
"history" => 39
),
"raj" => array (
"physics" => 30,
"maths" => 95,
"science" => 48
"history" => 39
),
"peter" => array (
"physics" => 31,
"maths" => 22,
"science" => 78
"history" => 85
)
"sunny" => array (
"physics" => 62,
"maths" => 75,
"history" => 80
)
);
echo "John marks in english : " ;
echo $marks['john']['english'] . "<br />";
echo "Raj marks in maths : ";
echo $marks['raj']['maths'] . "<br />";
echo "Peter marks in science : " ;
echo $marks['peter']['science'] . "<br />";
echo "Sunny marks in history : " ;
echo $marks['peter']['history'] . "<br />";
?>
Result:
John marks in english : 50
Raj marks in maths : 95
Peter marks in science : 78
Sunny marks in history : 80
Share:
Discount Coupons
Comments
Waiting for your comments
Leave a Reply
|
__label__pos
| 0.721998 |
blob: 945a976499987b58d7274652f70fb0a4dad66f7b [file] [log] [blame]
// Copyright (c) 2017, the Dart project authors. Please see the AUTHORS file
// for details. All rights reserved. Use of this source code is governed by a
// BSD-style license that can be found in the LICENSE file.
/// @assertion
/// void lockSync([
/// FileLock mode = FileLock.exclusive,
/// int start = 0,
/// int end = -1
/// ])
/// Synchronously locks the file or part of the file.
/// . . .
/// Locks the byte range from start to end of the file, with the byte at position
/// end not included. If no arguments are specified, the full file is locked, If
/// only start is specified the file is locked from byte position start to the
/// end of the file, no matter how large it grows. It is possible to specify an
/// explicit value of end which is past the current length of the file.
///
/// @description Checks that if no arguments are specified, the full file is
/// locked.
/// @author [email protected]
import "dart:async";
import "dart:io";
import "../../../Utils/expect.dart";
import "../file_utils.dart";
import "lock_check_1_lib.dart";
void check(int fLen) {
File file = getTempFileSync();
asyncStart();
var rf = file.openSync(mode: FileMode.write);
rf.writeFromSync(new List.filled(fLen, 1));
rf.lockSync(FileLock.exclusive);
String eScript = Platform.script.toString();
var tests = [
() => checkLocked(eScript, rf.path),
() => checkLocked(eScript, rf.path, fLen, fLen + 10)
];
Future.forEach(tests, (Function f) => f()).whenComplete(() {
asyncEnd();
rf.unlockSync();
rf.closeSync();
file.deleteSync();
});
}
runMain() {
check(10);
check(100);
check(1000);
}
main(List<String> args) {
if(args.length > 0)
runProcess(args);
else {
runMain();
}
}
|
__label__pos
| 0.828633 |
Write jpg files over network
hello,
v4 writes from 2 station to a directory on a third pc. i check with dir node how many files are in this directory and generate a new filename with an ongoing numbering.
how can I setup this pcs for the best performance? giga ethernet, is there an advantage integrate the shared folder as a network folder.
and how can I prevent that the file writing interrupts the loop for a few milliseonds?
regards
and how can I prevent that the file writing interrupts the loop for a few milliseonds.
Have a look at starting a second instance of vvvv and sharing the texture through SharedTexture or Screenshot.
yes giga ethernet will improve write time. doubt any advantage how you address that share. having a ‘quiet’ network would help too. fast hard drives won’t help.
as GG says, you’ll need to share the texture to another thread and save to prevent vvvv freeze.
Or use screen shot to capture the other instances output instead of shaded texture, if thats an issue, another idea is to write locally and then use copy to move the file across the network.
|
__label__pos
| 0.816904 |
Go Down
Topic: New Modbusmq project: Testers needed! (Read 11026 times) previous topic - next topic
marioquark
i'm working on Modbusmq, a modification of the libmodbus library (https://launchpad.net/libmodbus by Stephane Raimbault) for the Arduino platform. for now, it's a Modbus TCP slave.
any tester and debugger is appreciated to help me makin' a stable program.
that's the code: http://bazaar.launchpad.net/~marioquark-yahoo/modbusmq/modbusmq/files
thank you,
marioquark
marioquark
i am locked:
timing in TCP management doesn't work as it sould: replies are sent with a delay i don't want. this make timeout occur.
probably this happens due to a too big transmission buffer: i can't resize it! :( nor i can set the TCP_NODELAY option (library doesn't offer this).
any suggestion?
please help me, i think my project could be very useful!
marioquark
my libraty is finished and it seems to work perfectly... ;D
Arnonh
nice work
can you please explain a bit a about how to use it?
marioquark
ok. i'll write down a few lines, then i'll make a wiki for it:
modbusmq is a ModBus ( http://en.wikipedia.org/wiki/Modbus ) TCP slave (server): briefly, it holds a data map, which can be queried by Modbus clients (PC or other) with a standard protocol.
requirements:
- Arduino Ethernet Shield
usage:
Arduino side
- paste the code in a new project in the Arduino IDE
- change the network params, if needed: IP, Gateway.
- change the slave Address, if needed
- set or unset the DEBUG flag to have debug messages through serial line
PC side:
- start a modbus client to query Arduino IP, with the correct slave modbus address (there are many freeware, eg: ModScan)
TODO:
- test two slaves in the same network.
- test two masters (clients) for the same slave
chris.moe
Hello,
I just tested the Modbusmq and so far it works fine. But I'm struggling a little with all the pointers.
I already managed to send a modbus command from a PC, which sets different bits to true.
But now I want to switch relais, which are connected to a PCF8574. For this I normally use the command:
PCF8574Write(B11111110);
Now I tried to convert the "mb_mapping->tab_coil_status" to a value which I can send with PCF8574Write. But without success. It is too long ago, that I have done somethink with pointers. I tried the following:
for (n=0;n<8;n++)
{
PCFStatus1[n] = byte(*(mb_mapping->tab_coil_status + n));
DEBUG_PRINTLN(n);
delay(200);
}
This seems to me not very efficient and it is also not working.
Can someone help?
BR,
Chris
marioquark
PCF8574Write(B11111110);
what command is it? a function from you? from a library?
if you have to copy bytes to/from a vector, use the memcpy() standard C function. reference: http://www.cplusplus.com/reference/clibrary/cstring/memcpy/
maybe
memcpy(PCFStatus1, mb_mapping->tab_coil_status, n);
bye
quark
chris.moe
Sounds very promising!
But I still have some problems. I tried the following:
byte PCFStatus1;
memcpy(PCFStatus1, mb_mapping.tab_coil_status, 8);
PCF8574Write(PCFStatus1);
Here the function:
//Write data to PCF8574
void PCF8574Write(byte _data )
{
Wire.beginTransmission(PCF8574_ADDR);
Wire.send(_data);
Wire.endTransmission();
}
But I get the following error:
error: invalid conversion from 'uint8_t' to 'void*'
I do not understand it, why void*?
Any idea?
chris.moe
Hello,
can you give some explanations for the following (is there already some wiki):
If data are correct received, how to use them in the main loop?
I assume there are the following buffers/memorys:
mb_mapping->nb_coil_status
mb_mapping->nb_input_status
mb_mapping->nb_holding_registers
mb_mapping->nb_input_registers
I have also seen some functions like "get_byte_from_bits".
Is there somewhere a more detailed explanation? Maybe in the libmodbus library?
For my problem to write data to the PCF8574 can I use the "get_byte_from_bits" function? But how exactly?
Thank you for your help!
chris.moe
Got it running with the following code:
int PCFStatus1;
// Gets the byte value from many input/coil status.
PCFStatus1 = get_byte_from_bits(mb_mapping.tab_coil_status, 0, 8);
PCF8574Write(byte(PCFStatus1));
Jason Goldenberg
I am having trouble getting the modbus sketch to even work on my arduino mega and ethernet shield. Where do I edit the network settings. I tried following the example, but I cannot even ping the arduino. Is there some setup for the wiznet chip? Any help would be appreciated. I am trying to get RC servo control using modbus words.
Thanks
hgs75
Hello, it works fine (in my initial test), but the DEBUG flag has to be disabled because if it is enables the Arduino hangs :-/, so I can't debug or use the serial port for interchange data.
Any idea to make it work with debug flag?.
chris.moe
I got it working with the DEBUG flag, but it starts working only after having the serial monitor started.
I made the following network settings:
const byte gateway[] = { 192, 168, 2, 254 };
const byte mac[] = { 0xDE, 0xAD, 0xBE, 0xEF, 0xFE, 0xED };
const byte ip[] = { 192, 168, 2, 100 }; // Arduino has this IP
You have to use the same subnet (192.168.2.x) as your computer uses.
BR,
Chris
Loren.R
Can anyone give me an example how to read an input on the Arduino, and map the I/O status to be polled by the Master?
I am not sure how the mapping is handled.
LAVco
wow great to see MODBUS/TCP functional. this library will certainly come in handy.
Common sense is not so common...
Go Up
Please enter a valid email to subscribe
Confirm your email address
We need to confirm your email address.
To complete the subscription, please click the link in the email we just sent you.
Thank you for subscribing!
Arduino
via Egeo 16
Torino, 10131
Italy
|
__label__pos
| 0.949027 |
RK3588移植-ffmpeg交叉编译
1.下载ffmpeg
git clone https://git.ffmpeg.org/ffmpeg.git ffmpeg
2.交叉编译
进入下载目录，将ffmpeg编译成arm64平台的版本，编译后的文件存放于./instal_arm64中。
sudo ./configure --prefix=./instal_arm64 --enable-shared --disable-static --enable-cross-compile --arch=arm64 --disable-stripping --target-os=linux --cc=aarch64-linux-gnu-gcc
验证：进入instal_arm64/lib文件夹，使用file命令查看文件格式，若显示为ARM aarch64即成功。
在这里插入图片描述
3.修改cmakelist.txt
需要将交叉编译好的lib/include 文件夹导入项目中，修改cmakelist.txt如下图所示。
在这里插入图片描述
在这里插入图片描述
为方便复制这里将改动内容的文字贴在下面，路径需要改成自己的下载路径
#ffmpeg
# 设置ffmpeg依赖库及头文件所在目录，并存进指定变量
set(ffmpeg_libs_DIR /home/yi/Downloads/ffmpeg/instal_arm64/lib)
set(ffmpeg_headers_DIR /home/yi/Downloads/ffmpeg/instal_arm64/include)
add_library( avcodec SHARED IMPORTED)
add_library( avfilter SHARED IMPORTED )
add_library( swresample SHARED IMPORTED )
add_library( swscale SHARED IMPORTED )
add_library( avformat SHARED IMPORTED )
add_library( avutil SHARED IMPORTED )
#指定所添加依赖库的导入路径
set_target_properties( avcodec PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libavcodec.so )
set_target_properties( avfilter PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libavfilter.so )
set_target_properties( swresample PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libswresample.so )
set_target_properties( swscale PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libswscale.so )
set_target_properties( avformat PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libavformat.so )
set_target_properties( avutil PROPERTIES IMPORTED_LOCATION ${ffmpeg_libs_DIR}/libavutil.so )
# 添加头文件路径到编译器的头文件搜索路径下，多个路径以空格分隔
include_directories(${ffmpeg_headers_DIR})
link_directories(${ffmpeg_libs_DIR})
--------省略若干字。。。。。----------------------
target_link_libraries(rknn_yolov5_demo
${RKNN_RT_LIB}
${RGA_LIB}
${OpenCV_LIBS}
${ffmpeg_libs_DIR}
avcodec avformat avutil swresample swscale swscale avfilter
)
4.将lib文件复制到install目录下的lib目录
这里也许可以只复制项目中需要的lib文件.
sudo cp ~/Downloads/ffmpeg/instal_arm64/lib/* install/rknn_yolov5_demo_Linux/lib/
5.测试文件
这里用一个ffpeg解析rtsp流的案例来测试是否成功。因为源码篇幅过长，放到本文最后。程序大意是：读取rtsp流，将rtsp流解析成帧，并且输入到输出文件，同时统计帧数。
6.运行测试样例
sudo ./build-rk3588…
adb push …
这里省略
最终可得到如图所示的结果，表示编译运行成功。
在这里插入图片描述
7.错误
运行时出现了Failed to resolve hostname …无法解析域名的错误。
在这里插入图片描述
解决方案：
有两种可能的原因，1.没有配置域名服务器114.114.114.114 。2.开发板没有网络。
1. 配置域名服务器
在这里插入图片描述
2. 连接网络
可以尝试ping www.baidu.com，若没有问题的话，则不是网络问题。
1、安装nmcli
sudo apt-get install nmcli
2、查看网络设备
sudo nmcli dev
3、开启wifi
sudo nmcli r wifi on
4、扫描wifi
sudo nmcli dev wifi
5、连接wifi
sudo nmcli dev wifi connect “wifi名” password “密码”
n.测试文件源码
案例源码：
#ifndef INT64_C
#define INT64_C(c) (c ## LL)
#define UINT64_C(c) (c ## ULL)
#endif
extern "C" {
/*Include ffmpeg header file*/
#include <libavformat/avformat.h>
#include <libavcodec/avcodec.h>
#include <libswscale/swscale.h>
#include <libavutil/imgutils.h>
#include <libavutil/opt.h>
#include <libavutil/mathematics.h>
#include <libavutil/samplefmt.h>
}
#include <iostream>
using namespace std;
int main(void)
{
AVFormatContext* ifmt_ctx = NULL, * ofmt_ctx = NULL;
const char* in_filename, * out_filename;
//in_filename = "rtsp://admin:WY@[email protected]/h264/ch1/main/av_stream";
in_filename="rtsp://wowzaec2demo.streamlock.net/vod/mp4:BigBuckBunny_115k.mp4";
out_filename = "output.flv";
avformat_network_init();
AVDictionary* avdic = NULL;
char option_key[] = "rtsp_transport";
char option_value[] = "tcp";
av_dict_set(&avdic, option_key, option_value, 0);
char option_key2[] = "max_delay";
char option_value2[] = "5000000";
av_dict_set(&avdic, option_key2, option_value2, 0);
AVPacket pkt;
AVOutputFormat* ofmt = NULL;
int video_index = -1;
int frame_index = 0;
int i;
//打开输入流
int ret;
if ((ret = avformat_open_input(&ifmt_ctx, in_filename, 0, &avdic)) < 0)
{
cout<<"Could not open input file."<<endl;
goto end;
}
if ((ret = avformat_find_stream_info(ifmt_ctx, 0)) < 0)
{
cout<<"Failed to retrieve input stream information"<<endl;
goto end;
}
//nb_streams代表有几路流
for (i = 0; i < ifmt_ctx->nb_streams; i++)
{
if (ifmt_ctx->streams[i]->codecpar->codec_type == AVMEDIA_TYPE_VIDEO)
{
//视频流
video_index = i;
cout << "get videostream." << endl;
break;
}
}
av_dump_format(ifmt_ctx, 0, in_filename, 0);
//打开输出流
avformat_alloc_output_context2(&ofmt_ctx, NULL, NULL, out_filename);
if (!ofmt_ctx)
{
printf("Could not create output context\n");
ret = AVERROR_UNKNOWN;
goto end;
}
ofmt = ofmt_ctx->oformat;
for (i = 0; i < ifmt_ctx->nb_streams; i++)
{
AVStream* in_stream = ifmt_ctx->streams[i];
AVStream* out_stream = avformat_new_stream(ofmt_ctx, in_stream->codec->codec);
if (!out_stream)
{
printf("Failed allocating output stream.\n");
ret = AVERROR_UNKNOWN;
goto end;
}
//将输出流的编码信息复制到输入流
ret = avcodec_copy_context(out_stream->codec, in_stream->codec);
if (ret < 0)
{
printf("Failed to copy context from input to output stream codec context\n");
goto end;
}
out_stream->codec->codec_tag = 0;
if (ofmt_ctx->oformat->flags & AVFMT_GLOBALHEADER)
out_stream->codec->flags |= AV_CODEC_FLAG_GLOBAL_HEADER;
}
av_dump_format(ofmt_ctx, 0, out_filename, 1);
if (!(ofmt->flags & AVFMT_NOFILE))
{
ret = avio_open(&ofmt_ctx->pb, out_filename, AVIO_FLAG_WRITE);
if (ret < 0)
{
printf("Could not open output URL '%s'", out_filename);
goto end;
}
}
//写文件头到输出文件
ret = avformat_write_header(ofmt_ctx, NULL);
if (ret < 0)
{
printf("Error occured when opening output URL\n");
goto end;
}
//while循环中持续获取数据包，不管音频视频都存入文件
while (1)
{
AVStream* in_stream, * out_stream;
//从输入流获取一个数据包
ret = av_read_frame(ifmt_ctx, &pkt);
if (ret < 0)
break;
in_stream = ifmt_ctx->streams[pkt.stream_index];
out_stream = ofmt_ctx->streams[pkt.stream_index];
//copy packet
//转换 PTS/DTS 时序
pkt.pts = av_rescale_q_rnd(pkt.pts, in_stream->time_base, out_stream->time_base, (enum AVRounding)(AV_ROUND_NEAR_INF | AV_ROUND_PASS_MINMAX));
pkt.dts = av_rescale_q_rnd(pkt.dts, in_stream->time_base, out_stream->time_base, (enum AVRounding)(AV_ROUND_NEAR_INF | AV_ROUND_PASS_MINMAX));
//printf("pts %d dts %d base %d\n",pkt.pts,pkt.dts, in_stream->time_base);
pkt.duration = av_rescale_q(pkt.duration, in_stream->time_base, out_stream->time_base);
pkt.pos = -1;
//此while循环中并非所有packet都是视频帧，当收到视频帧时记录一下
if (pkt.stream_index == video_index)
{
printf("Receive %8d video frames from input URL\n", frame_index);
frame_index++;
}
//将包数据写入到文件。
ret = av_interleaved_write_frame(ofmt_ctx, &pkt);
if (ret < 0)
{
if (ret == -22) {
continue;
}
else {
printf("Error muxing packet.error code %d\n", ret);
break;
}
}
//av_free_packet(&pkt); //此句在新版本中已deprecated 由av_packet_unref代替
av_packet_unref(&pkt);
}
//写文件尾
av_write_trailer(ofmt_ctx);
end:
av_dict_free(&avdic);
avformat_close_input(&ifmt_ctx);
//Close input
if (ofmt_ctx && !(ofmt->flags & AVFMT_NOFILE))
avio_close(ofmt_ctx->pb);
avformat_free_context(ofmt_ctx);
if (ret < 0 && ret != AVERROR_EOF)
{
cout<<"Error occured."<<endl;
return -1;
}
return 0;
}
• 0
点赞
• 7
收藏
• 打赏
打赏
• 2
评论
“相关推荐”对你有帮助么？
• 非常没帮助
• 没帮助
• 一般
• 有帮助
• 非常有帮助
提交
评论 2
添加红包
请填写红包祝福语或标题
红包个数最小为10个
红包金额最低5元
当前余额3.43前往充值 >
需支付：10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
打赏作者
Sahm5k
你的鼓励将是我创作的最大动力
¥2 ¥4 ¥6 ¥10 ¥20
输入1-500的整数
余额支付 (余额：-- )
扫码支付
扫码支付：¥2
获取中
扫码支付
您的余额不足，请更换扫码支付或充值
打赏作者
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0
抵扣说明：
1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、C币套餐、付费专栏及课程。
余额充值
|
__label__pos
| 0.862756 |
Winter 2024 Midterm Exam
← return to practice.dsc10.com
Instructor(s): Suraj Rampure, Janine Tiefenbruck
This exam was administered in-person. The exam was closed-notes, except students were provided a copy of the DSC 10 Reference Sheet. No calculators were allowed. Students had 50 minutes to take this exam.
Video 🎥
Here’s a walkthrough video of some of the problems on the exam.
Clue is a murder mystery game where players use the process of elimination to figure out the details of a crime. The premise is that a murder was committed inside a large home, by one of 6 suspects, with one of 7 weapons, and in one of 9 rooms.
The game comes with 22 cards, one for each of the 6 suspects, 7 weapons, and 9 rooms. To set up the game, one suspect card, one weapon card, and one room card are chosen randomly, without being looked at, and placed aside in an envelope. The cards in the envelope represent the details of the murder: who did it, with what weapon, and in what room.
The remaining 19 cards are randomly shuffled and dealt out to the players (as equally as possible). Players then look at the cards they were dealt and can conclude that any cards they see were not involved in the murder. In the gameplay, players take turns moving around to different rooms of the house on the gameboard, which gives them opportunities to see cards in other players’ hands and further eliminate suspects, weapons, and rooms. The first player to narrow it down to one suspect, with one weapon, and in one room can make an accusation and win the game!
Suppose Janine, Henry, and Paige are playing a game of Clue. Janine and Paige are each dealt 6 cards, and Henry is dealt 7. The DataFrame clue has 22 rows, one for each card in the game. clue represents Janine’s knowledge of who is holding each card. clue is indexed by “Card”, which contains the name of each suspect, weapon, and room in the game. The “Category” column contains “suspect”, “weapon”, or “room”. The “Cardholder” column contains “Janine”, “Henry”, “Paige”, or “Unknown”.
Since Janine’s knowledge is changing throughout the game, the “Cardholder” column needs to be updated frequently. At the beginning of the game, the “Cardholder” column contains only “Janine” and “Unknown” values. We’ll assume throughout this exam that clue contains Janine’s current knowledge at an arbitrary point in time, not necessarily at the beginning of the game. For example, clue may look like the DataFrame below.
Note: Throughout the exam, assume we have already run import babypandas as bpd and import numpy as np.
Problem 1
Each of the following expressions evaluates to an integer. Determine the value of that integer, if possible, or circle “not enough information."
Important: Before proceeding, make sure to read the page called Clue: The Murder Mystery Game.
Problem 1.1
(clue.get("Cardholder") == "Janine").sum()
Answer: 6
This code counts the number of times that Janine appears in the Cardholder column. This is because clue.get("Cardholder") == "Janine" will return a Series of True and False values of length 22 where True corresponds to a card belonging to Janine. Since 6 cards were dealt to her, the expression evaluates to 6.
Difficulty: ⭐️⭐️
The average score on this problem was 78%.
Problem 1.2
np.count_nonzero(clue.get("Category").str.contains("p"))
Answer: 13
This code counts the number of cells that contain that letter "p" in the Category column. clue.get("Category").str.contains("p") will return a Series that contains True if "p" is part of the entry in the "Category" column and False otherwise. The words "suspect" and "weapons" both contain the letter "p" and since there are 6 and 7 of each respectively, the expression evaluates to 13.
Difficulty: ⭐️⭐️
The average score on this problem was 75%.
Problem 1.3
clue[(clue.get("Category") == "suspect") & (clue.get("Cardholder") == "Janine")].shape[0]
Answer: not enough information
This code first filters only for rows that contain both "suspect" as the category and "Janine" as the cardholder and returns the number of rows of that DataFrame with .shape[0]. However, from the information given, we do not know how many "suspect" cards Janine has.
Difficulty: ⭐️⭐️
The average score on this problem was 83%.
Problem 1.4
len(clue.take(np.arange(5, 20, 3)).index)
Answer: 5
np.arange(5, 20, 3) is the arary np.array([5, 8, 11, 14, 17]). Recall that .take will filter the DataFrame to contain only certain rows, in this case rows 5, 8, 11, 14, and 17. Next, .index extracts the index of the DataFrame, so the length of the index is the same as the number of rows contained in the DataFrame. There are 5 rows.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 69%.
Problem 1.5
len(clue[clue.get("Category") >= "this"].index)
Answer: 7
Similarly to the previous problem, we are getting the number of rows of the DataFrame clue after filtering it. clue.get("Category") >= "this" returns a Boolean Series where True is returned when a string in "Category" is greater than alphabetically than "this". This only happens when the string is "weapon", which occurs 7 times.
Difficulty: ⭐️⭐️⭐️⭐️⭐️
The average score on this problem was 29%.
Problem 1.6
clue.groupby("Cardholder").count().get("Category").sum()
Answer: 22
groupby("Cardholder").count() will return a DataFrame indexed by "Cardholder" where each column contains the number of cards that each "Cardholder" has. Then we sum the values in the "Category" column, which evaluates to 22 because the sum of the total number of cards each cardholder has is the total number of cards in play!
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 52%.
Problem 2
Since Janine’s knowledge of who holds each card will change throughout the game, the clue DataFrame needs to be updated by setting particular entries.
Suppose more generally that we want to write a function that changes the value of an entry in a DataFrame. The function should work for any DataFrame, not just clue.
What parameters would such a function require? Say what each parameter represents.
Answer: We would need four parameters:
• df, the DataFrame to change.
• row, the row label or row number of the entry to change.
• col, the column label of the entry to change.
• val, the value that we want to store at that location.
Difficulty: ⭐️⭐️⭐️⭐️
The average score on this problem was 43%.
Problem 3
An important part of the game is knowing when you’ve narrowed it down to just one suspect with one weapon in one room. Then you can make your accusation and win the game!
Suppose the DataFrames grouped and filtered are defined as follows.
grouped = (clue.reset_index()
.groupby(["Category", "Cardholder"])
.count()
.reset_index())
filtered = grouped[grouped.get("Cardholder") == "Unknown"]
Problem 3.1
Fill in the blank below so that "Ready to accuse" is printed when Janine has enough information to make an accusation and win the game.
if filtered.get("Card").______ == 3:
print("Ready to accuse")
What goes in the blank?
Answer: sum()
It is helpful to first visualize how both the grouped (left) and filtered (right) DataFrames could look:
Image 1
grouped DataFrame contains the the number of cards for a certain "Category"/"Cardholder" combination
Image 2
filtered DataFrame contains the the number of cards that are Unknown by "room", "suspect" and "weapon"
Now, let’s think about the scenario presented. We want a method that will return 3 from filtered.get("Card").___. We do not use count() because that is an aggregation function that appears after a .groupby, and there is no grouping here.
According to the instructions, we want to know when we narrowed it down to just one suspect with one weapon in one room. This means for filtered DataFrame, each row should have 1 in the "Card" column when you are already to accuse. sum() works because when you have only 1 unknown card for each of the three categories, that means you have a sum of 3 unknown cards in total. You can make an accusation now!
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 50%.
Problem 3.2
Now, let’s look at a different way to do the same thing. Fill in the blank below so that "Ready to accuse" is printed when Janine has enough information to make an accusation and win the game.
if filtered.get("Card").______ == 1:
print("Ready to accuse")
What goes in the blank?
Answer: max()
This problem follows the same logic as the first except we only want to accuse when filtered.get("Card").___ == 1. As we saw in the previous part, we only want to accuse when all the numbers in the "Card" column are 1, as this represents one unknown in each category. This means the largest number in the "Card" column must be 1, so we can fill in the blank with max().
Difficulty: ⭐️⭐️⭐️⭐️
The average score on this problem was 40%.
Problem 4
When someone is ready to make an accusation, they make a statement such as:
“It was Miss Scarlett with the dagger in the study"
While the suspect, weapon, and room may be different, an accusation will always have this form:
“It was ______ with the ______ in the ______"
Suppose the array words is defined as follows (note the spaces).
words = np.array(["It was ", " with the ", " in the "])
Suppose another array called answers has been defined. answers contains three elements: the name of the suspect, weapon, and room that we would like to use in our accusation, in that order. Using words and answers, complete the for-loop below so that accusation is a string, formatted as above, that represents our accusation.
accusation = ""
for i in ___(a)___:
accusation = ___(b)___
Problem 4.1
What goes in blank (a)?
Answer: [0, 1, 2]
answers could potentially look like this array np.array(['Mr. Green', 'knife', 'kitchen']). We want accusation to be the following: It was Mr. Green with the knife in the kitchen” where the underline represent the string from the words array and the nonunderlined parts represent the string from the answers array. In the for loop, we want to iterate through words and answers simultaneously, so we can use [0, 1, 2] to represent the indices of each array we will be iterating through.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 52%.
Problem 4.2
What goes in blank (b)?
Answer: accusation + words[i] + answers[i]
We are performing string concatenation here. Using the example from above, we want to add to the string accusation in order of accusation, words, answer. After all, we want “It was” before “Janine”.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 56%.
Problem 5
Recall that the game Clue comes with 22 cards, one for each of the 6 suspects, 7 weapons, and 9 rooms. One suspect card, one weapon card, and one room card are chosen randomly, without being looked at, and placed aside in an envelope. The remaining 19 cards (5 suspects, 6 weapons, 8 rooms) are randomly shuffled and dealt out, splitting them as evenly as possible among the players. Suppose in a three-player game, Janine gets 6 cards, which are dealt one at a time.
Answer the probability questions that follow. Leave your answers unsimplified.
Problem 5.1
Cards are dealt one at a time. What is the probability that the first card Janine is dealt is a weapon card?
Answer: \frac{6}{19}
The probability of getting a weapon card is just the number of weapon cards divided by the total number of cards. There are 6 weapon cards and 19 cards total, so the probability has to be \frac{6}{19}. Note that it does not matter how the cards were dealt. Though each card is dealt one at a time to each player, Janine will always end up with a randomly selected 6 cards, out of the 19 cards available.
Difficulty: ⭐️⭐️
The average score on this problem was 80%.
Problem 5.2
What is the probability that all 6 of Janine’s cards are weapon cards?
Answer: \frac{6}{19} \cdot \frac{5}{18} \cdot \frac{4}{17} \cdot \frac{3}{16} \cdot \frac{2}{15} \cdot \frac{1}{14}
We can calculate the answer using the multiplication rule. The probability of getting Janine getting all the weapon cards is the probability of getting a dealt a weapon card first multiplied by the probability of getting a weapon card second multiplied by continuing probabilities of getting a weapon card until probability of getting a weapon card on the sixth draw. The denominator of each subsequent probability decreases by 1 because we remove one card from the total number of cards on each draw. The numerator also decreases by 1 because we remove a weapon card from the total number of available weapon cards on each draw.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 62%.
Problem 5.3
Determine the probability that exactly one of the first two cards Janine is dealt is a weapon card. This probability can be expressed in the form \frac{k \cdot (k + 1)}{m \cdot (m + 1)} where k and m are integers. What are the values of k and m?
Hint: There is no need for any sort of calculation that you can’t do easily in your head, such as long division or multiplication.
Answer: k = 12, m = 18
m has to be 18 because the denominator is the number of cards available during the first and second draw. We have 19 cards on the first draw and 18 on the second draw, so the only way to get that is for m = 18.
The probability that exactly one of the cards of your first two draws is a weapon card can be broken down into two cases: getting a weapon card first and then a non-weapon card, or getting a non-weapon card first and then a weapon card. We add the probabilities of the two cases together in order to calculate the overall probability, since the cases are mutually exclusive, meaning they cannot both happen at the same time.
Consider first the probability of getting a weapon card followed by a non-weapon card. This probability is \frac{6}{19} \cdot \frac{13}{18}. Similarly, the probability of getting a non-weapon card first, then a weapon card, is \frac{13}{19} \cdot \frac{6}{18}. The sum of these is \frac{6 \cdot 13}{19 \cdot 18} + \frac{13 \cdot 6}{19 \cdot 18}.
Since we want the numerator to look like k \cdot (k+1), we want to combine the terms in the numerator. Since the fractions in the sum are the same, we can represent the probability as 2 \cdot \frac{6}{19} \cdot \frac{13}{18}. Since 2\cdot 6 = 12, we can express the numerator as 12 \cdot 13, so k = 12.
Difficulty: ⭐️⭐️⭐️⭐️
The average score on this problem was 31%.
Problem 6
Which of the following probabilities could most easily be approximated by writing a simulation in Python? Select the best answer.
Answer: The probability that Janine has three or more suspect cards.
Let’s explain each choice and why it would be easy or difficult to simulate in Python. The first choice is difficult because these simulations depend on Janine’s strategies and decisions in the game. There is no way to simulate people’s choices. We can only simulate randomness. For the second choice, we are not given information on how long each part of the gameplay takes, so we would not be able to simulate the length of a game. The third choice is very plausible to do because when cards are dealt out to Janine, this is a random process which we can simulate in code, where we keep track of whether she has three of more suspect cards. The fourth choice follows the same reasoning as the first choice. There is no way to simulate Janine’s moves in the game, as it depends on the decisions she makes while playing.
Difficulty: ⭐️⭐️
The average score on this problem was 83%.
Problem 7
Part of the gameplay of Clue involves moving around the gameboard. The gameboard has 9 rooms, arranged on a grid, and players roll dice to determine how many spaces they can move.
The DataFrame dist contains a row and a column for each of the 9 rooms. The entry in row r and column c represents the shortest distance between rooms r and c on the Clue gameboard, or the smallest dice roll that would be required to move between rooms r and c. Since you don’t need to move at all to get from a room to the same room, the entries on the diagonal are all 0.
dist is indexed by "Room", and the room names appear exactly as they appear in the index of the clue DataFrame. These same values are also the column labels in dist.
Problem 7.1
Two of the following expressions are equivalent, meaning they evaluate to the same value without erroring. Select these two expressions.
Explain in one sentence why these two expressions are the same.
Answer: dist.get("kitchen").loc["library"] and dist.get("library").loc["kitchen"]
dist.get("kitchen").iloc["library"] and dist.get("library").iloc["kitchen"] are both wrong because they uses iloc inappropriately. iloc[] takes in an integer number representing the location of column, row, or cell you would like to extract and it does not take a column or index name.
dist.get("kitchen").loc["library"] and dist.get("library").loc["kitchen"] lead to the same answer because the DataFrame has a unique property! The entry at r, c is the same as the entry at c, r because both are the distances for the same two rooms. The distance from the kitchen to library is the same as the distance from the library to kichen.
Difficulty: ⭐️⭐️
The average score on this problem was 84%.
Problem 7.2
On the Clue gameboard, there are two “secret passages." Each secret passage connects two rooms. Players can immediately move through secret passages without rolling, so in dist we record the distance as 0 between two rooms that are connected with a secret passage.
Suppose we run the following code.
nonzero = 0
for col in dist.columns:
nonzero = nonzero + np.count_nonzero(dist.get(col))
Determine the value of nonzero after the above code is run.
Answer: nonzero = 68
The nonzero variable represents the entries in the DataFrame where the distance between two rooms is not 0. There are 81 entries in the DataFrame because there are 9 rooms and 9 \cdot 9 = 81. Since the diagonal of the DataFrame is 0 (due to the distance from a room to itself being 0), we know there are at most 72 = 81 - 9 nonzero entries in the DataFrame.
We are also told that there are 2 secret passages, each of which connects 2 different rooms, meaning the distance between these rooms is 0. Each secret passage will cause 2 entries in the DataFrame to have a distance of 0. For instance, if the secret passage was between the kitchen and dining room, then the distance from the kitchen to the dining room would be 0, but also the distance from the dining room to the kitchen would be 0. Since there are 2 secret passages and each gives rise to 2 entries that are 0, this is 4 additional entries that are 0. This means there are 68 nonzero entries in the DataFrame, coming from 81 - 9 - 4 = 68.
Difficulty: ⭐️⭐️⭐️⭐️⭐️
The average score on this problem was 28%.
Problem 7.3
Fill in blanks so that the expression below evaluates to a DataFrame with all the same information as dist, plus one extra column called "Cardholder" containing Janine’s knowledge of who holds each room card.
dist.merge(___(a)___, ___(b)___, ___(c)___)
1. What goes in blank (a)?
2. What goes in blank (b)?
3. What goes in blank (c)?
Answer:
• (a): clue.get(["Cardholder"])
• (b): left_index=True
• (c): right_index=True
Since we want to create a DataFrame that looks like dist with an extra column of "Cardholder", we want to extract just that column from clue to merge with dist. We do this with clue.get(["Cardholder"]). This is necessary because when we merge two DataFrames, we get all columns from either DataFrame in the end result.
When deciding what columns to merge on, we need to look for columns from each DataFrame that share common values. In this case, the common values in the two DataFrames are not in columns, but in the index, so we use left_index=True and right_index=True.
Difficulty: ⭐️⭐️⭐️⭐️⭐️
The average score on this problem was 28%.
Problem 7.4
Suppose we generate a scatter plot as follows.
dist.plot(kind="scatter", x="kitchen", y="study");
Suppose the scatterplot has a point at (4, 6). What can we conclude about the Clue gameboard?
Answer: Another room besides the kitchen is 6 spaces away from the study.
Let’s explain each choice and why it is correct or incorrect. The scatterplot shows how far a room is from the kitchen (as shown by values on the x-axis) and how far a room is from the study (as shown by the values on the y-axis). Each room is represented by a point. This means there is a room that is 4 units away from the kitchen and 6 units away from the study. This room can’t be the kitchen or study itself, since a room must be distance 0 from itself. Therefore, we conclude, based on the y-coordinate, that there is a room besides the kitchen that is 6 units away from the study.
Difficulty: ⭐️⭐️⭐️⭐️
The average score on this problem was 47%.
Problem 8
The histogram below shows the distribution of game times in minutes for both two-player and three-player games of Clue, with each distribution representing 1000 games played.
Problem 8.1
How many more three-player games than two-player games took at least 50 minutes to play? Give your answer as an integer, rounded to the nearest multiple of 10.
Answer: 80
First, calculate the number of three-player games that took at least 50 minutes. We can calculate this number by multiplying the area of that particular histogram bar (from 50 to 60) by the total number of three player games(1000 games) total. This results in (60-50) \cdot 0.014 \cdot 1000 = 140. We repeat the same process to find the number of two-player games that took at least 50 minutes, which is (60-50) \cdot 0.006 \cdot 1000 = 60. Then, we find the difference of these numbers, which is 140 - 60 = 80.
An easier way to calculate this is to measure the difference directly. We could do this by finding the area of the highlighted region below and then multiplying by the number of games. This represents the difference between the number of three-player games and the number of two player games. This, way we need to do just one calculation to get the same answer: (60 - 50) \cdot (0.014 - 0.006) \cdot 1000 = 80.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 61%.
Problem 8.2
Calculate the approximate area of overlap of the two histograms. Give your answer as a proportion between 0 and 1, rounded to two decimal places.
Answer: 0.74
To find the area of overlap of the two histograms, we can directly calculate the area of overlap in each bin and add them up as shown below. However, this requires a lot of calculation, and is not advised.
• From 10-20: (20-10) \cdot 0.006 = 0.06
• From 20-30: (30-20) \cdot 0.018 = 0.18
• From 30-40: (40-30) \cdot 0.028 = 0.28
• From 40-50: (50-40) \cdot 0.016 = 0.16
• From 50-60: (60-50) \cdot 0.006 = 0.06
The summation of the overlap here is 0.74!
A much more efficient way to do this problem is to find the area of overlap by taking the total area of one distribution (which is 1) and subtracting the area in that distribution that does not overlap with the other. In the picture below, the only area in the two-player distribution that does not overlap with three-player distribution is highlighted. Notice there are only two regions to find the area of, so this is much easier. The calculation comes out the same: 1 - ((20 - 10) \cdot (0.022-0.006) + (30 - 20) \cdot (0.028 - 0.018) = 0.74.
Difficulty: ⭐️⭐️⭐️
The average score on this problem was 56%.
👋 Feedback: Find an error? Still confused? Have a suggestion? Let us know here.
|
__label__pos
| 0.868298 |
Reader Level:
Articles
XML Serializer Class for Reading and Writing XML
By Sam Hobbs on Apr 01, 2010
This article introduces the XmlSerializer class for reading and writing XML.
• 1
• 0
• 24680
Download Files:
There are many ways to read and write XML.
The advantage of the XmlSerializer class is that you can read and/or write XML with very little code. Most of the code required is simply the definition of the data. In other words, if our data is a list of Links consisting of a HREF or URL, a title and a category, then that data could be defined in the following manner:
public
class LinkObject
{
string ThisCategory;
string ThisHRef;
string ThisTitle;
public string Category
{
get { return ThisCategory; }
set { ThisCategory = value; }
}
public string HRef
{
get { return ThisHRef; }
set { ThisHRef = value; }
}
public string Title
{
get { return ThisTitle; }
set { ThisTitle = value; }
}
}
Using the XmlSerializer class, we use Serialize.Deserialize to read the data and XmlSerializer.Serialize to write the data. An instance of the XmlSerializer class could be created using:
XmlSerializer
Serializer = new XmlSerializer(typeof(LinkObjectsList));
Then the data could be written using:
TextWriter
Writer = new StreamWriter(Filename);
Serializer.Serialize(Writer, LinksList);
Writer.Close();
Data could be read using:
TextReader
Reader = new StreamReader(Filename);
LinksList = (LinkObjectsList)Serializer.Deserialize(Reader);
Reader.Close();
It is nearly that easy. Note that when the data is as simple of the above data, it is possible to read and write the data using a DataTable. If however the data is more complicated than what a single DataTable is capable of, then the XmlSerializer class can be easier (see below).
Note that the LinkObject class above represents one link. We are writing and reading a list of links, where list could be called an array or a collection or a table or something else. We can create a list of links using:
List
<LinkObject> LinksList = new List<LinkObject>();
If we do that, then in the XML the list of Link items will have the element name "ArrayOfLink" by default (that is how XmlSerializer works). If we want the element name to be something else, such as "Links", then we need to use an attribute (A C# attribute) to specify the name we want to be used. The XmlRoot attribute can be used except it can only be used with a class, structure, enumeration, or interface; it cannot be used with a declaration of an instance of an object. So the following shows how to create a class for our list of LinkObject items:
[XmlRoot("Links")]
public
class LinkObjectsList : List<LinkObject> { }
Note that all items to be serialized and/or deserialized must have Public access.
A XmlType attribute can be used to specify that each LinkObject item is to be serialized and deserialized as a "Link" object; in other words, to use Link as the XML element name instead of LinkObject.
Fields (members of the class to be serialized and deserialized) are attributes of the element that contains them if the "XmlAttribute" is used for the item in the class.
More Complex Data
Members of the class to be serialized and deserialized can be an instance of another class or a list or array or other collection. This is where the power of the XmlSerializer class is, since you can read and write complex data simply by writing the data definitions. Use the XmlElement attribute to override the name used in the XML for items in a class.
Sam Hobbs
Programmer / Analyst initially for the Unites States Army in 1972. Subsequent employers and clients have included Carte Blanche, Bank of America, Parsons Corporation, Lockheed Corporation and SHL Systemhouse.
Personal Blog: http://simplesamples.info
COMMENT USING
|
__label__pos
| 0.897021 |
clippy.exe
Process name: clippy
Application using this process: Clippy
clippy.exe
Process name: clippy
Application using this process: Clippy
clippy.exe
Click here to run a scan if you are experiencing issues with this process.
Process name: clippy
Application using this process: Clippy
Recommended: Scan your system for invalid registry entries.
What is clippy.exe doing on my computer?
clippy.exe is a clippy belonging to Clippy from Way Out There Software Non-system processes like clippy.exe originate from software you installed on your system. Since most applications store data in your system's registry, it is likely that your registry has suffered fragmentation and accumulated invalid entries which can affect your PC's performance. It is recommended that you check your registry to identify slowdown issues.
clippy.exe
Is clippy.exe harmful?
clippy.exe has not been assigned a security rating yet. Check your computer for viruses or other malware infected files.
clippy.exe is unrated
Can I stop or remove clippy.exe?
Most non-system processes that are running can be stopped because they are not involved in running your operating system. Scan your system now to identify unused processes that are using up valuable resources. clippy.exe is used by 'Clippy'.This is an application created by 'Way Out There Software'. To stop clippy.exe permanently uninstall 'Clippy' from your system. Uninstalling applications can leave invalid registry entries, accumulating over time. Run a free scan to find out how to optimize software and system performance.
Is clippy.exe CPU intensive?
This process is not considered CPU intensive. However, running too many processes on your system may affect your PC’s performance. To reduce system overload, you can use the Microsoft System Configuration Utility to manually find and disable processes that launch upon start-up. Alternatively, download SpeedUpMyPC to automatically scan and identify any unused processes.
Why is clippy.exe giving me errors?
Process related issues are usually related to problems encountered by the application that runs it. The safest way to stop these errors is to uninstall the application and run a system scan to automatically identify any unused processes and services that are using up valuable resources.
The safest way to stop these errors is to uninstall the application and run a scan to identify any system issues including invalid registry entries that have accumulated over time.
Process Library is the unique and indispensable process listing database since 2004 Now counting 140,000 processes and 55,000 DLLs. Join and subscribe now!
System Tools
SpeedUpMyPC
Toolbox
ProcessQuicklink
|
__label__pos
| 0.975114 |
Microsoft
Sum the top n values in a range using dynamic table behavior in Excel
Summing values is simple, summing the n top values in a range complicates the task, but it's certainly not impossible for Excel.
Counting the top n values is a common task, summing them complicates things a bit, but it's easier than you might think. You might consider sorting the values and referencing the appropriate cells, but doing so in inefficient.
If n varies and you enter and delete data, you need a dynamic formula in the form:
=SUMIF(range,">="&LARGE(range,n))
where range identifies the data you're summing (in a table) and n is the value that represents the number of values you want to sum. The sheet below shows a column of values that aren't in any particular order, except perhaps, data entry order. Using Excel's new table feature, I named the table in A4:C11 topn. The formula will work without the table, but you'll have to update the column references when you add data . Do yourself a favor and use a table.
In the sheet above, B1 is n in the following formula:
=SUMIF(topn[Sold],">="&LARGE(topn[Sold],B1))
The LARGE() function returns the nth largest number in the specified range. The SUMIF() function then sums all values in the range that are greater than or equal to the result of LARGE(). For example, in the above sheet, LARGE() returns 33. Therefore, SUMIF() sums the following values: 87, 73, 75, 33, and 82. The result is 350.
Changing the value in B1 changes the n factor and the formula updates accordingly. In addition, if you enter 0, the formula returns 0, not an error. If you enter a number that's greater than the actual number of values in the column, the formula returns 0. You'll know something's wrong, but you could use ISERROR() to return a more meaningful message.
I told you it was easy! Using Excel's dynamic table behavior and this simple nesting formula, you can quickly return the n top records. To sum the n lowest values, replace LARGE() with SMALL().
About Susan Harkins
Susan Sales Harkins is an IT consultant, specializing in desktop solutions. Previously, she was editor in chief for The Cobb Group, the world's largest publisher of technical journals.
Editor's Picks
|
__label__pos
| 0.927739 |
Representation Theory
157
浏览
0
关注
Representation theory is a branch of mathematics that studies abstract algebra|abstract algebraic structures by representing their element (set theory)|elements as linear transformations of vector spaces, and studiesModule (mathematics)|modules over these abstract algebraic structures. In essence, a representation makes an abstract algebraic object more concrete by describing its elements by matrix (mathematics)|matrices and the algebraic operations in terms of matrix addition and matrix multiplication. The algebraic objects amenable to such a description include group (mathematics)|groups, associative algebras and Lie algebras. The most prominent of these (and historically the first) is the group representation|representation theory of groups, in which elements of a group are represented by invertible matrices in such a way that the group operation is matrix multiplication. Representation theory is a powerful tool because it reduces problems in abstract algebra to problems in linear...
[展开]
相关概念
Lie Algebra
Lower Bound
Finite Group
Indexation
Lie Group
主要的会议/期刊
演化趋势
Chart will load here
Representation Theory文章数量变化趋势
Feedback
Feedback
Feedback
我想反馈：
排行榜
|
__label__pos
| 0.997608 |
Commit c992f570 authored by Dyangol's avatar Dyangol
Update accel-ppp.md
parent 748a9d8e
# Instal·lació
Considerem la instal·lació del servidor L2TP/PPPoE/IPoE accel-ppp en una distribució
Debian 9. Aquest servidor no compta de moment amb paqueria oficial per Debian. Caldrà
compilar-lo.
## Preparatius i compilació
Per compilar correctament aquest paquet caldrà instal·lar els paquets:
`apt install git build-essential cmake libsnmp-dev linux-headers-amd64 libpcre3-dev libssl-dev liblua5.1-0-dev`
ens situem al directory `/usr/local/src` i clonem el git del codi oficial. En el nostre
cas hem triat el tag corresponent a la darrera versió estable.
```
# git clone https://git.code.sf.net/p/accel-ppp/code accel-ppp
# cd accel-ppp
./accel-ppp# git checkout 1.11.1
```
creem el directori `build` i executem `cmake` per tal d'afegir a la compilació
diferents opcions (IPoE, VLAN-mon, etc.)
```
./accel-ppp/build# cmake -DKDIR=/usr/src/linux-headers-$(uname -r) -DCPACK_TYPE=Debian9 -DBUILD_IPOE_DRIVER=TRUE -DBUILD_VLAN_MON_DRIVER=TRUE -DRADIUS=TRUE -DNETSNMP=TRUE -DLUA=TRUE ..
./accel-ppp/build# make
```
Si no obtenim cap error en la compilació procedim a instal·lar els nous
mòduls de kernel que s'han compilat:
```
./accel-ppp/build# cp drivers/ipoe/driver/ipoe.ko /lib/modules/$(uname -r)
./accel-ppp/build# depmod-a
./accel-ppp/build# modprobe vlan_mon
./accel-ppp/build# modprobe ipoe
./accel-ppp/build# echo "vlan_mon" >> /etc/modules
./accel-ppp/build# echo "ipoe" >> /etc/modules
```
## Instal·lació del paquet
Ara ja podem crear el paquet gràcies a la comanda `cpack`:
```
./accel-ppp/build# cpack -G DEB
./accel-ppp/build# apt install ./accel-ppp.deb
```
Habilitem el servei en el sistema operatiu:
```
./accel-ppp/build# systemctl enable accel-ppp
```
# installation
guide inspired by [Compilando Accel-PPP - Debian 9 Stretch](https://www.youtube.com/watch?v=DpWAidcy5KY). Did not follow the cmake arguments in 18:05, and got a compilation error because it was missing `libssl1.0-dev` (so not `libssl-dev` as stated in youtube video guide) [source of solution](https://accel-ppp.org/forum/viewtopic.php?t=756).
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment
|
__label__pos
| 0.975421 |
Geting Table column data type : Table Column « Database ADO.net « C# / C Sharp
Geting Table column data type
Geting Table column data type
using System;
using System.Data;
using System.Data.OleDb;
public class DatabaseInfo {
public static void Main () {
String connect = "Provider=Microsoft.JET.OLEDB.4.0;data source=.\\Employee.mdb";
OleDbConnection con = new OleDbConnection(connect);
con.Open();
Console.WriteLine("Made the connection to the database");
Console.WriteLine("Information for each table contains:");
DataTable tables = con.GetOleDbSchemaTable(OleDbSchemaGuid.Tables,new object[]{null,null,null,"TABLE"});
foreach(DataColumn col in tables.Columns)
Console.WriteLine("{0}\t{1}", col.ColumnName, col.DataType);
con.Close();
}
}
Related examples in the same category
1.Read column values as Sql* types using the GetSql* methods
2.Read column values as C# types using the 'Get' methods
3.Refernece column name in SqlDataReader
|
__label__pos
| 0.999066 |
Export to pdf ( Create pdf on fly ) using ASP.net and C#
Consider following code for creating pdf document on fly using ASP.net in C#.
using System;
using System.Data;
using System.Configuration;
using System.Collections;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;
using System.IO;
using CrystalDecisions.CrystalReports.Engine;
using CrystalDecisions.Shared;
namespace WebApplication1
{
public partial class WebForm1 : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
}
protected void Button1_Click(object sender, EventArgs e)
{
MemoryStream oStream;
Response.Clear();
Response.Buffer = true;
ReportDocument rd = new ReportDocument();
rd.Load(Server.MapPath("cr1.rpt"));
oStream = (MemoryStream)rd.ExportToStream(ExportFormatType.PortableDocFormat);
Response.ContentType = "application/pdf";
Response.BinaryWrite(oStream.ToArray());
Response.End();
oStream.Flush();
oStream.Close();
oStream.Dispose();
}
}
}
Another way to Open report as a PDF Format
At first need to add following namespace
using CrystalDecisions.CrystalReports.Engine;
Then need to use following code to open it on pdf format
using (ReportClass rptH = new ReportClass())
{
rptH.FileName = Server.MapPath("~/")+ @"Views/Report/crJournal.rpt"; //Your rpt file path
rptH.Load();
rptH.SetDataSource(GetStudents());// Define/Set data Source for report
rptH.ExportToHttpResponse(ExportFormatType.PortableDocFormat, System.Web.HttpContext.Current.Response, false, "crReport");
}
|
__label__pos
| 0.912168 |
Noise: white, pink, brown, black, and buffers
Posted by on Nov 4, 2012 in Tutorial | 0 comments
Noise: white, pink, brown, black, and buffers
This post is a continuation of the previous one on network congestion.
A white noise is one in which each sample is uncorrelated with any other. A white Gaussian noise is one in which the samples are distributed according to a normal curve with a stable mean and variance. The power spectrum of a white noise process is uniformly flat at all frequencies. One way to think about this property is to say that there is equal power in equal frequency bands.
Another “brand” of noise is pink noise. The power spectrum of a pink noise falls off as 1/f and so this form of noise is often called “1/f noise”. As in the case of white noise, there are equal power frequency bands; however, in this case, the equal power bands are constant on a logarithmic scale. Pink noise often arise naturally due to what are called “relaxation processes”. In a typical relaxation process, some ensemble of elements (e.g., particles, dipoles, or whatever), is in an excited state and can “relax” from that state with an exponentially distributed time. Given just one such process, the power spectrum turns out to be that of a low pass filter with a cut-off frequency of 1/{\tau}. If there are several such relaxation processes active in a system, and if these are exponentially related to one another, the result is a pink noise. A way to simulate this kind of behavior can be done with three (or more) dice. The output of the process is the sum of the faces of the three dice. The first die is thrown at each stage. The second die is thrown at each second stage. The third die is thrown at each fourth stage. If there is a fourth die, it would be cast at each eighth stage; and so on. The average value is 10.5, which is naturally 3 times 3.5. The progression of relaxation times is simulated by the different degrees of persistence in the values of the dice.
A brown noise arises from the integration of white noise. In this case, the power spectrum falls off as 1/f^2. Beyond brown noise lies black noise. One example of a black noise would be a white noise that had been integrated twice, for a power spectrum of 1/f^4.
So, in general, we could consider any noise spectrum of the form f^{-\beta} where \beta is between 0 or greater. The specific cases we’ve considered so far at \beta = 0 for white noise, \beta = -1 for pink noise, \beta = -2 for brown noise, and \beta > 2 for black noise. In our previous post, we introduced the Hurst exponent, H. It turns out that there is a simple relationship between the spectral exponent \beta and the Hurst exponent, H; namely, \beta = 2H + 1. So, for white noise with \beta = 0, H = 1/2 and the result is a random walk. For the common Hurst exponent of H = 0.72, \beta = 2.44; and the result is a black noise.
For these brown and black noises, there are long range dependencies (LRD) between increasingly distant samples. Stated another way, the greater the value of \beta, the greater the likelihood of runs of positive or negative values. When such processes are discovered in queuing systems, the usual assumptions concerning convergence towards large-scale behaviors break down.
The classic paper that presented this to an engineering audience was Leland, W., M. Taqqu, W. Willinger, and D. Wilson, “On the Self-Similar Nature of Ethernet Traffic (Extended Version),” IEEE/ACM Transactions on Networking, 2(1), pp. 1-13, 1994. I can recall my own amazement upon reading this paper for the first time. Perhaps the best representation of the concepts is the following figure, taken from a follow-up paper, Willinger, W. and V. Paxson, “Where Mathematics Meets the Internet,” Notices of the American Mathematical Society, Vol. 45, No. 8, pp. 961-970, 1998.
Difference between classic and fractal traffic
Difference between classic and fractal traffic
In the case of the “fractal” traffic, the image looks equally bursty at any time scale, while the “classic” traffic becomes increasingly smooth as the observation window is increased. This is an expression of the fact that the Gaussian probability distribution that underlies the “classic” traffic models has a preferred scale; namely, its mean value. As the observation time is increased well beyond the mean arrival rate of traffic, plus or minus the standard deviation, then the fluctuations around the mean become less and less apparent.
On the other hand, with “fractal” traffic, there are fluctuations at all scales; and there is no benefit in attempting to “average them out”. Of course, that was the whole idea of a buffer in the first place wasn’t it? The point of the buffer was to hold, say, 100 samples of the traffic; and assuming that it was Poisson distributed, then the fluctuations would be about ±10. More generally, for a Poisson process, a buffer that was K deep would have fluctuations of about K^{1/2}. As buffer depth is increased, the relative fluctuations and the likelihood of buffer overflow is reduced.
This just doesn’t happen with “fractal” traffic. An excellent introductory article on this form of traffic is “A Short Tutorial on Fractals and Internet Traffic” by Thomas B. Fowler. In the same publication, there is an excellent follow-up article, “A Method for Analyzing Congestion in Pareto and Related Queues“, by Martin Fischer and Carl Harris. [Note that these two documents require Microsoft Word or some application that can open Word documents.]
If offered load is not Poisson-distributed, then what is it? As discussed in the articles I just referred to, the distributions tend to have “long tails”, which is to say that they do not fall off as rapidly as the normal distribution (or other distributions of the exponential family) for high values. More generally, these distributions are referred to as power-law distributions. A power law has for its mathematical form K x^{-\alpha} where K and \alpha are constants, at least for some part of the domain x.
One class of such distributions are the Pareto distributions; however, there are a broader class of these as well. The Pareto Type I distribution is given as follows:
\bar{F}(x) = \Pr (X>x) = \begin{cases} \left(\frac{x_m}{x}\right){}^{\alpha } & \text{for } x\geq x_m \\ 1 & \text{for } xhad to converge to nice, neat normal distributions. It turns out that the normal distribution is only one of a class of so-called Levy alpha-stable distributions for which a law of large numbers (central limit theorem) applies. Those deeply interested in the mathematics can read through the Wikipedia article that I just linked to and follow the references there.
The main point for my purposes here is to point out that these more general distributions have four parameters, one for "shift" or central location (think "mean"), one for asymmetry (think "skew"), one for spread (think "variance"), and one for asymptotic behavior (think "tail"). The normal, Lévy, Pareto, Cauchy and other distributions are all specific cases of the Lévy alpha-stable distribution. Also, each variant obeys a central limit theorem in the sense that the sum of random variables that are distributed as any member of this class is also distributed as the same distribution. Said another way, the sum of a number of Paretian random variables is also a Paretian random variable just as surely as the sum of a number of Gaussian random variables is a Gaussian.
I hope that this fact helps persuade you, gentle reader, that simply adding up some millions of subscribers, any group of which might obey some anomalous probability distribution, does not yield a normal curve due to a law of large numbers. Quite the contrary: add up millions of power-tail distributions and viola, you have a power-tail distribution.
Back to buffers... Because of the general applicability of these observations, consideration of Pareto-distributed traffic on queues has become important over the past several years. In particular, a Pareto/D/1/k model has been a frequent basis for study and simulation. A reasonable representative of this class of work is a paper entitled "Evaluation of Pareto/D/1/k queue by simulation" by Mirtchev and Goleva. The general result in this and similar studies is that the probability of congestion is often one or more orders of magnitude larger than for Poisson distributed traffic under similar assumptions of mean load.
And this brings me back to the broad concept of a congestion policy and how paging systems differ from other forms of data network. If we wrap our minds around the notion that any system given Paretian load will experience congestion at a significant rate from time to time, and especially under situations of emergency (hurricanes, floods, fire, terrorism, and et cetera), should a carrier that has accepted messages for delivery throw them in the bit bucket or retry them. In paging systems, especially in two-way paging, there is a dogged policy of message retry until delivery is confirmed. Under conditions of network congestion for any reason, messages are retained in input terminals and linked to the subscriber's account. Various agents within the network accept messages for delivery to the subscriber and either return success or failure to the account. Undelivered messages are retried by policy on both time-driven and event-driven bases.
One aspect of message delivery retry procedures that is of high importance in wireless systems is their impact on battery life at the mobile. TCP can make very strong demands on mobile battery because of its demands on end-to-end packet transmission and acknowledgement; that is, in a typical Internet environment, the onus for message confirmation lies with the sender. The intermediate network elements are free to discard packets due to congestion (or whatever) and push the responsibility for delivery back to the end points. In contrast, in a paging network, once a message is accepted, the sender is free to disconnect. In this connectionless model, the network assumes responsibility for message delivery and also assumes a certain responsibility for the efficiency of delivery; that is, not to arbitrarily consume the battery life of either sender or recipient in order to achieve delivery.
I'll continue with further thoughts on message delivery policy in the face of congestion issues, et al, in subsequent posts. Suffice it to say that I believe that any network that purports to provide service for first responders, emergency situations, or any critical response, must be designed around certain positive principles that do not just dismiss network congestion issues but adapt to them dynamically.
Leave a Reply
|
__label__pos
| 0.908468 |
Hardware monitoring
XE Product imageBE Product imageEE Product image
In this window, we can configure the hardware monitoring services.
Display temperatures in Fahrenheit
In the sensor icons, the OSD panel, the Desktop Gadget and the SensorPanel, temperature values are displayed in Fahrenheit (rather than Celsius) when this option is enabled.
Enable CPU throttling measurement
This option can be used to enable CPU throttling measurement on Intel processors. Measuring throttling may cause system instability. Throttling is a self-protection mechanism in Intel processors to prevent physical damage due to overheating.
Enable disk temperature measurement
Here we can enable or disable disk temperature monitoring.
Disk temperature polling frequency
With this option, we can configure the interval between disk temperature measurements. Setting this option to less than 20 seconds may result in data corruption on older hard disk drives. For modern HDDs and SSDs, it is safe to enter any value of our choice.
Decimal digits for voltage values
With this option, we can set the number of decimal digits to be displayed for voltage readings. For modern computers, selecting at least 3 digits is recommended.
Decimal digits for clock speeds
With this option, we can set the number of decimal digits to be displayed for clock speeds, e.g. CPU core clock, FSB clock and memory clock.
Tjmax temperature
With this option, we can set the Tjmax temperature that is used to calculate core temperature readings for Intel processors. When “Automatic” is selected, AIDA64 will use the default values as defined in the Intel Digital Thermal Sensors (DTS) specifications.
Degree symbol
With this option, we can configure the character to be used as a degree symbol for temperature readings.
|
__label__pos
| 0.674948 |
Ansible Pilot
Ansible troubleshooting - Error 305: command-instead-of-shell
How to Solve the Ansible Error 305 command-instead-of-shell
November 1, 2023
Access the Complete Video Course and Learn Quick Ansible by 200+ Practical Lessons
Introduction
Ansible, the popular automation tool, provides a powerful platform for managing and configuring IT infrastructure efficiently. While Ansible offers various modules to execute tasks, it’s crucial to choose the right module for specific use cases. Ansible-Lint, a widely-used linting tool for Ansible playbooks, enforces rules to help you follow best practices. In this article, we’ll delve into Rule 305, “command-instead-of-shell,” which encourages the use of the command module over the shell module when it’s not necessary to employ shell features.
Understanding Rule 305
Rule 305, “command-instead-of-shell,” is a valuable guideline that promotes efficiency and best practices in Ansible playbooks. It suggests that the command module should be preferred over the shell module unless specific shell features, such as environment variable expansion or chaining multiple commands using pipes, are required.
Problematic Code
Let’s explore a piece of problematic code that Rule 305 can identify in your playbooks:
---
- name: Problematic example
hosts: all
tasks:
- name: Echo a message
ansible.builtin.shell: echo hello # <-- command is better in this case
changed_when: false
In this code snippet, the playbook utilizes the shell module (ansible.builtin.shell) to execute a simple command that echoes a message. While this code will work, it’s not the most efficient or recommended approach.
Output:
WARNING Listing 1 violation(s) that are fatal
command-instead-of-shell: Use shell only when shell functionality is required.
305.yml:5 Task/Handler: Echo a message
Read documentation for instructions on how to ignore specific rule violations.
Rule Violation Summary
count tag profile rule associated tags
1 command-instead-of-shell basic command-shell, idiom
Failed: 1 failure(s), 0 warning(s) on 1 files. Last profile that met the validation criteria was 'min'.
The Best Resources For Ansible
Certifications
Video Course
Printed Book
eBooks
Correct Code
The corrected code that adheres to Rule 305 is as follows:
---
- name: Correct example
hosts: all
tasks:
- name: Echo a message
ansible.builtin.command: echo hello
changed_when: false
In this improved version, the command module (ansible.builtin.command) is used to execute the same “echo” command. This approach aligns with best practices and is more efficient.
Why Choose the Command Module Over the Shell Module
The command module is preferred over the shell module in many cases due to several advantages:
1. Efficiency: The command module is considerably faster than the shell module, making your playbooks run more quickly.
2. Predictability: Using the command module ensures more predictable and reliable execution, as it doesn’t involve shell-specific behavior.
3. Idempotence: The command module is idempotent by design, meaning it only makes changes when necessary, enhancing playbook reliability.
4. Security: By avoiding unnecessary shell access, you reduce potential security risks and vulnerabilities.
Exception Handling
While Rule 305 encourages the use of the command module, there may be situations where you genuinely need the features offered by the shell module. In such cases, it’s essential to carefully consider whether the use of shell features justifies the performance trade-off. If you find that the shell module is indeed necessary, you can continue using it with an understanding of the potential efficiency implications.
Conclusion
Rule 305, “command-instead-of-shell,” is a valuable guideline within Ansible-Lint that promotes efficiency and best practices in Ansible playbooks. By choosing the command module over the shell module when shell-specific features are not required, you can enhance the speed, predictability, and reliability of your automation tasks. This practice contributes to a more efficient and secure Ansible workflow, ultimately ensuring that your playbooks perform optimally and effectively manage your IT infrastructure.
Subscribe to the YouTube channel, Medium, Website, Twitter, and Substack to not miss the next episode of the Ansible Pilot.
Academy
Learn the Ansible automation technology with some real-life examples in my
My book Ansible By Examples: 200+ Automation Examples For Linux and Windows System Administrator and DevOps
BUY the Complete PDF BOOK to easily Copy and Paste the 250+ Ansible code
Want to keep this project going? Please donate
Access the Complete Video Course and Learn Quick Ansible by 200+ Practical Lessons
Follow me
Subscribe not to miss any new releases
|
__label__pos
| 0.952999 |
German
Sprachen
English
Bengali
French
German
Japanese
Korean
Portuguese
Spanish
Tamil
Operators (qiskit.opflow)
Operators and State functions are the building blocks of Quantum Algorithms.
A library for Quantum Algorithms & Applications is more than a collection of algorithms wrapped in Python functions. It needs to provide tools to make writing algorithms simple and easy. This is the layer of modules between the circuits and algorithms, providing the language and computational primitives for QA&A research.
We call this layer the Operator Flow. It works by unifying computation with theory through the common language of functions and operators, in a way which preserves physical intuition and programming freedom. In the Operator Flow, we construct functions over binary variables, manipulate those functions with operators, and evaluate properties of these functions with measurements.
The Operator Flow is meant to serve as a lingua franca between the theory and implementation of Quantum Algorithms & Applications. Meaning, the ultimate goal is that when theorists speak their theory in the Operator Flow, they are speaking valid implementation, and when the engineers speak their implementation in the Operator Flow, they are speaking valid physical formalism. To be successful, it must be fast and physically formal enough for theorists to find it easier and more natural than hacking Matlab or NumPy, and the engineers must find it straightforward enough that they can learn it as a typical software library, and learn the physics naturally and effortlessly as they learn the code. There can never be a point where we say „below this level this is all hacked out, don’t come down here, stay in the interface layer above.“ It all must be clear and learnable.
Before getting into the details of the code, it’s important to note that three mathematical concepts unpin the Operator Flow. We derive most of the inspiration for the code structure from John Watrous’s formalism (but do not follow it exactly), so it may be worthwhile to review Chapters I and II, which are free online, if you feel the concepts are not clicking.
1. An n-qubit State function is a complex function over n binary variables, which we will often refer to as n-qubit binary strings. For example, the traditional quantum „zero state“ is a 1-qubit state function, with a definition of f(0) = 1 and f(1) = 0.
2. An n-qubit Operator is a linear function taking n-qubit state functions to n-qubit state functions. For example, the Pauli X Operator is defined by f(Zero) = One and f(One) = Zero. Equivalently, an Operator can be defined as a complex function over two n-qubit binary strings, and it is sometimes convenient to picture things this way. By this definition, our Pauli X can be defined by its typical matrix elements, f(0, 0) = 0, f(1, 0) = 1, f(0, 1) = 1, f(1, 1) = 0.
3. An n-qubit Measurement is a functional taking n-qubit State functions to complex values. For example, a Pauli Z Measurement can be defined by f(Zero) = 0 and f(One) = 1.
Bemerkung
While every effort has been made to make programming the Operator Flow similar to mathematical notation, in some places our hands are tied by the design of Python. In particular, when using mathematical operators such as + and ^ (tensor product), beware that these follow Python operator precedence rules. For example, I^X + X^I will actually be interpreted as I ^ (X+X) ^ I == 2 * I^X^I. In these cases, you should use extra parentheses, like (I ^ X) + (X ^ I), or use the relevant method calls.
Below, you’ll find a base class for all Operators, some convenience immutable global variables which simplify Operator construction, and two groups of submodules: Operators and Converters.
Operator Base Class
The OperatorBase serves as the base class for all Operators, State functions and measurements, and enforces the presence and consistency of methods to manipulate these objects conveniently.
OperatorBase()
A base class for all Operators: PrimitiveOps, StateFns, ListOps, etc.
Operator Globals
The operator_globals is a set of immutable Operator instances that are convenient building blocks to reach for while working with the Operator flow.
One qubit Pauli operators:
X, Y, Z, I
Clifford+T, and some other common non-parameterized gates:
CX, S, H, T, Swap, CZ
One qubit states:
Zero, One, Plus, Minus
Submodules
Operators
The Operators submodules include the PrimitiveOp, ListOp, and StateFn class groups which represent the primary Operator modules.
primitive_ops
Primitive Operators (qiskit.opflow.primitive_ops)
list_ops
List Operators (qiskit.opflow.list_ops)
state_fns
State Functions (qiskit.opflow.state_fns)
Converters
The Converter submodules include objects which manipulate Operators, usually recursing over an Operator structure and changing certain Operators‘ representation. For example, the PauliExpectation traverses an Operator structure, and replaces all of the OperatorStateFn measurements containing non-diagonal Pauli terms into diagonalizing circuits following by OperatorStateFn measurement containing only diagonal Paulis.
converters
Converters (qiskit.opflow.converters)
evolutions
Operator Evolutions (qiskit.opflow.evolutions)
expectations
Expectations (qiskit.opflow.expectations)
gradients
Gradients (qiskit.opflow.gradients)
Utility functions
commutator(op_a, op_b)
Compute commutator of op_a and op_b.
anti_commutator(op_a, op_b)
Compute anti-commutator of op_a and op_b.
double_commutator(op_a, op_b, op_c[, sign])
Compute symmetric double commutator of op_a, op_b and op_c.
Exceptions
OpflowError(*message)
For Opflow specific errors.
|
__label__pos
| 0.940117 |
Uploaded image for project: 'JDK'
1. JDK
2. JDK-8209171
Simplify Java implementation of Integer/Long.numberOfTrailingZeros()
Details
• Type: Enhancement
• Status: Resolved
• Priority: P5
• Resolution: Fixed
• Affects Version/s: None
• Fix Version/s: 12
• Component/s: core-libs
• Labels:
• Subcomponent:
• Resolved In Build:
b09
Description
Currently, the Java implementation of numberOfTrailingZeros() is derived from HD:
public static int numberOfTrailingZeros(long i) {
// HD, Figure 5-14
int x, y;
if (i == 0) return 64;
int n = 63;
y = (int)i; if (y != 0) { n = n -32; x = y; } else x = (int)(i>>>32);
y = x <<16; if (y != 0) { n = n -16; x = y; }
y = x << 8; if (y != 0) { n = n - 8; x = y; }
y = x << 4; if (y != 0) { n = n - 4; x = y; }
y = x << 2; if (y != 0) { n = n - 2; x = y; }
return n - ((x << 1) >>> 31);
}
It can be simplified through re-use of numberOfLeadingZeros() as
public static int numberOfTrailingZeros_02(int i) {
if (i == 0) return 32;
return 31 - numberOfLeadingZeros(i & -i);
}
This will reduce bytecode size, but also this shows better performance with both c1 and c2 compilers.
Long.numberOfTrailingZeros() can also be simplified via delegating to Integer's version.
This variant also shows better performance.
Attachments
Activity
People
• Assignee:
igerasim Ivan Gerasimov
Reporter:
igerasim Ivan Gerasimov
• Votes:
0 Vote for this issue
Watchers:
1 Start watching this issue
Dates
• Created:
Updated:
Resolved:
|
__label__pos
| 0.99737 |
You may also like
problem icon
Golden Thoughts
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
problem icon
Ladder and Cube
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
problem icon
Days and Dates
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
Folding in Half
Stage: 3 and 4 Short Challenge Level: Challenge Level:1
Since the original triangle is isosceles and right-angled, folding it produces a smaller triangle, also isosceles and right-angled.
By Pythagoras' Theorem, the hypotenuse of the original triangle is $\sqrt{200}=10\sqrt{2}$ cm.
Hence the difference between the perimeters of the two triangles is $(10+10+10\sqrt{2})-(5\sqrt{2}+5\sqrt{2} +10)=10$ cm.
Alternatively: let the length of the shorter sides of the new triangle be x cm, shown below. Then the perimeter of the original triangle is $(20+2x)$ cm and the perimeter of the new triangle is $(10+2x)$cm. Hence the difference between the perimeters of the two triangles is $10$cm.
This problem is taken from the UKMT Mathematical Challenges.
View the previous week's solution
View the current weekly problem
|
__label__pos
| 0.745966 |
Aqua Blog
Intro to Fileless Malware in Containers
Intro to Fileless Malware in Containers
A fileless attack is a technique that takes incremental steps toward gaining control of your environment while remaining undetected. In a fileless attack, the malware is directly loaded into memory and executed, evading common defenses and static scanning.
Often, attackers may also use compression or encryption to cloak the malware file to avoid detection. Since fileless is most commonly used against Windows, we have recently seen a growing trend in its use against Linux, and, more specifically, within containers. In this guide, we will break down a fileless attack by creating our own fileless demo and show which tools are required to detect the activity we are seeing.
Malware is malicious code intended to damage your software, steal information, or take full control of your supply chain. Malware can take on several forms: Viruses, Worms, Trojans, Ransomware, Bots, Adware, Fileless, etc., some of which are very sophisticated.
Fileless malware is an advanced kind of attack, used for loading/executing the malware from memory rather than from the file system. It is one of the most dangerous security threats today. According to Ponemon Institute report in 2017, fileless attacks are ten times more successful than file-based attacks. In fact, up to 77% successful attacks can be attributed to fileless techniques or exploits. Further, a 2020 WatchGuard report noted that this technique has increased by nearly 900% since 2019. Fileless attacks are undetectable by most antivirus software, endpoint detection and response (EDR), and traditional security tools because these usually only discover compromises based on file systems, file descriptors in unix systems, and handles in Windows. A fileless attack is executed from a memory address, making it extremely hard to collect evidence or forensic clues about what happened. For more details about this form of attack, see here
Fileless attacks use common artifacts to hide themselves. They are often camouflaged within popular trusted software only inject malicious code into widely used applications. Quietly hidden, they are able to launch assaults on software supply chains and spread fileless attacks, exploiting trusted software relationships and networks to penetrate organizations.
In the past, most successful fileless attacks occurred in Windows via hijacked artifacts such as PowerShell, Microsoft Office macros, WMI, scripting languages (VBScript, Windows PowerShell), and other popular post-exploitation tools (PowerShell empire, Powersploit, Metasploit, cobaltstrike, etc).
Today we’re seeing a sharp increase in attacks in Linux as well as in containers, a technology based on the Linux kernel that uses namespaces and cgroups. Let’s take a look at one of the ways in which process injection works in Linux and see how easy it is for bad actors to perform fileless attacks in containers.
Fileless Malware Attacks on Linux
Fileless malware attacks targeting Linux systems follow a series of steps, starting with the infection and ending with the execution of malicious code. From there, the attacker can then compromise both the server and data. An attack might begin in several ways (phishing email, malicious link, etc.), but the most common and easiest way for it to be successful is by exploiting existing vulnerabilities. Using vulnerability scanning tools is the first step to identify and manage common vulnerabilities and exposures (CVE), but they are not enough to stop advanced attacks that take place in runtime.
Steps in a Linux fileless malware attack
Figure 1
Step 1: Infection via exploitation of a Vulnerability. First, the attack begins by discovering an unpatched vulnerability or breach such as a flaw in a network protocol. This is where you would find the greatest value in using tools for vulnerability scanning, misconfigurations, compliances, secrets.
Since exploitable vulnerabilities may be used as a gateway to gain access to the target system, vulnerability scanning is an important shift-left control in software lifecycle (Figure 2). Still, a fileless attack cannot be detected with static scanning tools because it happens in runtime.
Vulnerability scanning, an important shift-left control in a software lifecycle
Figure 2
Step 2: Modification of a Linux Process. Once access is gained through vulnerability exploitation, the malicious program could employ many techniques to perform process injection, including ptrace syscall, LD_PRELOAD environment, and more.
Step 3: Insertion of malicious code in Memory. Using a fileless technique, it’s possible to insert malicious code into memory without writing files. For example, the memfd_create create an anonymous descriptor to be used to insert in a running process.
Step 4: Execution of Malicious code. At this point, the system has been compromised after the malicious code has run. Whether through malware, crypto mining, or other malicious techniques, your system would become defenseless against the stealing of sensitive data, the damaging of servers, the encrypting critical files, and much more.
Accomplishing the Code Injection on Linux
Code injection is a common technique used by fileless attacks to modify a Linux process after gaining initial access, usually through a vulnerability or exploit. In the Linux context, a process is an instance of a program running, the ID for the process is known as PID. The code is then injected using the memory address of that active process, usually an ELF binary file. Executing code in the context of another process may require access to other resources (memory, networking, etc.). Code injection on Linux often uses syscalls such as ptrace and memfd_create, or environment variables such as LD _PRELOAD.
The Ptrace system call is used by debuggers (such as gdb and dbx) on Unix systems to inspect system calls invoked by the attached child process. The main actors are the tracer, the process that takes control of the execution of another process, and the tracee, the process controlled by the tracer. It is trickier to utilize ptrace because you need privileged access to use it.
The LD_PRELOAD environment variable loads all shared objects into the process before its main function is invoked, this is one way to inject code. However, to do this it does require that the target process be restarted. Preloading is a feature of the dynamic linker (ld) available on most Unix systems. It allows loading a user-defined shared library before any other libraries linked to an executable.
The memfd_create is a system call added in Linux kernel 3.17 (sys_356/319). This function allows for the creation of anonymous files that are located at RAM and have volatile storage. Because of this, it can be used to load arbitrary programs such as malwares and arbitratry binaries. It doesn’t require superuser privileges to use it.
In short, code injection can happen many ways, from a syscall hooking to a zero-day vulnerability, as in the case of Log4j CVE-2021-44228, which exploits a breach in the Java programming language allowing for the execution of fileless malware.
Simulating a Fileless Attack
1. Memfd_create, a native Linux syscall where everything begins.
There are many ways to execute a fileless injection. To keep things simple, we have chosen to perform an injection that involves the native capabilities of the Linux kernel using the memfd_create system call.
We can mirror the process of fileless malware by creating a program that uses the syscall memfd_create, which makes an anonymous descriptor and uses it to inject code.
Memfd_create is a system call Linux uses to create an anonymous file descriptor in /proc/PID/fd/ which may use execve to execute it. This means that there is no mounting device, temporary file storage (tmpfs), or any temporary RAM storage (/dev/shm) visible for security tools based only in file system scanning. As with any programming language for Linux developers (see Figure 1), a program can be created using this system call.
Programming languages supporting system call under the code 319
Figure 3 – Programming languages supporting system call under the code 319
2. A program to accomplish fileless execution.
By calling memfd_create, we obtain an anonymous descriptor that can be used to load arbitrary binaries, such as malware. Along with the execve syscall, the program is executed pointing to the anonymous descriptor created by memfd_create in the previous step. One of the many implementations found on GitHub with the described logic is memrun (Figure 4). This program contains all the necessary steps to perform a code injection into memory using the Golang programming language. The main parameters are the ELF binary path being injected and the live process whose address memory is being used. In the sample demo, you can see a running process “nginx” being used to inject the “invasive binary” as “./memrun nginx invasive_binary”.
Code injection into memory using the Golang programming language
Figure 4 – memrun code
3. Adding the program to image container.
For demo purposes, we’ve created a new image (Figure 5), adding the binary inside the official nginx image to be used when the container is running.
Image container with binary added while running
Figure 5 – Image container
Now that we have simulated how fileless malware injection is accomplished, let’s move on to analyzing the code injection itself. See the note about memrun for more details.
Detecting the attack with Tracee
Using Tracee, an open-source tool that identifies suspicious behavior in security runtime, we can detect this fileless execution technique (Figure 6). Tracee analyzes events collected at the kernel level in real time using eBPF technology. In the demo (Figure 6) the main syscalls used were execve, close, openat, memfd_create, etc., along with other key events. In Tracee, you can use term “Signatures” as an abstraction to analyze and identify the security threat such as code injection, dynamic code loading, fileless execution, etc. These signatures act as behavioral indicators developed by Team Nautilus, security research experts in cloud native software. See the complete signature list here.
1. Executing the demo container
2. Executing Tracee
3. Code injection detected by Tracee
Example of fileless execution using the nginx process to perform code injection
Figure 6 – Fileless execution using the nginx process to perform code injection.
View the complete instructions to reproduce by yourself in this repository:
https://github.com/krol3/demo-fileless
Detecting the Attack with Cloud Native Detection & Response
cndr-malware-fileless-02
Figure 7 – Detection and Response (CNDR) to detect Fileless Malware
We recommend using Cloud Native Detection and Response (CNDR) to detect Fileless Malware attacks and benefit from an improved UI over our open source project Tracee. CNDR is a part of Aqua Platform’s runtime capabilities and is built on top of Tracee, with a larger database of behavioral indicators and a comprehensive, easy user interface which includes enterprise level support. For example, with Tracee you have more than 10+ default rules whereas with CNDR you will have 100+ security signatures. These behavioral indicators are written by Aqua Nautilus Security Research based on actual observed cloud native attacks in the real world.
Scanning your artifacts such as code, container images, Kubernetes manifests, infra as code, etc. is the first step to avoid misconfigurations, hardcode secrets, and vulnerabilities. This limits attackers’ ability to gain an early foothold. Remember that at any moment a new threat could be discovered to hack your system, and advanced techniques such as fileless malware that are hard to detect can be present in your system. Thinking about the next log4j or spring4shell zero-day vulnerabilities? You need a way to detect sophisticated threats that can gain a foothold in your environment and evade detection.
Can you currently detect these threats? Pay attention to the Real-time Malware Protection control in the demo.
Related resource: View a demo video that shows the difference between detection and basic risk posture for your cloud environment (CSPM).
Conclusion
Fileless malware is a powerful attack technique that’s grown more in prominence because it’s incredibly difficult to detect and can be cleverly hidden from security tools. It’s vital to use security tools that will help you discover and respond to advanced attacks like fileless, identifying suspicious events during runtime in cloud native environments. Threats are constantly evolving and being discovered (Zero Days). That’s why it’s important to stay up-to-date in identifying malicious behavior (signatures) defined by security research experts to help you prioritize your real risks.
Trivy is your Swiss army knife for security scanning of vulnerabilities, misconfigurations, and secrets, which you can get more details about here. Tracee is a runtime security and forensics tool for Linux that, together with Aqua, gives you detection and response capabilities in a unique platform integrated into all phases of your application, from code to runtime. Get more details on how Aqua delivers a complete solution in security cloud-native here.
Carolina Valencia
Carolina Valencia was a software developer interested in best practices for securing cloud-native applications. A dedicated enthusiast and member of the open-source community, she is contributor to open source projects as well as Kubernetes documentation.
|
__label__pos
| 0.781226 |
0
estoy tratando de mejorar el código de mi primer proyecto que es un to do list usando iconos de fontawesome. En cada ítem yo tengo, entre otras cosas, un botón que sirve para marcar si la tarea esta completada o no. Este boton aparece con un icono por default que en el código esta guardado en la variable CIRCLE_INCON, pero lo que quiero es que cuando se complete la tarea cambie a otro icono que esta guardado en la variable CHECK_ICON. Lo que hice es lo siguiente (Solo muestro las partes del codigo relacionadas con el botón para completar la tarea que yo llamo checkBtn):
//ICONS
const CIRCLE_INCON = `<i class="far fa-circle fa-lg"></i>`;
const CHECK_ICON = `<i class="far fa-check-circle fa-lg"></i>`;
//LOS "..." SIGNIFICAN QUE HAY CODIGO ANTES DE ESO
//PUEDEN VER ABAJO EL CODIGO COMPLETO
function addTask(fromList){
//...
//CREATING COMPONENTS
const checkBtn= document.createElement("button");
//...
//SUB-COMPONENT CHECKBOX
checkBtn.classList.add("checkbox");
checkBtn.innerHTML = CIRCLE_INCON;
checkBtn.value= "not-checked";
//...
//ASI ES COMO LLAMO A LA FUNCION QUE SE ENCARGA DE CHEQUEAR O COMPLETAR LA TAREA
item.addEventListener("click", (element) => {
element = element.target;
let circleClass = element.classList.contains("fa-circle")
let checkClass= element.classList.contains("fa-check-circle");
(circleClass || checkClass) ? checkTask() : false;
});
//CHECK THE TASK
function checkTask (){
//SI EL VALOR DEL CHECKBUTTON NO ESTA CHEQUEADO EJECUTA LA FUNCION COMPLETEDTASK()
checkBtn.value ==="not-checked" ? completedTask(): incompletedTask();
function completedTask() {
checkBtn.setAttribute("value","checked");
checkBtn.innerHTML = CHECK_ICON;
saveValue(checkBtn, CHECK_ICON);
}
function incompletedTask() {
checkBtn.setAttribute("value","not-checked");
checkBtn.innerHTML = CIRCLE_INCON;
saveValue(checkBtn, CIRCLE_INCON);
}
//ACA ES COMO GUARDO EL VALOR DE COMPLETED AL LOCALSTORAGE
function saveValue(el, icon){
const dataLS = JSON.parse(localStorage.getItem('item.list'));
dataLS[id].completed = el.value;
dataLS[id].icon = icon
el.innerHTML = dataLS[id].icon; //AQUI ES COMO INTENTE GUARDAR EL ICONO AL LOCALSTORAGE
localStorage.setItem('item.list', JSON.stringify(dataLS));
}
toDoTxt.classList.toggle("completed");
};
Así el código funciona exactamente como quiero pero yo lo que quiero es simplificar la función checkTask() usando .toggle() a algo como esto:
//PRIMERO CAMBIO DE TAG DE "button" A UN "i"
const checkBtn= document.createElement("i");
//CHECK THE TASK
function checkTask (){
//SI EL VALOR DEL CHECKBUTTON NO ESTA CHEQUEADO EJECUTA LA FUNCION COMPLETEDTASK()
checkBtn.classList.toggle("fa-circle");
checkBtn.classList.toggle("fa-check-circle");
toDoTxt.classList.toggle("completed");
//ACA ES COMO GUARDO EL VALOR DE COMPLETED AL LOCALSTORAGE
function saveValue(el, icon){
const dataLS = JSON.parse(localStorage.getItem('item.list'));
dataLS[id].completed = el.value;
dataLS[id].icon = icon
el.innerHTML = dataLS[id].icon; //AQUI ES COMO INTENTE GUARDAR EL ICONO AL LOCALSTORAGE
localStorage.setItem('item.list', JSON.stringify(dataLS));
}
};
Pero el problema es que cuando hago eso eso no aparece el icono sino que aparece un cuadrado. Si alguien sabe lo que pasa que por favor me explique se lo agradecería mucho. A continuación dejo el código completo para que vean como esta compuesto.
const form = document.getElementById("list");
const input = document.getElementById("input");
const normalButton = document.getElementById("normal");
const finishedTask = document.getElementById("done");
const unfinishedTask = document.getElementById("not-done");
const refreshBtn = document.getElementById("refresh-btn");
const failBox = document.getElementById("fail");
const closeBtn = document.getElementById("close-btn");
const taskCategories = document.getElementById("categories")
//ICONS
const CIRCLE_INCON = `<i class="far fa-circle fa-lg"></i>`;
const CHECK_ICON = `<i class="far fa-check-circle fa-lg"></i>`;
const TRASH_ICON = `<i class="fas fa-trash-alt"></i>`;
const EDIT_ICON =`<i class="far fa-edit"></i>`;
let lists = [];
document.addEventListener("keyup",(event)=>{ if(event.keyCode === 13) addTask() });
refreshBtn.addEventListener("click",() => refreshPage());
// Obtener desde localStorage al cargar todo el DOM
window.addEventListener('load', function() {
lists = JSON.parse(localStorage.getItem("item.list")) || [];
// Agregar en HTML los elementos encontrados
lists.forEach((item) => addTask(item));
});
function addTask(fromList){
event.preventDefault();
let inputValue= (fromList) ? fromList.name : input.value;
if(inputValue === "" || inputValue === null) return failAlert();
function failAlert(){
failBox.style.display = "block";
closeBtn.addEventListener("click", ()=> failBox.style.display = "none")
}
//CREATING COMPONENTS
const item = document.createElement("li");
const deleteBtn = document.createElement("button");
const toDoTxt = document.createElement("p");
const editBtn = document.createElement("button");
const checkBtn= document.createElement("i");
const btnContainer = document.createElement("div");
btnContainer.classList.add("buttons");
//ITEM COMPONENT
item.classList.add("item");
let id = item.dataset. id;
id = (fromList) ? fromList.id : lists.length;
//APPEND COMPONENTS TO THE ITEM
item.appendChild(checkBtn);
item.appendChild(toDoTxt);
form.appendChild(item);
item.appendChild(btnContainer);
//SUB-COMPONENT TODO
toDoTxt.classList.add("text");
const text= document.createTextNode(inputValue);
toDoTxt.appendChild(text);
//SUB-COMPONENT CHECKBOX
checkBtn.classList.add("checkbox");
checkBtn.innerHTML = CIRCLE_INCON;
checkBtn.value= "not-checked";
//SUB-COMPONENT EDIT BUTTON
editBtn.classList.add("edit");
editBtn.innerHTML = EDIT_ICON;
btnContainer.appendChild(editBtn);
//SUB COMPONENT DELETE BUTTON
deleteBtn.classList.add("delete");
deleteBtn.innerHTML = TRASH_ICON;
btnContainer.appendChild(deleteBtn);
item.addEventListener("click", (element) => {
element = element.target;
let circleClass = element.classList.contains("fa-circle")
let checkClass= element.classList.contains("fa-check-circle");
let trashClass = element.classList.contains("fa-trash-alt");
let editClass = element.classList.contains("fa-edit");
(circleClass || checkClass) ? checkTask() : false;
(trashClass) ? deleteTask() : false;
(editClass) ? editTask() : false;
});
//CHECK THE TASK
function checkTask (){
//SI EL VALOR DEL CHECKBUTTON NO ESTA CHEQUEADO EJECUTA LA FUNCION COMPLETEDTASK()
checkBtn.value ==="not-checked" ? completedTask(): incompletedTask();
checkBtn.classList.toggle("fa-circle");
checkBtn.classList.toggle("fa-check-circle");
function completedTask() {
checkBtn.setAttribute("value","checked");
checkBtn.innerHTML = CHECK_ICON;
saveValue(checkBtn, CHECK_ICON);
}
function incompletedTask() {
checkBtn.setAttribute("value","not-checked");
checkBtn.innerHTML = CIRCLE_INCON;
saveValue(checkBtn, CIRCLE_INCON);
}
//ACA ES COMO GUARDO EL VALOR DE COMPLETED AL LOCALSTORAGE
function saveValue(el, icon){
const dataLS = JSON.parse(localStorage.getItem('item.list'));
dataLS[id].completed = el.value;
dataLS[id].icon = icon
el.innerHTML = dataLS[id].icon; //AQUI ES COMO INTENTE GUARDAR EL ICONO AL LOCALSTORAGE
localStorage.setItem('item.list', JSON.stringify(dataLS));
}
toDoTxt.classList.toggle("completed");
};
// EDIT THE TASK
function editTask (){
//ACA YO ES DONDE EDITO LAS TAREAS
toDoTxt.innerHTML = `<div class=".edit-container" id = "edit-container"></div>`;
let editContainer = document.getElementById("edit-container");
let editInput = document.createElement("input");
let submitEdit = document.createElement("button");
editInput.classList.add("edit-input");
submitEdit.classList.add("submit-edit");
submitEdit.innerHTML = `<i class="fas fa-plus-circle fa-lg"></i>`
editContainer.appendChild(editInput);
editContainer.appendChild(submitEdit);
submitEdit.addEventListener("click",() => editTask());
//ACA ES COMO GUARDO EL NUEVO NOMBRE EDITADO AL LOCALSTORAGE
function saveNewTask(){
const dataLS = JSON.parse(localStorage.getItem('item.list'));
dataLS[parseInt(id)].name = editInput.value;
localStorage.setItem('item.list', JSON.stringify(dataLS));
}
function editTask (){
toDoTxt.innerHTML = editInput.value;
saveNewTask();
};
};
//DELETE THE TASK
function deleteTask (){
//DONDE ELIMINO LAS TAREAS
const dataLS = JSON.parse(localStorage.getItem('item.list'));
form.removeChild(item);
deleteBtn.parentNode.parentNode
dataLS.splice(dataLS.id, 1);
localStorage.setItem('item.list', JSON.stringify(dataLS));
};
taskCategories.addEventListener("click", (element) => {
element = element.target;
console.log(element);
if (element === normalButton) return goToNormal();
if (element === finishedTask) return seeFinishedTasks();
if (element === unfinishedTask) return seeUnfinishedTask();
});
let goToNormal = () => item.style.display = "flex";
let seeFinishedTasks = () => {
checkBtn.value==="checked" ? item.style.display = "flex" : item.style.display = "none";
};
let seeUnfinishedTask = () => {
checkBtn.value==="not-checked" ? item.style.display = "flex" : item.style.display = "none";
};
//UPLOADING THE DATA
let data = createDataList(inputValue, checkBtn.value, CIRCLE_INCON);
if(!fromList) {
lists.push(data);
save();
}
function save(){
localStorage.setItem("item.list", JSON.stringify(lists));
}
function createDataList(name, completed, icon){
return {id: lists.length, name: name, completed: completed, icon: icon};
}
input.value = "";
}
function refreshPage(){
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<link rel="stylesheet" href="css/style.css">
<link href="https://fonts.googleapis.com/css?family=Roboto+Slab&display=swap" rel="stylesheet">
<link href="https://fonts.googleapis.com/css?family=Roboto+Condensed:300&display=swap" rel="stylesheet">
<link href="https://stackpath.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css" rel="stylesheet" integrity="sha384-wvfXpqpZZVQGK6TAh5PVlGOfQNHSoD2xbE+QkPxCAFlNEevoEH3Sl0sibVcOQVnN" crossorigin="anonymous">
<title>To Do List</title>
</head>
<body>
<div class="container">
<div class="header">
<i class="fas fa-sync-alt" id="refresh-btn"></i>
<div class="title-container">
<h1 class="title">To Do List App</h1>
</div>
<div class="date" id="date"></div>
<div class="categories" id="categories">
<button id="normal" class="normal">NORMAL</button>
<button id="done" class="done">DONE</button>
<button id="not-done" class="not-done">NOT-DONE</button>
</div>
</div>
<div class="fail" id="fail">
<i class="fas fa-times-circle" style="color: #fff;" id="close-btn"></i>
<p class="failed-text">Please type a valid to do</p>
</div>
<div class="content">
<ul id="list" class="list"></ul>
</div>
<div class="add-item">
<input type="text" class="input" id="input" placeholder="Add something to do">
<i class="fas fa-plus-circle fa-lg" id="button" onclick="addTask()"></i>
</div>
</div>
<script defer src="https://kit.fontawesome.com/09faf5376a.js" crossorigin="anonymous"></script>
<script defer src="app.js/app.js"></script>
</body>
</html>
3
• 1
La ruta de tu href del fontawesome es incorrecta, revísala porque no es una hoja de estilos válida. Commented el 11 jun. 2020 a las 13:06
• Pero si estuviera mal la ruta ¿no tendrian que aparecer todos los iconos en mi pagina en forma de cuadrado o no tendria que aparecer un error 404 en la consola?. Porque tengo otros iconos que uso en la pagina y si aparecen esos iconos.
– FERRE FACU
Commented el 11 jun. 2020 a las 15:42
• Igual gracias ya una de las respuestas en la publicación soluciono mi problema.
– FERRE FACU
Commented el 11 jun. 2020 a las 15:43
2 respuestas 2
3
Para que fontawesome funcione es necesario que varias cosas estén correctas. 1. Que la url de la librería esté correctamente embedeada
<link href=""> en la cabecera.
1. El elemento que contendrá la tipografía (El icono) debe de contener la clase fa y la clase del icono en concreto
class="fa fa-check-circle"
Todos los iconos contienen fa-nombreIcono 3. En algunos casos, embedear solo una librería como por ejemplo "regular" ocasiona que debas usar otro tipo de clases, por ejemplo
class="fad fa-check-circle"
En todos los casos, para que un icono se muestre solo debes cambiar agregar las clases del icono en cuestión.
El tema de los "cuadritos" se debe a que la tipografía de la que está compuesta fontawesome( De ahi su nombre "Fuente increible" ) no está cargando correctamente. Revisa la consola del navegador y verifica que no existan errores 404 (Esto indicaría que si añadiste la url pero esta no es correcta o el archivo no existe).
Por otro lado, sería mejor si añadieras más información y el código que usas en el html.
1
• Con respecto a si hay errores 404, no me ha aparecido ningún error en consola y los links se que esta bien puestos porque hay otros icono que estoy cuando que si me estar cargando correctamente. Igual ahora voy a agregar el html a la publicación
– FERRE FACU
Commented el 9 jun. 2020 a las 0:54
2
+50
Tienes que agregar primero la clase far o fas al elemento i después de crearlo. Sin eso te aparecera un cuadro y no el icono cuando intentes cambiarlo.
La corrección sería así:
//PRIMERO CAMBIO DE TAG DE "button" A UN "i"
const checkBtn= document.createElement("i");
// Agrega la clase 'far fa-circle' al elemento i cuando se crea
checkBtn.className = 'far fa-circle';
// ...
//CHECK THE TASK
function checkTask (){
//SI EL VALOR DEL CHECKBUTTON NO ESTA CHEQUEADO EJECUTA LA FUNCION COMPLETEDTASK()
checkBtn.classList.toggle("fa-circle");
checkBtn.classList.toggle("fa-check-circle");
toDoTxt.classList.toggle("completed");
//ACA ES COMO GUARDO EL VALOR DE COMPLETED AL LOCALSTORAGE
function saveValue(el, icon){
const dataLS = JSON.parse(localStorage.getItem('item.list'));
dataLS[id].completed = el.value;
dataLS[id].icon = icon
el.innerHTML = dataLS[id].icon; //AQUI ES COMO INTENTE GUARDAR EL ICONO AL LOCALSTORAGE
localStorage.setItem('item.list', JSON.stringify(dataLS));
}
};
Espero te sirva.
1
• Si gracias por la respuesta y la explicación. Ambas me sirvieron.
– FERRE FACU
Commented el 11 jun. 2020 a las 15:49
Tu Respuesta
By clicking “Publica tu respuesta”, you agree to our terms of service and acknowledge you have read our privacy policy.
¿No es la respuesta que buscas? Examina otras preguntas con la etiqueta o formula tu propia pregunta.
|
__label__pos
| 0.761666 |
Library providing syntactic sugar for creating variant forms of a canonical function
Switch branches/tags
Nothing to show
Clone or download
Fetching latest commit…
Cannot retrieve the latest commit at this time.
Permalink
Type Name Latest commit message Commit time
Failed to load latest commit information.
changelog.d
docs
src/variants
tests
.editorconfig
.gitignore
.travis.yml
AUTHORS.rst
CONTRIBUTING.rst
LICENSE
MANIFEST.in
NEWS.rst
README.rst
pyproject.toml
requirements_dev.txt
setup.cfg
setup.py
tox.ini
README.rst
variants
Documentation Status
variants is a library that provides syntactic sugar for creating alternate forms of functions and other callables, in the same way that alternate constructors are class methods that provide alternate forms of the constructor function.
To create a function with variants, simply decorate the primary form with @variants.primary, which then adds the .variant decorator to the original function, which can be used to register new variants. Here is a simple example of a function that prints text, with variants that specify the source of the text to print:
import variants
@variants.primary
def print_text(txt):
print(txt)
@print_text.variant('from_file')
def print_text(fobj):
print_text(fobj.read())
@print_text.variant('from_filepath')
def print_text(fpath):
with open(fpath, 'r') as f:
print_text.from_file(f)
@print_text.variant('from_url')
def print_text(url):
import requests
r = requests.get(url)
print_text(r.text)
print_text and its variants can be used as such:
print_text('Hello, world!') # Hello, world!
# Create a text file
with open('hello_world.txt', 'w') as f:
f.write('Hello, world (from file)')
# Print from an open file object
with open('hello_world.txt', 'r') as f:
print_text.from_file(f) # Hello, world (from file)
# Print from the path to a file object
print_text.from_filepath('hello_world.txt') # Hello, world (from file)
# Print from a URL
hw_url = 'https://ganssle.io/files/hello_world.txt'
print_text.from_url(hw_url) # Hello, world! (from url)
Differences from singledispatch
While variants and singledispatch are both intended to provide alternative implementations to a primary function, the overall aims are slightly different. singledispatch transparently dispatches to variant functions based on the type of the argument, whereas variants provides explicit alternative forms of the function. Note that in the above example, both print_text.from_filepath and print_text.from_url take a string, one representing a file path and one representing a URL.
Additionally, the variants is compatible with singledispatch, so you can have the best of both worlds; an example that uses both:
@variants.primary
@singledispatch
def add(x, y):
return x + y
@add.variant('from_list')
@add.register(list)
def add(x, y):
return x + [y]
Which then automatically dispatches between named variants based on type:
>>> add(1, 2)
3
>>> add([1], 2)
[1, 2]
But also exposes the explicit variant functions:
>>> add.from_list([1], 2)
[1, 2]
>>> add.from_list()
7 @add.register(list)
8 def add(x, y):
----> 9 return x + [y]
TypeError: unsupported operand type(s) for +: 'int' and 'list'
It is important to note that the variants decorators must be the outer decorators.
Installation
To install variants, run this command in your terminal:
$ pip install variants
Requirements
This is a library for Python, with support for versions 2.7 and 3.4+.
|
__label__pos
| 0.96461 |
Q
Part III: What's possible in the realm of sockets when communicating with CICS?
I have a general question about what is possible in the realm of sockets when communicating with a CICS mainframe.
We have a piece of middleware software that collects and consolidates data from various sources and then passes that data on to various external systems such as ODBC databases, etc. We have provided a generic XML protocol over a TCP/IP socket connection to allow flexible connections to any backend host computers. The problem we are now hitting is a client who claims our topology/protocol isn't possible for them to handle in their environment.
Here is our system in a nutshell: We open a socket connection to the remote port and send what amounts to a logon message. Assuming we receive a favorable response we then send data messages up, each of which has a response that can contain error information or possible result from the remote processing. If a specified amount of idle time elapses between these data messages, we send a kind of keep-alive message that again has a response. Finally, when the overall processing is complete send the equivalent of a logoff message.
A few additional details: Because our software has no guarantee of how fast the host program will process any individual data message, we support a small fixed number of simultaneous connections so that long-running tasks won't totally block other shorter-running ones. The volume of these data messages varies from many-per-second to periods where minutes (or even perhaps hours) can go by without any new data. This approach works fine for the likes of a Java program running on a Unix box, but does it really present such a problem in the CICS world?
Continued from part II.
The problem is that hanging around on an open TCP/IP socket requires a CICS transaction on the end to get the flow when it arrives. You cannot use a listener-type techniques because that's only processing initial binds -- not just something that takes ages to arrive.
The only way to accomplish what you want is to use a CICS waiting mechanism, and poll the socket at frequent (as often as you think prudent bearing in mind that while waiting you will not be processing the flow from the client) intervals. This could go on forever, and so you are in danger of ending up in a situation whereby all your CICS active transaction slots are filled with transactions just polling away and not doing anything of interest.
One way around this is to use the CMAXTASK facilities to limit the number of the second service transactions in the region, but that means that the clients will be hanging around in cases of stress not getting anything back from CICS. You can have multiple CICS regions servicing the flows (using MVS TCP/IP port sharing) to alleviate the conditions, but you still may end up in an apparent application wait state.
Whether or not the client has been coded to cope with a volume-related stall (somewhat unlikely as Unix applications tend always to assume that infinite resource is available and so trivial volume-related stalls never occur) the first service transaction can use XC INQUIRE facilities to detect the number of active second service transactions and then simply refuse to start the second service transaction if capacity is exceeded (it might be nice to return an error message).
The connection may also TCP/IP timeout at any point -- so ensure that you CLOSE() the socket before XC RETURNing.
So, the last bit of code in the second service transaction should look like:
CURTRY = MAXTRY (whatever you decide is the maximum number of intervals)
waitloop:
RECV(PEEK) ; if non-zero, reloop on the receive logic
IF ( CURTRY <= 0 ) then end by CLOSE()ing the socket ; XC RETURN
XC DELAY INTERVAL(n)
reloop to waitloop
You will have to decide upon how long a client can keep the connection going by sending an heartbeat; is it valid to have an inifinite number of these? If this is the case, then each socket/connection will have an associated CICS transaction and this will potentially eat up the number of CICS AMAXTASK slots -- so you could get into a stall whereby everything is not running as they are just awaiting a next flow from the client.
Similarly, think about defining the maximum number of heartbeats (both per socket and per not-doing-anything) and policing that count.
I'd hope that as this XML protocol is under you control, you could include on the initial logon-type of flow some sort of control information that would set a maximum time interval for the lifetime of the socket and the Heartbeat properties. The second service transaction could use that info to limit the volumes of inactive sockets.
Similarly, you should be able to code up the client so that its receipt of the security identification/lgon response can respond to a too-many-active-at-the-moment-retry-later can be coded around. It already should be able to cope with an error because the Socket is no longer open (because the server/CICS has closed it).
So, it all comes down to restricting the number of 2nd Service Transactions hanging around just awaiting for something interesting from the Client.
You can tell your doubting customer that I've said it's OK to use your software within CICS! But you should ensure that your protocol is sufficiently flexable first.
Click to return to part I.
This was first published in December 2002
This Content Component encountered an error
Pro+
Features
Enjoy the benefits of Pro+ membership, learn more and join.
Have a question for an expert?
Please add a title for your question
Get answers from a TechTarget expert on whatever's puzzling you.
You will be able to add details on the next page.
0 comments
Oldest
Forgot Password?
No problem! Submit your e-mail address below. We'll send you an email containing your password.
Your password has been sent to:
-ADS BY GOOGLE
SearchWindowsServer
SearchEnterpriseLinux
SearchServerVirtualization
SearchCloudComputing
Close
|
__label__pos
| 0.56859 |
IL.Create/Update Legal Entity
Lucidchart link
Authorize user
1. Validate MIS API Token
2. Check MIS scopes legal_entity:write in order to perform this action
1. In case error - generate 401 response
Digital signature
1. Validate signature
2. Extract signer Certificate details
3. Save Signed Content
Validate Tax ID
1. Check that EDRPOU in Certificate details exists and not empty
1. Check that EDRPOU in Certificate details is valid according to ^[0-9]{8,10}$
2. Check that EDRPOU in Certificate details is equal to EDRPOU in Legal Entity payload
1. In case validation fails - generate 422 error
2. If EDRPOU in Certificate details is empty check that DRFO exists and not empty
1. Check that DRFO in Certificate details is valid according to ^[0-9]{9,10}$
2. Check that DRFO in Certificate details is equal to EDRPOU in Legal Entity payload
1. In case validation fails - generate 422 error
3. In case EDRPOU and DRFO is empty return error 422, msg "EDRPOU and DRFO is empty in digital sign"
Validate Legal Entity request
1. Validate request using JSON schema
1. In case validation fails - generate 422 error
Validate KVEDS
Validate KVEDs allowed for registration according to Legal Entity type
1. If Request: $.type = MSP, check that at least one KVED Request: $.kveds is in KVEDS_ALLOWED_MSP list
2. If Request: $.type = PHARMACY, check that at least one KVED Request: $.kveds is in KVEDS_ALLOWED_PHARMACY list
Validate owners position
owners positions should be one of the list - env. configuration OWNER_POSITIONS.
At the moment it is - "P1, P2, P3, P32, P4, P6, P5, P18, P19, P22, P23, P24, P25, P26, P229, P230, P231, P232, P233, P234, P235, P236, P240"
EDR Validation
The data provided by MIS during LE registration must matches with the data stored in EDR
General rules
1. LE must be registered in EDR
2. LE must be active in EDR
3. If new LE type is created then no active or suspended LE type should exist in e-Health with same EDRPOU and different edr_id as active record in EDR
Implementation
TBC
Save signed content
Generate RESOURCE_NAME for signed content
Generate RESOURCE_ID for Legal Entity
Invoke Media Content Storage to get upload URL for the document
ParameterSource
action'PUT'
bucket'LEGAL_ENTITIES'
resource_id:RESOURCE_NAME
resource_name:RESOURCE_ID
Upload file via Google Cloud API
Refresh EDR Data
1. Get detailed data from EDR for active EDR record: call EDR data validation#Getdetailedinformation using id, received on step EDR Validation.
2. Create or Update record in prm.edr_data for active EDR record data.
Mapping
eHealthEDR
edr_id$.id
name$.names.name
short_name$.names.short
public_name$.names.display
legal_form$.olf_code
kveds$.activity_kinds
registration_address$.address
state$.state
Search Legal Entity using EDRPOU from Signer Certificate
Search Legal Entity in prm.legal_entities using EDRPOU from Signer Certificate where (legal_entities.type = $.type or legal_entities.type = 'PRIMARY_CARE' if $.type ='MSP') and status = active or suspended
1. In case record found go to Update Client in Auth
2. In case record not found go to Generate client and legal entity identifier
Request parameters:
ParameterSourceDescription
EDRPOUSigner certificateExtracted from public signer certificate
See Service specification
Update Client in Auth
Update Redirect URI
Update Legal Entity
Update legal_entities (see Table specs)
1. Change the record according to received request
1. Some of the fields must not be changed. This fields must be excluded from PATCH request: EDRPOU, TYPE
2. set accreditations = Request: $.medical_services_providers.accreditations
3. save residence address into legal_entities.residence_address
2. If nhs_reviewed = true reset it to false
Update licenses (see Table specs)
1. Search for a record in prm.licenses for current legal entity
1. Check if any attribute has changed comparing request and prm data. If yes, then:
1. For each record in prm.legal_entities related to the license:
1. If nhs_reviewed = true reset it to false
2. If nhs_verified = true
1. reset it to false
2. set `nhs_unverified_at` to current date
2. Update license data according to the request
See Service specification
Update LE official name
Extract official name from Digital signature.
• Update `legal_entities.public_name` in prm with official name from Digital signature
• Update `clients.name` in mithril with official name from Digital signature
Update Contract Request
1. Find contract requests related to current legal entity in status = NEW, APPROVED, PENDING_NHS_SIGN, NHS_SIGNED, set status = 'TERMINATED', status_reason "Legal Entity Data has changed"
Reset flags
Update Contract
In case Legal Entity already exists in DB we need to check whether next fields weren't changed:
• name
• addresses (any field in array)
• status
Compare fields in prm.legal_entities and in request. In case any of the fields above were changed
find contracts by contractor_legal_entity_id=$legal_entities.id and status='VERIFIED'. Set ops.contracts.is_suspended=true
Generate client and legal entity identifier
Generate UUID as client identifier and legal entity identifier
Create client and connection in Auth
1. Invoke Auth API (idempotent PUT) to register new client and generate client secret key
2. Determine MIS_ID (consumer_id) using api-key
3. Invoke Put client connection in order to UpSet connection in Mithril for this client and consumer
Create new Legal Entity
Populate licenses table (see Table specs)
1. select licenses.id from prm.licenses where type = $.type and id in (select license_id from legal_entities where edr_data_id in (select id from edr_data where edr_id = active edr id))
1. If record found then
1. check if any attribute has changed comparing request and prm data. If yes, then:
1. For each record in prm.legal_entities related to the license:
1. If nhs_reviewed = true reset it to false
2. If nhs_verified = true
1. reset it to false
2. set `nhs_unverified_at` to current date
2. Update license data according to the request
2. if record not found then create new record:
1. set 'licenses.type' = MSP if Request: $.type = MSP
2. set 'licenses.type' = PHARMACY if Request: $.type = PHARMACY
3. set 'licenses.is_active' = true and populate inserted_by, inserted_at, updated_by, updated_at fields
4. set issued_by, issued_date, expiry_date, active_from_date, what_licensed, order_no according to Request: $.medical_services_providers.licenses data
5. Link created record and edr_data record
Populate legal_entities table (see Table specs)
1. Create new record according to received request
1. Link created record and edr_data record
2. Link created record and license
3. set accreditations = Request: $.medical_services_providers.accreditations
4. save registration_address separately
5. set nhs_verified, nhs_reviewed to false
6. set `nhs_unverified_at` to current date
7. set status = active suspended
See Service specification
Save LE official name
Extract official name from Digital signature.
• Save LE official name to prm DB = legal_entities.public_name
• Save LE official name to mithril DB = clients.name
Check UKR_MED_REGISTRY
Manual verification is not required if Legal Entity is in Registry
1. Search EDRPOU in UKR_MED_REGISTRY with corresponding type
See Service specification
Set Legal Entity mis_verified to VERIFIED
Set mis_verified to VERIFIED for PRM.legal_entity if Legal Entity is in Registry
Request parameters:
ParameterSourceDescription
idCreate new Legal Entity, Search Legal Entity using Tax ID from Signer Certificate
mis_verified
Const: VERIFIEDConstant value
Set Legal Entity mis_verified to NOT_VERIFIED
Set mis_verified to NOT_VERIFIED for PRM.legal_entity if Legal Entity is not in Registry
Request parameters:
ParameterSourceDescription
idCreate new MSP, Search MSP using Tax ID from Signer Certificate
mis_verifiedConst: NOT_VERIFIEDConstant value
Register new employee
Init IL.Create employee request
Determine employee_type
Mapping
ParameterSource
legal_entity_idGenerated previously legal_entities.id
employee_type
Variable: Employee_type
position
Request: $.owner.position
status
Const: PENDING_VERIFICATION
start_date
now()
party
Request: $.owner
Except: $.owner.position
|
__label__pos
| 0.961205 |
Using Retina MacBook Pro with external display
Discussion in 'MacBook Pro' started by C4FF, Jun 12, 2012.
1. C4FF macrumors member
Joined:
Jun 12, 2012
#1
Quick question.
In theory, if I connected a Retina MacBook Pro to an external screen and closed it (with mouse and keybord connected obviously), would the retina display stop completely?
I ask because if I wanted to game on a HDTV, would the machine being retina but CLOSED use more gpu than a regular MacBook pro doing the same task?
That was hard to explain :p
2. GGJstudios macrumors Westmere
GGJstudios
Joined:
May 16, 2008
#2
Yes, when you operate in clamshell mode, the internal display is off.
3. surjavarman macrumors 6502a
Joined:
Nov 24, 2007
#3
Yes cause it will go in to sleep mode as soon as you close it
4. C4FF thread starter macrumors member
Joined:
Jun 12, 2012
Share This Page
|
__label__pos
| 0.974367 |
Support
Account
Home Forums General Issues Use get_sub_field outside have_rows Reply To: Use get_sub_field outside have_rows
• Hey, here it is:
$content_blocks = get_field('content_blocks', $project->ID);
if( $content_blocks ):
foreach ($content_blocks as $key => $content_block):
$layout = $content_block['layout'];
$block_type = $classes[ $content_block['acf_fc_layout'] ];
$start = $content_block['start_at'];
$stop = $content_block['stop_at'];
if( $block_type == 'block__image' ):
$image = $content_block['image'];
$image_url = $image['sizes']['post-thumbnail'];
...
elseif( $block_type == 'block__video' ):
// Here is the problem... can't get the video the same way as with the image
$video = $content_block['video'];
// $video_url = get_sub_field('video', false, false);
...
endif;
endforeach;
endif;
|
__label__pos
| 0.996352 |
s.p. digital icon s.p. Digital slogan
Single Sign-On (SSO) is an authentication process that enables users to access multiple applications or systems with a single set of credentials, such as a username and password. Instead of requiring users to log in separately to each application, SSO allows them to authenticate once and gain access to all authorized resources seamlessly. This enhances user experience by reducing the need for multiple logins and simplifying the authentication process. Additionally, SSO improves security by centralizing user authentication and enforcing consistent access policies across different platforms or services.
|
__label__pos
| 0.945763 |
Replacing Cartopy’s Background Image
metadata
• keywords:
• software
• python
• cartopy
• published:
• RSS Feed
Whilst discussing a friend’s summary of 2017 I found it difficult to place parts of Michigan that I had visited as The Great Lakes were missing from the state boundaries. My friend then countered that my own maps did not feature The Great Lakes either. Disbelieving, I went away and checked; he was correct - Cartopy’s default background image does not show The Great Lakes (shown below).
Cartopy’s default background image
Cartopy’s default background image
If you would like to check for yourself then the following Python code will show you where the default image is located for your installation of Cartopy.
As of January 2018, that folder only contains one image, called 50-natural-earth-1-downsampled.png, which has dimensions 720x360 and size 327K. The sidecar JSON file (shown below) describes the contents of the folder.
I decided to create my own folder of background images, using the same specification that Cartopy uses, so that I could have both higher resolution background images and ones that contained large inland bodies of water. The script that I wrote to download the ZIP archives from Natural Earth, extract the TIFFs, create the PNGs and create the JSON is shown below. It runs both convert and optipng during its execution.
The JSON file that it creates is shown below for comparison with the original one.
The PNG files that it creates are shown below for comparison with the original one too. Note that The Great Lakes and Lake Victoria are easily visible (to name a few).
Download:
1. 512x256
2. 1024x512
3. 2048x1024
4. 4096x2048
5. 21600x10800
Great - but how am I going to use this new folder that has the same organisation as Cartopy’s folder? I have written a replacement function, called pyguymer.add_map_background(), that I can use so that my scripts use this new folder. Now all that I need to do is set the environment variable $CARTOPY_USER_BACKGROUNDS to the path that my new images are in and then replace all instances of ax.stock_img() in all of my Python scripts with either pyguymer.add_map_background(ax) (for a low-resolution background image for testing purposes) or pyguymer.add_map_background(ax, resolution = "large4096px") (for a high-resolution background image for publishing purposes).
You can see the fruits of this work by looking at the example image in my Flight Map Creator project on GitHub. The following web pages have been useful during this project:
|
__label__pos
| 0.632196 |
SAP Basis SP01 Output control: spool requests - SAP Corner
Direkt zum Seiteninhalt
SP01 Output control: spool requests
Do you have any questions?
Daily checks are still commonplace for many SAP customers today; with Avantra, they are a thing of the past. These are manual checks that a bot can perform hundreds of times each day. Similarly, a bot can create incidents or notifications when something goes wrong.
Basis includes a client/server architecture and configuration, a relational database management system (DBMS), and a graphical user interface (GUI). In addition to the interfaces between system elements, Basis includes a development environment for R/3 applications, a data directory, and user and system administration and monitoring tools.
SAP Business Objects: CMCRegister Card Configuration Permissions
The presentation layer is based on the software components, collectively called "SAP GUI". This includes several possible implementation variants: for example, SAP GUI for HTML (Web GUI) and Web Dynpro for ABAP (WDA). Since the respective GUI depends entirely on the concrete application, the presentation layer looks very different in practice.
SAP Basis is the cornerstone of your SAP system and failures can lead to significant and annoying problems. For assistance in building and expanding SAP Basis, SAP Basis consultants can help. Certified SAP consultants enable tailored solutions for any business landscape.
"Shortcut for SAP Systems" makes it easier and quicker to complete a number of SAP basis tasks.
The coverage of old core tasks (such as security or compliance) and new core tasks (such as cloud or mobility) must be increased in the sense of a holistic consideration.
Understanding the structure and functioning of the system is especially important for IT administration. It is not for nothing that "SAP Basis Administrator" is a separate professional field. On the page www.sap-corner.de you will find useful information on this topic.
The service catalogue must be structured in such a way that the criteria, which cannot be answered clearly, can be identified and subjected to continuous consideration.
SAP Corner
Zurück zum Seiteninhalt
|
__label__pos
| 0.687796 |
Importing
To import a spreadsheet containing site data into the Workbench, follow these steps:
1. Make sure the name of the spreadsheet you want to import is ‘content.xlsx’.
2. Make a zip archive containing the .xlsx file and any other external files used for include tokens (e.g. an ‘include’ folder containing static html content). Make sure that you make the zip archive from the files themselves, not from a folder containing those files (This means when you open the zip archive, you should immediately see the ‘content.xlsx’ file rather than a folder).
3. Click Export/Import > Import, select the zip file, and click Upload.
You will instantly see changes you made in the spreadsheet appear on the Workbench dashboard.
|
__label__pos
| 0.981095 |
What I Learned About volume
What I Learned
I learned that volume is the filling of a 3-D object w/unit cubes. Volume for any type of shape is V=Bh
Some tips, tricks, and instructions on volume
How you find volume for some types of 3-D shapes
· How to find volume of a triangular prism is (½bh)h
· How to find volume of a rectangular prism is V=(L* w)h
· How to find volume of a cylinder is (πr2)
· How to find volume of a cube is L*w*h
Things you might want to know or use for finding volume.
h= height of a prism or cylinder. B= the # of cubic units in the first layer. h= # of layers. To find the area of a circle use πr². This also can help π=3.14 or 22/7
Hope that helped you!
|
__label__pos
| 0.778324 |
Changing the name on Automation emails?
Everytime a user gets an email through an automation it says it's from our sys admin.
So it'd say eg: John Smith via Smartsheet ([email protected])
This is an issue as our sys admin 'John smith' is nothing to do with that specific email automation, but because it was set up by him or he is the sys admin, it's his name that it uses.
This seems to happen with every automation (we have a quite a few) and we'd like to change it to something more generic. We've just gone to an Enterprise plan, so there may be some extra functionality that could do this that we've not discovered yet.
How would we go about doing this?
Thanks!
Best Answer
Answers
|
__label__pos
| 0.507039 |
文档章节
一篇写数据时间序列分析的不错的文章
o
osc_fmg49rzg
发布于 2019/03/20 09:58
字数 5148
阅读 114
收藏 0
转自
https://www.cnblogs.com/foley/p/5582358.html#undefined
什么是时间序列
时间序列简单的说就是各时间点上形成的数值序列，时间序列分析就是通过观察历史数据预测未来的值。在这里需要强调一点的是，时间序列分析并不是关于时间的回归，它主要是研究自身的变化规律的（这里不考虑含外生变量的时间序列）。
为什么用python
两个字总结“情怀”，爱屋及乌，个人比较喜欢python，就用python撸了。能做时间序列的软件很多，SAS、R、SPSS、Eviews甚至matlab等等，实际工作中应用得比较多的应该还是SAS和R，前者推荐王燕写的《应用时间序列分析》，后者推荐“基于R语言的时间序列建模完整教程”这篇博文（翻译版）。python作为科学计算的利器，当然也有相关分析的包:statsmodels中tsa模块，当然这个包和SAS、R是比不了，但是python有另一个神器：pandas！pandas在时间序列上的应用，能简化我们很多的工作。
环境配置
python推荐直接装Anaconda，它集成了许多科学计算包，有一些包自己手动去装还是挺费劲的。statsmodels需要自己去安装，这里我推荐使用0.6的稳定版，0.7及其以上的版本能在github上找到，该版本在安装时会用C编译好，所以修改底层的一些代码将不会起作用。
时间序列分析
1.基本模型
自回归移动平均模型(ARMA(p，q))是时间序列中最为重要的模型之一，它主要由两部分组成： AR代表p阶自回归过程，MA代表q阶移动平均过程，其公式如下：
依据模型的形式、特性及自相关和偏自相关函数的特征，总结如下：
在时间序列中，ARIMA模型是在ARMA模型的基础上多了差分的操作。
2.pandas时间序列操作
大熊猫真的很可爱，这里简单介绍一下它在时间序列上的可爱之处。和许多时间序列分析一样，本文同样使用航空乘客数据（AirPassengers.csv）作为样例。
数据读取：
复制代码
# -*- coding:utf-8 -*-
import numpy as np
import pandas as pd
from datetime import datetime
import matplotlib.pylab as plt
# 读取数据，pd.read_csv默认生成DataFrame对象，需将其转换成Series对象
df = pd.read_csv('AirPassengers.csv', encoding='utf-8', index_col='date')
df.index = pd.to_datetime(df.index) # 将字符串索引转换成时间索引
ts = df['x'] # 生成pd.Series对象
# 查看数据格式
ts.head()
ts.head().index
复制代码
查看某日的值既可以使用字符串作为索引，又可以直接使用时间对象作为索引
ts['1949-01-01']
ts[datetime(1949,1,1)]
两者的返回值都是第一个序列值：112
如果要查看某一年的数据，pandas也能非常方便的实现
ts['1949']
切片操作：
ts['1949-1' : '1949-6']
注意时间索引的切片操作起点和尾部都是包含的，这点与数值索引有所不同
pandas还有很多方便的时间序列函数，在后面的实际应用中在进行说明。
3. 平稳性检验
我们知道序列平稳性是进行时间序列分析的前提条件，很多人都会有疑问，为什么要满足平稳性的要求呢？在大数定理和中心定理中要求样本同分布（这里同分布等价于时间序列中的平稳性），而我们的建模过程中有很多都是建立在大数定理和中心极限定理的前提条件下的，如果它不满足，得到的许多结论都是不可靠的。以虚假回归为例，当响应变量和输入变量都平稳时，我们用t统计量检验标准化系数的显著性。而当响应变量和输入变量不平稳时，其标准化系数不在满足t分布，这时再用t检验来进行显著性分析，导致拒绝原假设的概率增加，即容易犯第一类错误，从而得出错误的结论。
平稳时间序列有两种定义：严平稳和宽平稳
严平稳顾名思义，是一种条件非常苛刻的平稳性，它要求序列随着时间的推移，其统计性质保持不变。对于任意的τ，其联合概率密度函数满足：
严平稳的条件只是理论上的存在，现实中用得比较多的是宽平稳的条件。
宽平稳也叫弱平稳或者二阶平稳（均值和方差平稳），它应满足：
• 常数均值
• 常数方差
• 常数自协方差
平稳性检验：观察法和单位根检验法
基于此，我写了一个名为test_stationarity的统计性检验模块，以便将某些统计检验结果更加直观的展现出来。
复制代码
# -*- coding:utf-8 -*-
from statsmodels.tsa.stattools import adfuller import pandas as pd import matplotlib.pyplot as plt import numpy as np from statsmodels.graphics.tsaplots import plot_acf, plot_pacf
# 移动平均图 def draw_trend(timeSeries, size): f = plt.figure(facecolor='white') # 对size个数据进行移动平均 rol_mean = timeSeries.rolling(window=size).mean() # 对size个数据进行加权移动平均 rol_weighted_mean = pd.ewma(timeSeries, span=size) timeSeries.plot(color='blue', label='Original') rolmean.plot(color='red', label='Rolling Mean') rol_weighted_mean.plot(color='black', label='Weighted Rolling Mean') plt.legend(loc='best') plt.title('Rolling Mean') plt.show() def draw_ts(timeSeries):
f = plt.figure(facecolor='white') timeSeries.plot(color='blue') plt.show() '''
Unit Root Test The null hypothesis of the Augmented Dickey-Fuller is that there is a unit root, with the alternative that there is no unit root. That is to say the bigger the p-value the more reason we assert that there is a unit root ''' def testStationarity(ts): dftest = adfuller(ts) # 对上述函数求得的值进行语义描述 dfoutput = pd.Series(dftest[0:4], index=['Test Statistic','p-value','#Lags Used','Number of Observations Used']) for key,value in dftest[4].items(): dfoutput['Critical Value (%s)'%key] = value return dfoutput # 自相关和偏相关图，默认阶数为31阶 def draw_acf_pacf(ts, lags=31): f = plt.figure(facecolor='white') ax1 = f.add_subplot(211) plot_acf(ts, lags=31, ax=ax1) ax2 = f.add_subplot(212) plot_pacf(ts, lags=31, ax=ax2) plt.show()
复制代码
观察法，通俗的说就是通过观察序列的趋势图与相关图是否随着时间的变化呈现出某种规律。所谓的规律就是时间序列经常提到的周期性因素，现实中遇到得比较多的是线性周期成分，这类周期成分可以采用差分或者移动平均来解决，而对于非线性周期成分的处理相对比较复杂，需要采用某些分解的方法。下图为航空数据的线性图，可以明显的看出它具有年周期成分和长期趋势成分。平稳序列的自相关系数会快速衰减，下面的自相关图并不能体现出该特征，所以我们有理由相信该序列是不平稳的。
单位根检验：ADF是一种常用的单位根检验方法，他的原假设为序列具有单位根，即非平稳，对于一个平稳的时序数据，就需要在给定的置信水平上显著，拒绝原假设。ADF只是单位根检验的方法之一，如果想采用其他检验方法，可以安装第三方包arch，里面提供了更加全面的单位根检验方法，个人还是比较钟情ADF检验。以下为检验结果，其p值大于0.99，说明并不能拒绝原假设。
3. 平稳性处理
由前面的分析可知，该序列是不平稳的，然而平稳性是时间序列分析的前提条件，故我们需要对不平稳的序列进行处理将其转换成平稳的序列。
a. 对数变换
对数变换主要是为了减小数据的振动幅度，使其线性规律更加明显（我是这么理解的时间序列模型大部分都是线性的，为了尽量降低非线性的因素，需要对其进行预处理，也许我理解的不对）。对数变换相当于增加了一个惩罚机制，数据越大其惩罚越大，数据越小惩罚越小。这里强调一下，变换的序列需要满足大于0，小于0的数据不存在对数变换。
ts_log = np.log(ts)
test_stationarity.draw_ts(ts_log)
b. 平滑法
根据平滑技术的不同，平滑法具体分为移动平均法和指数平均法。
移动平均即利用一定时间间隔内的平均值作为某一期的估计值，而指数平均则是用变权的方法来计算均值
test_stationarity.draw_trend(ts_log, 12)
从上图可以发现窗口为12的移动平均能较好的剔除年周期性因素，而指数平均法是对周期内的数据进行了加权，能在一定程度上减小年周期因素，但并不能完全剔除，如要完全剔除可以进一步进行差分操作。
c. 差分
时间序列最常用来剔除周期性因素的方法当属差分了，它主要是对等周期间隔的数据进行线性求减。前面我们说过，ARIMA模型相对ARMA模型，仅多了差分操作，ARIMA模型几乎是所有时间序列软件都支持的，差分的实现与还原都非常方便。而statsmodel中，对差分的支持不是很好，它不支持高阶和多阶差分，为什么不支持，这里引用作者的说法：
作者大概的意思是说预测方法中并没有解决高于2阶的差分，有没有感觉很牵强，不过没关系，我们有pandas。我们可以先用pandas将序列差分好，然后在对差分好的序列进行ARIMA拟合，只不过这样后面会多了一步人工还原的工作。
diff_12 = ts_log.diff(12)
diff_12.dropna(inplace=True)
diff_12_1 = diff_12.diff(1)
diff_12_1.dropna(inplace=True)
test_stationarity.testStationarity(diff_12_1)
从上面的统计检验结果可以看出，经过12阶差分和1阶差分后，该序列满足平稳性的要求了。
d. 分解
所谓分解就是将时序数据分离成不同的成分。statsmodels使用的X-11分解过程，它主要将时序数据分离成长期趋势、季节趋势和随机成分。与其它统计软件一样，statsmodels也支持两类分解模型，加法模型和乘法模型，这里我只实现加法，乘法只需将model的参数设置为"multiplicative"即可。
from statsmodels.tsa.seasonal import seasonal_decompose
decomposition = seasonal_decompose(ts_log, model="additive") trend = decomposition.trend seasonal = decomposition.seasonal residual = decomposition.resid
得到不同的分解成分后，就可以使用时间序列模型对各个成分进行拟合，当然也可以选择其他预测方法。我曾经用过小波对时序数据进行过分解，然后分别采用时间序列拟合，效果还不错。由于我对小波的理解不是很好，只能简单的调用接口，如果有谁对小波、傅里叶、卡尔曼理解得比较透，可以将时序数据进行更加准确的分解，由于分解后的时序数据避免了他们在建模时的交叉影响，所以我相信它将有助于预测准确性的提高。
4. 模型识别
在前面的分析可知，该序列具有明显的年周期与长期成分。对于年周期成分我们使用窗口为12的移动平进行处理，对于长期趋势成分我们采用1阶差分来进行处理。
rol_mean = ts_log.rolling(window=12).mean()
rol_mean.dropna(inplace=True)
ts_diff_1 = rol_mean.diff(1)
ts_diff_1.dropna(inplace=True)
test_stationarity.testStationarity(ts_diff_1)
观察其统计量发现该序列在置信水平为95%的区间下并不显著，我们对其进行再次一阶差分。再次差分后的序列其自相关具有快速衰减的特点，t统计量在99%的置信水平下是显著的，这里我不再做详细说明。
ts_diff_2 = ts_diff_1.diff(1)
ts_diff_2.dropna(inplace=True)
数据平稳后，需要对模型定阶，即确定p、q的阶数。观察上图，发现自相关和偏相系数都存在拖尾的特点，并且他们都具有明显的一阶相关性，所以我们设定p=1, q=1。下面就可以使用ARMA模型进行数据拟合了。这里我不使用ARIMA(ts_diff_1, order=(1, 1, 1))进行拟合，是因为含有差分操作时，预测结果还原老出问题，至今还没弄明白。
from statsmodels.tsa.arima_model import ARMA
model = ARMA(ts_diff_2, order=(1, 1))
result_arma = model.fit( disp=-1, method='css')
5. 样本拟合
模型拟合完后，我们就可以对其进行预测了。由于ARMA拟合的是经过相关预处理后的数据，故其预测值需要通过相关逆变换进行还原。
复制代码
predict_ts = result_arma.predict()
# 一阶差分还原
diff_shift_ts = ts_diff_1.shift(1)
diff_recover_1 = predict_ts.add(diff_shift_ts)
# 再次一阶差分还原 rol_shift_ts = rol_mean.shift(1) diff_recover = diff_recover_1.add(rol_shift_ts) # 移动平均还原 rol_sum = ts_log.rolling(window=11).sum() rol_recover = diff_recover*12 - rol_sum.shift(1) # 对数还原 log_recover = np.exp(rol_recover) log_recover.dropna(inplace=True)
复制代码
我们使用均方根误差（RMSE）来评估模型样本内拟合的好坏。利用该准则进行判别时，需要剔除“非预测”数据的影响。
复制代码
ts = ts[log_recover.index] # 过滤没有预测的记录
plt.figure(facecolor='white') log_recover.plot(color='blue', label='Predict') ts.plot(color='red', label='Original') plt.legend(loc='best') plt.title('RMSE: %.4f'% np.sqrt(sum((log_recover-ts)**2)/ts.size)) plt.show()
复制代码
观察上图的拟合效果，均方根误差为11.8828，感觉还过得去。
6. 完善ARIMA模型
前面提到statsmodels里面的ARIMA模块不支持高阶差分，我们的做法是将差分分离出来，但是这样会多了一步人工还原的操作。基于上述问题，我将差分过程进行了封装，使序列能按照指定的差分列表依次进行差分，并相应的构造了一个还原的方法，实现差分序列的自动还原。
复制代码
# 差分操作
def diff_ts(ts, d):
global shift_ts_list # 动态预测第二日的值时所需要的差分序列 global last_data_shift_list shift_ts_list = [] last_data_shift_list = [] tmp_ts = ts for i in d: last_data_shift_list.append(tmp_ts[-i]) print last_data_shift_list shift_ts = tmp_ts.shift(i) shift_ts_list.append(shift_ts) tmp_ts = tmp_ts - shift_ts tmp_ts.dropna(inplace=True) return tmp_ts # 还原操作 def predict_diff_recover(predict_value, d): if isinstance(predict_value, float): tmp_data = predict_value for i in range(len(d)): tmp_data = tmp_data + last_data_shift_list[-i-1] elif isinstance(predict_value, np.ndarray): tmp_data = predict_value[0] for i in range(len(d)): tmp_data = tmp_data + last_data_shift_list[-i-1] else: tmp_data = predict_value for i in range(len(d)): try: tmp_data = tmp_data.add(shift_ts_list[-i-1]) except: raise ValueError('What you input is not pd.Series type!') tmp_data.dropna(inplace=True) return tmp_data
复制代码
现在我们直接使用差分的方法进行数据处理，并以同样的过程进行数据预测与还原。
diffed_ts = diff_ts(ts_log, d=[12, 1])
model = arima_model(diffed_ts)
model.certain_model(1, 1)
predict_ts = model.properModel.predict()
diff_recover_ts = predict_diff_recover(predict_ts, d=[12, 1]) log_recover = np.exp(diff_recover_ts)
是不是发现这里的预测结果和上一篇的使用12阶移动平均的预测结果一模一样。这是因为12阶移动平均加上一阶差分与直接12阶差分是等价的关系，后者是前者数值的12倍，这个应该不难推导。
对于个数不多的时序数据，我们可以通过观察自相关图和偏相关图来进行模型识别，倘若我们要分析的时序数据量较多，例如要预测每只股票的走势，我们就不可能逐个去调参了。这时我们可以依据BIC准则识别模型的p, q值，通常认为BIC值越小的模型相对更优。这里我简单介绍一下BIC准则，它综合考虑了残差大小和自变量的个数，残差越小BIC值越小，自变量个数越多BIC值越大。个人觉得BIC准则就是对模型过拟合设定了一个标准（过拟合这东西应该以辩证的眼光看待）。
复制代码
def proper_model(data_ts, maxLag):
init_bic = sys.maxint
init_p = 0
init_q = 0 init_properModel = None for p in np.arange(maxLag): for q in np.arange(maxLag): model = ARMA(data_ts, order=(p, q)) try: results_ARMA = model.fit(disp=-1, method='css') except: continue bic = results_ARMA.bic if bic < init_bic: init_p = p init_q = q init_properModel = results_ARMA init_bic = bic return init_bic, init_p, init_q, init_properModel
复制代码
相对最优参数识别结果：BIC: -1090.44209358 p: 0 q: 1 ， RMSE:11.8817198331。我们发现模型自动识别的参数要比我手动选取的参数更优。
7.滚动预测
所谓滚动预测是指通过添加最新的数据预测第二天的值。对于一个稳定的预测模型，不需要每天都去拟合，我们可以给他设定一个阀值，例如每周拟合一次，该期间只需通过添加最新的数据实现滚动预测即可。基于此我编写了一个名为arima_model的类，主要包含模型自动识别方法，滚动预测的功能，详细代码可以查看附录。数据的动态添加：
复制代码
from dateutil.relativedelta import relativedelta
def _add_new_data(ts, dat, type='day'):
if type == 'day': new_index = ts.index[-1] + relativedelta(days=1) elif type == 'month': new_index = ts.index[-1] + relativedelta(months=1) ts[new_index] = dat def add_today_data(model, ts, data, d, type='day'): _add_new_data(ts, data, type) # 为原始序列添加数据 # 为滞后序列添加新值 d_ts = diff_ts(ts, d) model.add_today_data(d_ts[-1], type) def forecast_next_day_data(model, type='day'): if model == None: raise ValueError('No model fit before') fc = model.forecast_next_day_value(type) return predict_diff_recover(fc, [12, 1])
复制代码
现在我们就可以使用滚动预测的方法向外预测了，取1957年之前的数据作为训练数据，其后的数据作为测试，并设定模型每第七天就会重新拟合一次。这里的diffed_ts对象会随着add_today_data方法自动添加数据，这是由于它与add_today_data方法中的d_ts指向的同一对象，该对象会动态的添加数据。
复制代码
ts_train = ts_log[:'1956-12']
ts_test = ts_log['1957-1':] diffed_ts = diff_ts(ts_train, [12, 1]) forecast_list = []
for i, dta in enumerate(ts_test): if i%7 == 0: model = arima_model(diffed_ts) model.certain_model(1, 1) forecast_data = forecast_next_day_data(model, type='month') forecast_list.append(forecast_data) add_today_data(model, ts_train, dta, [12, 1], type='month') predict_ts = pd.Series(data=forecast_list, index=ts['1957-1':].index)
log_recover = np.exp(predict_ts)
original_ts = ts['1957-1':]
复制代码
动态预测的均方根误差为：14.6479，与前面样本内拟合的均方根误差相差不大，说明模型并没有过拟合，并且整体预测效果都较好。
8. 模型序列化
在进行动态预测时，我们不希望将整个模型一直在内存中运行，而是希望有新的数据到来时才启动该模型。这时我们就应该把整个模型从内存导出到硬盘中，而序列化正好能满足该要求。序列化最常用的就是使用json模块了，但是它是时间对象支持得不是很好，有人对json模块进行了拓展以使得支持时间对象，这里我们不采用该方法，我们使用pickle模块，它和json的接口基本相同，有兴趣的可以去查看一下。我在实际应用中是采用的面向对象的编程，它的序列化主要是将类的属性序列化即可，而在面向过程的编程中，模型序列化需要将需要序列化的对象公有化，这样会使得对前面函数的参数改动较大，我不在详细阐述该过程。
总结
与其它统计语言相比，python在统计分析这块还显得不那么“专业”。随着numpy、pandas、scipy、sklearn、gensim、statsmodels等包的推动，我相信也祝愿python在数据分析这块越来越好。与SAS和R相比，python的时间序列模块还不是很成熟，我这里仅起到抛砖引玉的作用，希望各位能人志士能贡献自己的力量，使其更加完善。实际应用中我全是面向过程来编写的，为了阐述方便，我用面向过程重新罗列了一遍，实在感觉很不方便。原本打算分三篇来写的，还有一部分实际应用的部分，不打算再写了，还请大家原谅。实际应用主要是具体问题具体分析，这当中第一步就是要查询问题，这步花的时间往往会比较多，然后再是解决问题。以我前面项目遇到的问题为例，当时遇到了以下几个典型的问题：1.周期长度不恒定的周期成分，例如每月的1号具有周期性，但每月1号与1号之间的时间间隔是不相等的；2.含有缺失值以及含有记录为0的情况无法进行对数变换；3.节假日的影响等等。
o
粉丝 0
博文 500
码字总数 0
作品 0
私信 提问
加载中
请先登录后再评论。
暂无文章
pycurl libcurl link-time ssl backend (nss)
pip uninstall pycurlecho 'pycurl==7.19.5.1 --global-option="--with-nss"' > requires.pypip install -r requires.py...
小红手
38分钟前
17
0
计算机网络性能衡量
1、速率 单位时间(s)内传输信息(bit)量 单位：KB/s, MB/s, Gb/s K = 10^3 ,M = 10^6, G=10^9 一般表示的是理想的传输速率 2、带宽 计算机网络中的带宽和通信等领域的带宽概念不一样，计算机网...
osc_np3y0rbq
38分钟前
3
0
互联网掀起农家乐，巨头上演AI掘金战
配图来自Canva **前有网易、阿里AI养猪，后有腾讯AI养鹅，互联网大佬们纷纷玩起了“农家乐”，互联网的生意在尖端技术的引领之下频频跨界，巨头之间的较量也从线上延伸至线下。**自古“民以食...
osc_5cok9i01
40分钟前
9
0
原来！我在4年前就开始体验雾游戏了！
前有云游戏后有雾游戏，游戏的方式看来起来越来越多种多样。那么“震撼业界”的雾游戏到底是什么来头？它依靠什么改变游戏界？它的原理又是什么？ 本月月初，著名的日本游戏杂志《Fami通》表...
osc_j34n26zn
41分钟前
11
0
活动预告｜田溯宁与你相约GSMA Thrive·万物生晖，分享5G风口下的创新与投资洞察
在万物互联的时代背景下，5G+AI+IoT的技术变革与融合，正在引发一场深刻的全产业创新与变革。5G技术创新、行业应用及投资机遇已成为科技行业所瞩目的焦点。 6月30日，宽带资本董事长田溯宁将...
osc_0qnrwmy3
42分钟前
7
0
没有更多内容
加载失败，请刷新页面
加载更多
返回顶部
顶部
|
__label__pos
| 0.540559 |
hcoop-backups: Further tweak permissions.
[clinton/scripts.git] / rsync-shell
1 #!/usr/bin/perl
2
3 use strict;
4 use warnings FATAL => 'all';
5
6 use constant LOGFILE => '/var/log/rsync-shell.log';
7
8 my %commands = (
9 "backup" => \&backup,
10 "rsync" => \&rsync,
11 );
12
13 sub backup {
14 exec '/usr/bin/sudo',
15 '/afs/hcoop.net/common/etc/scripts/hcoop-backup-wrapper';
16 }
17
18 sub rsync {
19 my ($cmdline) = @_;
20
21 if ( $cmdline !~ m!^--server --sender -vre\.iL --bwlimit=325 \. /vicepa/hcoop-backups/files/[0-9]{4}\.[0-9]{2}\.[0-9]{2}$!s ) {
22 die "Incorrect arguments to rsync.\n";
23 }
24
25 exec '/usr/bin/rsync', split(' ', $cmdline)
26 or die "Could not run rsync command.\n";
27 }
28
29 sub main {
30 -f LOGFILE && open (LOG, '>>', LOGFILE)
31 or die "Can't open log file.\n";
32
33 print LOG "Session started on ", `date`;
34 print LOG "Commands: ", map { "<$_> " } @ARGV;
35 print LOG "\n";
36
37 $ARGV[1] =~ /^([^ ]+) *(.*?)$/s;
38 my $cmd = $commands{$1}
39 or die "Unsupported command.\n";
40
41 $cmd->($2);
42 }
43
44 main()
45
|
__label__pos
| 0.991058 |
Help with....Moving a prop?
Discussion in 'Mapping Questions & Discussion' started by Vilzzex (ActionDevVito), Aug 27, 2016.
1. Vilzzex (ActionDevVito)
Vilzzex (ActionDevVito) L2: Junior Member
Messages:
78
Positive Ratings:
62
So uh,i just GOT INTO hammer.Like,i am trying my best to get it,because i want to make both Garry's Mod and Team Fortress maps.I watched all tutorials made by UEAKCrash.I understand the input/output system and such,but i have a question.
I want to make a prop_dynamic move forward,but that it has a Kill Trigger on the front of it,much like UEAKCrash did in his "trainsawlaser" map where the train that lands into the control point has a kill trigger at the front so it can kill the player.I want to make a prop horizontally land into a certain place on the map,with it killing everyone that stands on it's way.
So how do i make it run into the certain place on the map,
and how do i make the kill trigger stick on the front of the prop?
Thanks in advance guys.
2. zahndah
aa zahndah professional letter
Messages:
718
Positive Ratings:
627
To make the prop_dynamic move, you would want to parent the prop to a func_tracktrain. Func_tracktrains are a brush entity that have the capability of following a path that you have laid down for them to follow.
You then would put down a path_track at where you want it to start, and name it path_1. Then shift+drag it to the next corner it takes / the end location if it wont need to be taking any corners. Next in the func_tracktrain, set the 'first stop target' to path_1, and the speed to whatever you want. Speed is measured in hammer units per second, so judge how fast you want it there. First stop target determines where the tracktrain and the prop you have parented to it start. Also, in the outputs tab of the last path_track in the path, create and output of 'OnPass - (name of your func_tracktrain) - TeleportToPathTrack - Path_1 (in parameter override)' Then create another output of 'OnPass - (name of func_tracktrain) - Stop'
Next, you need something to trigger the tracktrain to follow the path. The best option is likely a logic_timer for what you want to do. A logic_timer can trigger some outputs every x amount of seconds. So you can have it set to, say 30 seconds, then every 30 seconds the train would go along the path then teleport back to the first path. Set the 'refire interval' to however long you want the gap to be, then in the outputs tab create an output 'OnTimer - (name of func_tracktrain) - StartForward'
Now all you need to do is to create a trigger_hurt that is a big bigger than the func_tracktrain and the prop_dynamic, and set its damage to something high like 10000, and make sure that the func_tracktrain, prop_dynamic and trigger_hurt are all correctly in position ontop of eachother. If all is well and good, you should have a functioning train rain.
• Thanks Thanks x 2
3. Vilzzex (ActionDevVito)
Vilzzex (ActionDevVito) L2: Junior Member
Messages:
78
Positive Ratings:
62
Thanks a lot!
4. iiboharz
aa iiboharz eternally tired
Messages:
739
Positive Ratings:
1,077
Just a heads up! If your object only moves between two points, it might be worth looking into using func_movelinear or func_door for its movement instead!
5. Crowbar
aa Crowbar perfektoberfest
Messages:
1,443
Positive Ratings:
1,199
Func_tracktrains tend to be a nodraw brush approximately the size of the prop. Is some cases they also act as a rough collision box.
|
__label__pos
| 0.94901 |
Skip to content
Related Articles
Open in App
Not now
Related Articles
Hello World in Tkinter
Improve Article
Save Article
• Difficulty Level : Basic
• Last Updated : 17 May, 2020
Improve Article
Save Article
Tkinter is the Python GUI framework that is build into the Python standard library. Out of all the GUI methods, tkinter is the most commonly used method as it provides the fastest and the easiest way to create the GUI application.
Creating the Hello World program in Tkinter
Lets start with the ‘hello world’ tutorial. Here is the explanation for the first program in tkinter:
• from tkinter import *
In Python3 firstly we import all the classes, functions and variables from the tkinter package.
• root=Tk()
Now we create a root widget, by calling the Tk(). This automatically creates a graphical window with the title bar, minimize, maximize and close buttons. This handle allows us to put the contents in the window and reconfigure it as necessary.
• a = Label(root, text="Hello, world!")
Now we create a Label widget as a child to the root window. Here root is the parent of our label widget. We set the default text to “Hello, World!”
Note: This gets displayed in the window. A Label widget can display either text or an icon or other image.
• a.pack()
Next, we call the pack() method on this widget. This tells it to size itself to fit the given text, and make itself visible.It just tells the geometry manager to put widgets in the same row or column. It’s usually the easiest to use if you just want one or a few widgets to appear.
• root.mainloop()
The application window does not appear before you enter the main loop. This method says to take all the widgets and objects we created, render them on our screen, and respond to any interactions. The program stays in the loop until we close the window.
Below is the implementation.
# Python tkinter hello world program
from tkinter import *
root = Tk()
a = Label(root, text ="Hello World")
a.pack()
root.mainloop()
Output:
python-tkinter-hello-world
My Personal Notes arrow_drop_up
Related Articles
Start Your Coding Journey Now!
|
__label__pos
| 0.670487 |
mygrad.amin#
mygrad.amin(x: ArrayLike, axis: Union[None, int, Tuple[int, ...]] = None, keepdims: bool = False, *, constant: Optional[bool] = None) Tensor#
Return the minimum of a tensor or minimum along its axes.
Parameters
axisOptional[int, Tuple[int, …]]
Axis or axes along which to operate. By default, flattened input is used.
keepdimsbool, optional
If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the original arr.
constantOptional[bool]
If True, this tensor is treated as a constant, and thus does not facilitate back propagation (i.e. constant.grad will always return None).
Defaults to False for float-type data. Defaults to True for integer-type data.
Integer-type tensors must be constant.
Returns
minmygrad.Tensor
Minimum of a. If axis is None, the result is a 0-D tensor.
Examples
>>> import mygrad as mg
>>> import numpy as np
>>> a = mg.arange(4).reshape((2,2))
>>> a
Tensor([[0, 1],
[2, 3]])
>>> mg.amin(a) # Minimum of the flattened array
Tensor(0)
>>> mg.amin(a, axis=0) # Minima along the first axis
Tensor([0, 1])
>>> mg.amin(a, axis=1) # Minima along the second axis
Tensor([0, 2])
>>> b = mg.arange(5, dtype=float)
>>> b[2] = np.NaN
>>> mg.amin(b)
Tensor(nan)
|
__label__pos
| 0.713208 |
Describe different steps of object-oriented design process, Database Management System
Describe different steps of the object-oriented design process.
The broad steps of the object-oriented design process are:
a. Define context and modes of use of the system
b. Design system architecture
c. Recognize the principal system objects
d. Identification of the concurrency in problem
e. Handling the boundary condition
f. Developing design models
g. Specify the object interfaces.
Posted Date: 8/30/2013 6:01:30 AM | Location : United States
Related Discussions:- Describe different steps of object-oriented design process, Assignment Help, Ask Question on Describe different steps of object-oriented design process, Get Answer, Expert's Help, Describe different steps of object-oriented design process Discussions
Write discussion on Describe different steps of object-oriented design process
Your posts are moderated
Related Questions
You will be analysing a set of financial data of your choice. It is not necessary to collect any primary data of your own, and publicly available secondary data will be sufficient
Column Constraints: NOT NULL, UNIQUE, CHECK, PRIMARY KEY, DEFAULT, REFERENCES, On delete Cascade : Using this key whenever a parent row is removed in a referenced
A full discription of the new, a discription of its components, and the bebefit it will provide a company. A disscussion of the information usedin the system. A disscussion of any
After the completion of the logical model the physical model is to be made. The oracle 10 G will be used as a database server and all the tables with their column are created and i
What is an ER diagram? Why is it used in Database management system (DBMS)?
Automated system for appointment with sms and call confirmation with service providers and user has to be initiate automatically. After selecting time and date to use service, s
Explain five duties of Database Administrator? 1. DBA administers the three stages of the database and, in consultation along with the whole user community, sets up an defin
Write a query to define tables students and teachers as sub tables of people? Create table students of student under people Create table teachers of teacher under people
need to get answer for questions 3&4 on this page in part 1
|
__label__pos
| 0.909755 |
Take the 2-minute tour ×
WordPress Development Stack Exchange is a question and answer site for WordPress developers and administrators. It's 100% free, no registration required.
I've never used the Transients API before and was wondering if anyone has guidance on when to use it. The Codex article implies that as a theme developer I might want to set each new WP_Query() as a transient; I assume the same might be said for direct $wpdb queries and query_posts(). Is that overkill? And/Or are there other places I should default to using it?
I do often use caching plugins on my site (W3 Total Cache usually) and it sounds like using Transients might bump up the plugin's effectiveness but I don't want to go crazy wrapping everything in transients if that's not a best practice.
share|improve this question
Thanks everyone, I wish I could mark more than one answer as the 'solution'. Great info, much appreciated! – Michelle Mar 12 '12 at 14:13
5 Answers 5
up vote 14 down vote accepted
Transients are great when you're doing complex queries in your themes and plugins. I tend to use transients for things like menus and showing other things like Tweets from Twitter in a sidebar for example. I wouldn't use them for absolutely everything more-so just temporary pieces of data that can be cached.
Keep in mind that if you use something like Memcached with transients, then you will notice a massive performance gain. The rule with transients is to not use them for data that should not expire as they're really only for temporary data and keep in mind transients are not always stored in the database.
Some uses for transients:
• Complex and custom database queries
• Wordpress navigation menus
• Sidebar widgets that display info like; tweets, a list of recent site visitors or a Flickr photo stream
• Caching tag clouds
This article is a very informative one with quick benchmarks showing just how transients can speed up your site and even has a couple of examples. This other article also has a few great examples of using transients which might help you understand what to use them for as well.
share|improve this answer
2
Another use: caching external HTTP requests. Like hitting the twitter API. Don't do it on every page load, cache the results with a transient. – chrisguitarguy Mar 10 '12 at 18:18
2
+1 for the great tutorial links. – Tom Auger Mar 14 '12 at 12:46
There are several caching mechanisms in WordPress and their mechanics differ, depending on choice of object cache (native or not):
+-----------+-------------------------+---------------------+
| | Native | Object cache plugin |
+-----------+-------------------------+---------------------+
| Option | Persistent (database) | N/A |
| Transient | Persistent (database) | Persistent (varies) |
| Cache | Non-persistent (memory) | Persistent (varies) |
+-----------+-------------------------+---------------------+
In a nutshell what this means is that transient is always persistent (will survive between page loads unlike Cache natively), but it will make use of customized storage if provided (unlike Options).
This makes transients most versatile choice for caching.
However with flexibility comes undercurrent complexity and there are quite a few of nuances with them (such as limit on name length, different behavior with and without expiration, lack of garbage collection) that make them more complex than they seem.
Overall:
• use Options for saving things that must be persistent
• use Transients for caching anything else
• use Cache when you have very good grasp of all three and know that Cache fits use case better than others (which won't be often)
share|improve this answer
I think the code from Sterling could be improved by not call the get_transient function twice. Instead store the first result in a temporary variable. Because the idea behind the Transient API ist speed ;-)
private function _get_data( $query) {
$result = get_transient( $query );
if ( $result ) {
return $result;
} else {
return $this->_get_query( $query );
}
}
share|improve this answer
Short Answer: You should use it when/where you can.
Long Answer:
The Transients API is for caching. So you want to use it as much as you can. You can write a function that does this for you.
It's not overkill and if done properly ends up being pretty elegant:
// If the transient isn't false, then you can just get the cached version.
// If not, we'll call another function that runs a database query.
private function _get_data( $query) {
return
( get_transient( $query ) ) ?
get_transient( $query ) :
$this->_get_query( $query );
}
// After you make the query, set the transient so you can leverage on caching plugins.
private function _get_query( $query ) {
// Database logic. Results go to $results.
...
set_transient( $query, $results , 0 ); // 0 Means cache for as long as you can.
}
share|improve this answer
Transients API is really useful when you're fetching data from external sources like Facebook, Twitter.
To get more clear idea of what Transients API is and what is difference with Cache WordPress function, I recommend to watch Otto's and Nanic's talk from WordCamp San Francisco 2011
share|improve this answer
Your Answer
discard
By posting your answer, you agree to the privacy policy and terms of service.
Not the answer you're looking for? Browse other questions tagged or ask your own question.
|
__label__pos
| 0.639439 |
Category:
What is a Virtual Circuit?
Article Details
• Written By: Jessica Reed
• Edited By: Heather Bailey
• Last Modified Date: 20 September 2016
• Copyright Protected:
2003-2016
Conjecture Corporation
• Print this Article
Free Widgets for your Site/Blog
Although Stonehenge is the most famous, there are over 1,000 ancient stone circles standing in the British Isles. more...
September 26 , 1960 : The first televised US Presidential debate took place. more...
A virtual circuit, abbreviated VC and also known as a virtual connection or virtual channel, provides a connection between points in a network in both telecommunications and computer networks. The virtual circuit allows packets of information to pass between the two connections. Typically, these circuits are used in networks with fast transfer speeds, such as asynchronous transfer mode (ATM) connections. While the virtual circuit may appear to be a physical path that connects two points in the network, it actually switches back and forth between various circuits to create different paths as needed.
When used in telecommunications, circuits carry signals along the path between the starting point and ending point. The network is simply a collection of various circuits, or paths, to connect all the different users to the network or other connection point. When referring to circuits in the context of electronics, such as use in a computer, the circuit still runs between two points but is more likely to connect more than two points to carry signals in the form of electrical currents. These circuits are permanent, whereas a virtual circuit can create different paths through a collection of various circuits and has no fixed, permanent path it must follow.
Ad
Two types of virtual circuits exist: permanent virtual circuits (PVC) and switched virtual circuits (SVC). As the names suggest, a PVC stays connected at all times while a SVC only connects when in use and disconnects afterward. Typically PVCs are used on frame relay networks, which connect local networks with wider area networks. A SVC can be used on frame relay networks but must maintain a constant connection during the transfer.
Virtual circuits can also be referred to as logical circuits, and it is important to keep in mind that while the circuit can change its path and connect to different networks or points, it still only connects two points at one time. It determines what two connections it needs to make and sets up the best path for a smooth and fast transfer. For this reason it appears to be a normal circuit connection that stays in place. The difference lies in how the circuit can choose two different points to create a new connection when necessary. This allows for fast transfers among various networks using fewer resources.
Ad
You might also Like
Recommended
Discuss this Article
Post your comments
Post Anonymously
Login
username
password
forgot password?
Register
username
password
confirm
email
|
__label__pos
| 0.966134 |
Question: How Far Back Do Emails Go On IPhone?
Why have emails disappeared from my inbox?
Typically, emails go missing when an email is accidentally deleted.
It can also happen if the email system incorrectly flags an incoming message as spam, which would mean the message never reached your inbox.
Less frequently, an email can go missing if it’s archived and you don’t realize it..
How do I retrieve old emails from Gmail?
Recover messages from the TrashOn your computer, go to Gmail.On the left side of the page, scroll down, then click More. Trash.Check the box next to any messages you want to move.Click Move to .Choose where you want to move the message to.
How do I get my old emails on my new phone?
If you use Google, you can access your old emails on the new phone by logging in to your gmail account either using the Gmail App or using another Google-friendly inbox app.
Will deleting emails free up storage on iPhone?
Delete Emails Sooner The items that are waiting to be deleted take up space on your phone, so if you delete them sooner, you’ll free up space faster. To change that setting: Open the Settings app and select Passwords & Accounts. Then, tap the email account whose setting you want to change.
How do I manage email on my iPhone?
How to manage email accountsLaunch the Settings app from your Home screen.Scroll down in the Settings menu until you see Mail.Tap on Accounts.Tap on the email account you want to manage. Source: iMore.Oct 20, 2020
How do I find older emails on my iPhone 11?
How to Sync More or Less Emails on iPhoneOpen Settings.Tap Passwords & Accounts. … Tap the email account you want to change the settings for.Choose Mail Days to Sync, then choose how many recent days of email you want to download to Mail automatically. … Your mail is synchronized to your preferences.More items…•Jan 5, 2021
Why have my emails disappeared from my iPhone?
If iPhone emails keep disappearing after force reboot, there may be an error with the email account. … Then go back to “Mail, Contacts, Calendars” and choose “Add Account” option. Type your mail address and password to re-add mail account on your iPhone. Then you can check your inbox to see if you can find your emails.
Does deleting emails free up space on your phone?
Deleting emails from Gmail can free up storage space in the cloud. This is done in two steps: first you have to move the emails into the Trash, then you have to delete them from the Trash.
Do deleted emails take up space?
Anything that the Cloud sees as being “deleted”, either on your computer or directly in the Cloud, is moved to this Drive Trash Bin but not permanently deleted. Until it’s permanently deleted it still counts to your total storage usage.
Why can’t I see older emails on my iPhone?
If you are unable to see older messages in your Mail app for any email account that you are looking in, try checking the Mail preferences in your Settings app. Tap Settings > Mail, Contacts & Calendars > Organize by Thread. your emails showing as it did previously.
How do I stop emails disappearing from my iPhone?
Fix 2. Delete and re-add email account after rebooting iPhoneRemove or delete the email account from iPhone > Reboot iPhone.Re-add email account and log in with password > Check and get disappeared emails back.Go to Settings > Go to Mail, Contacts, Calendars.Tab on your account and look down for Mail Days to Sync;More items…•Feb 4, 2021
How do I transfer my old emails to my new iPhone?
You add the mail accounts to the new phone, and that will restore your mail, unless you are using a POP mail account. In that case, your mail is deleted from the server, and is just present on the old device.
How do I see old emails on iPhone?
Tap the “Settings” icon, select “Mail, Contacts, Calendars,” and then select the email account. Tap the “Archive Messages” ON/OFF switch to “ON.”
Do old emails take up space on iPhone?
You’ll still have your address book and calendar, but email won’t take up any space. Unfortunately, Apple’s Mail app–both on mobile devices and Macs–doesn’t offer a lot of control over how much storage it uses. The best solution may be using another email app if this is a problem for you.
How do I recover an old email account from years ago?
Recovering Old Email Accounts Most email providers have a way for you to recover access to your account. Many email providers support a way to send a recovery link to a predesignated email address or phone number. When you click this link, you can select a new password and log back into your account.
How do I view Outlook emails older than 12 months?
To temporarily view messages older than 12 months you can scroll to the end of an email list in a folder and click ‘Click here to view more on Microsoft Exchange’ or ‘More’. To change your view permanently you can change the setting in account settings.
How do I save all my emails on my iPhone?
On the iPhone there is an option called “Mail Days To Sync.” This option allows you to save space on your iPhone by telling the server to stop pushing emails to your phone. There are several options to select: “No Limit” — This option will send all the emails in your inbox and folders to your iPhone.
How do I find old emails?
How to Access Old EmailsLog into your account and take note of the left-hand navigation. … Check your “Inbox.” This might seem elementary, but not everyone adheres to the zero-inbox philosophy. … Click on the “All Mail” link and scroll through the pages of emails until you find the ones you are looking for. … Click on the “Trash” link.More items…
How do emails just disappear?
Emails might skip your inbox if they were accidentally archived, deleted, or marked as spam. Tip: To filter your search results even more, you can also use search operators. You may have created a filter that automatically archives or deletes certain emails.
|
__label__pos
| 0.990512 |
New Feature: Playlist Center! Pick a topic and let our playlists guide the way.
Start learning with our library of video tutorials taught by experts. Get started
Collaborative Workflows with InDesign and InCopy
Illustration by
Opening linked InCopy stories directly
From:
Collaborative Workflows with InDesign and InCopy
with Anne-Marie Concepción
Video: Opening linked InCopy stories directly
Another way that you can use InCopy in standalone mode is to open up a workflow story directly. In other words to break all the rules that I've been telling you and instead of using InCopy's open a file to open up the INDD file or to open up an assignment file, you go directly to the stories folder or to the content folder if you using an assignments workflow and open up these ICML stories directly. So let's see what happens. I'm going to open up the actual INDD files, so we are doing it correctly and we will look at the formatting for one of the stories.
Expand all | Collapse all
1. 3m 57s
1. Welcome
1m 25s
2. Using the exercise files
2m 32s
2. 25m 58s
1. Overview of this course
3m 2s
2. Understanding the parallel workflow
6m 54s
3. Rewards and challenges in the new workflow
9m 3s
4. Requirements and recommendations
6m 59s
3. 32m 52s
1. Setting up projects and users
3m 32s
2. Understanding stories and frames
7m 1s
3. Making stories editable for InCopy from InDesign
7m 25s
4. Editing workflow stories in InCopy
7m 32s
5. Checking stories in and out
4m 48s
6. Completing a project in InDesign
2m 34s
4. 32m 34s
1. Three main views of a file
8m 37s
2. Becoming familiar with default panels
6m 4s
3. Customizing the interface
9m 4s
4. Navigating stories and views
8m 49s
5. 43m 18s
1. Working with the Assignments panel
5m 15s
2. Editing in Layout view
8m 44s
3. Editing in Story or Galley view
10m 49s
4. Copyfitting text
5m 49s
5. Inserting special characters
6m 39s
6. Importing text
3m 34s
7. Working with read-only layouts
2m 28s
6. 32m 6s
1. Applying styles for copyfit
7m 37s
2. Applying local character formatting
6m 53s
3. Applying local paragraph formatting
7m 10s
4. Splitting and spanning columns
5m 7s
5. Using the Eyedropper tool to copy/paste formatting
5m 19s
7. 40m 27s
1. Checking spelling
4m 51s
2. Using the language dictionaries
3m 23s
3. Using the thesaurus
1m 46s
4. Using Find/Change
10m 34s
5. Working with the Autocorrect feature
2m 59s
6. Building text macros
4m 55s
7. Using inline notes
6m 22s
8. Working with built-in scripts
5m 37s
8. 25m 36s
1. Adding footnotes
2m 22s
2. Using conditional text
6m 16s
3. Creating hyperlinks
3m 33s
4. Inserting cross-references
7m 29s
5. Working with tables
5m 56s
9. 14m 25s
1. Setting up and using Track Changes
6m 4s
2. Customizing the markup
4m 7s
3. Accepting and rejecting changes
4m 14s
10. 27m 30s
1. Using the Position tool
5m 14s
2. Using the Object menu
5m 58s
3. Importing and replacing images
6m 36s
4. Inserting images into the story
5m 22s
5. Using Mini Bridge and Bridge
4m 20s
11. 25m 45s
1. Creating new InCopy documents
6m 54s
2. Creating InCopy templates
6m 10s
3. Opening linked InCopy stories directly
3m 20s
4. Opening Word files in InCopy
2m 59s
5. Placing Buzzword files in InCopy
6m 22s
12. 23m 37s
1. Exporting stories to Word, RTF, and Buzzword
5m 2s
2. Exporting layouts to PDF
4m 36s
3. Exporting galleys and stories to PDF
7m 11s
4. Printing from InCopy
6m 48s
13. 48m 17s
1. Exporting stories from the layout
10m 2s
2. Working with the Assignments panel in InDesign
7m 8s
3. Editing and updating files
7m 37s
4. Using inline notes
7m 39s
5. Workflow features in the Links panel
6m 0s
6. Placing new InCopy files
4m 15s
7. Closing out of a project
5m 36s
14. 23m 29s
1. Layout workflow overview
8m 11s
2. Updating stories and designs
11m 38s
3. Tips for successful layout workflows
3m 40s
15. 27m 16s
1. Creating assignments in InDesign
12m 19s
2. Working with assignments in InCopy
5m 22s
3. Keeping layout files local
2m 42s
4. Solving common assignment issues
6m 53s
16. 19m 0s
1. Creating assignment packages in InDesign
4m 42s
2. Working with assignment packages in InCopy
5m 20s
3. Keeping packages up to date
2m 33s
4. Using DropBox with an InCopy workflow
6m 25s
17. 4m 27s
1. Community help and resources
4m 11s
2. Goodbye
16s
Watch this entire course now—plus get access to every course in the library. Each course includes high-quality videos taught by expert instructors.
Become a member
please wait ...
Collaborative Workflows with InDesign and InCopy
7h 30m Intermediate Sep 23, 2010
Viewers: in countries Watching now:
In Collaborative Workflows with InDesign and InCopy Anne-Marie Concepción shows how Adobe InCopy and InDesign work together, helping editors and designers collaborate on publications, and save time and money, with no additional hardware, software, or expensive publication management systems. This course shows how to set up for the workflow, how to address cross-platform Mac and Windows issues when working in a mixed environment, how to work with remote writers and designers, and how to integrate with Microsoft Word. Exercise files are included with the course.
Topics include:
• Setting up projects and users on a local network
• Using e-mail-based assignments and Dropbox to manage remote users
• Copyfitting and formatting text
• Using advanced editing tools
• Working with paragraph, character, and table styles
• Tracking changes in InCopy and InDesign
• Creating cross-references and hyperlinks
• Creating InCopy templates
• Combining InCopy with Microsoft Word
• Inserting and formatting images
• Reviewing features specific to InDesign
Subject:
Design
Software:
InCopy InDesign
Author:
Anne-Marie Concepción
Opening linked InCopy stories directly
Another way that you can use InCopy in standalone mode is to open up a workflow story directly. In other words to break all the rules that I've been telling you and instead of using InCopy's open a file to open up the INDD file or to open up an assignment file, you go directly to the stories folder or to the content folder if you using an assignments workflow and open up these ICML stories directly. So let's see what happens. I'm going to open up the actual INDD files, so we are doing it correctly and we will look at the formatting for one of the stories.
Like let's say this one, the big story, the main story. It starts out bold and then it flows over here and then it goes over here and it's talking about different pictures and things like that. All right, if press Ctrl or Command+A you can see it's one threaded story going throughout, and the story is called main story. So now I'm going to close this, and instead of opening up the layout, I am going to go right into the stories folder and open up main story.
There is no clue that you're doing things a little weird, so I want you to be aware of this and maybe you can use it to your advantage. When you open up a workflow story directly, you see the actual content in that story. It's kind of like opening up a placed Photoshop image directly in Photoshop rather than trying to edit it through InDesign. So we are seeing the actual story and this is actually what gets edited when you check out this story from the layout or from the assignment. But let me zoom in here a bit. What happens is that the editor will often say, "Where's the rest of the document?" And I've gotten calls from new users saying, "I thought the whole point that we moved to InCopy so we could see the entire layout and I opened up the file and InCopy and it's not working." So I will always say, "Look in the tab that has the name of the file. What is the extension? Is it INDD or ICML for an assignment?" And they will say, "No, it's this one, ICML." So that's what happened, is that they have opened this up directly. Now if somebody has the layout or assignment open that has a link to this story, it will be checked out by this user.
So there is no danger of two more people working on the same story. Notice I see a little pencil icon because it automatically checks it out to me as soon as I open it, and I can use any of the three views to edit the story. So in some cases, this might be something you want to do. If you have an editor that just has a really bad network connection that day and they really need to get in and edit a story that you have placed, have them open the actual InCopy file. It opens a lot faster over the network than the entire layout or even an assignment and they can go ahead and edit it to their heart's content.
Now what they can't see is copyfit progress info, because InCopy is not aware of how much space this thing has. So you are not going to see any copyfit progress info and it's highly likely that the margins being used here does not equal the text area of the column in InDesign. So they are not really going to be able to proof line breaks. But they will be able to actually edit the story and apply style. So oftentimes, opening up an InCopy story directly is a savvy move, but in case you did it by mistake now you know what happened and what you should do, close this document and open up the actual INDD or assignment file.
There are currently no FAQs about Collaborative Workflows with InDesign and InCopy.
Share a link to this course
What are exercise files?
Exercise files are the same files the author uses in the course. Save time by downloading the author's files instead of setting up your own files, and learn by following along with the instructor.
Can I take this course without the exercise files?
Yes! If you decide you would like the exercise files later, you can upgrade to a premium account any time.
Become a member Download sample files See plans and pricing
Please wait... please wait ...
Upgrade to get access to exercise files.
Exercise files video
How to use exercise files.
Learn by watching, listening, and doing, Exercise files are the same files the author uses in the course, so you can download them and follow along Premium memberships include access to all exercise files in the library.
Upgrade now
Exercise files
Exercise files video
How to use exercise files.
For additional information on downloading and using exercise files, watch our instructional video or read the instructions in the FAQ.
This course includes free exercise files, so you can practice while you watch the course. To access all the exercise files in our library, become a Premium Member.
join now Upgrade now
Are you sure you want to mark all the videos in this course as unwatched?
This will not affect your course history, your reports, or your certificates of completion for this course.
Mark all as unwatched Cancel
Congratulations
You have completed Collaborative Workflows with InDesign and InCopy.
Return to your organization's learning portal to continue training, or close this page.
OK
Become a member to add this course to a playlist
Join today and get unlimited access to the entire library of video courses—and create as many playlists as you like.
Get started
Already a member?
Become a member to like this course.
Join today and get unlimited access to the entire library of video courses.
Get started
Already a member?
Exercise files
Learn by watching, listening, and doing! Exercise files are the same files the author uses in the course, so you can download them and follow along. Exercise files are available with all Premium memberships. Learn more
Get started
Already a Premium member?
Exercise files video
How to use exercise files.
Ask a question
Thanks for contacting us.
You’ll hear from our Customer Service team within 24 hours.
Please enter the text shown below:
The classic layout automatically defaults to the latest Flash Player.
To choose a different player, hold the cursor over your name at the top right of any lynda.com page and choose Site preferencesfrom the dropdown menu.
Continue to classic layout Stay on new layout
Exercise files
Access exercise files from a button right under the course name.
Mark videos as unwatched
Remove icons showing you already watched videos if you want to start over.
Control your viewing experience
Make the video wide, narrow, full-screen, or pop the player out of the page into its own window.
Interactive transcripts
Click on text in the transcript to jump to that spot in the video. As the video plays, the relevant spot in the transcript will be highlighted.
Are you sure you want to delete this note?
No
Notes cannot be added for locked videos.
Thanks for signing up.
We’ll send you a confirmation email shortly.
Sign up and receive emails about lynda.com and our online training library:
Here’s our privacy policy with more details about how we handle your information.
Keep up with news, tips, and latest courses with emails from lynda.com.
Sign up and receive emails about lynda.com and our online training library:
Here’s our privacy policy with more details about how we handle your information.
submit Lightbox submit clicked
Terms and conditions of use
We've updated our terms and conditions (now called terms of service).Go
Review and accept our updated terms of service.
|
__label__pos
| 0.737678 |
大学IT网 - 最懂大学生的IT学习网站! QQ资料交流群:367606806
当前位置:大学IT网 > Java技巧 > struts2文件上传下载
struts2文件上传下载(1)
关键词：struts文件上传文件下载 阅读(1067) 赞(20)
[摘要]本文是对struts2文件上传下载的讲解，对学习Java编程技术有所帮助,与大家分享。
之前没写过struts2上传下载，只是知道大概怎么搞，最近项目中用到，正巧赶上了，编码过程中遇到过一些问题，如上传文件类型，中文乱码，兼容性问题等，不过都一一解决了，下面附上自己的代码。以便将来查阅...
也希望这篇随便能够帮助很多朋友少走些弯路把！
1.文件上传
首先是jsp页面的代码
在jsp页面中定义一个上传标签
<tr>
<td align="right" bgcolor="#F5F8F9"><b>附件：</b></td>
<td bgcolor="#FFFFFF">
<input type="file" name="upload" />
</td>
<td bgcolor="#FFFFFF"> </td>
</tr>
然后是BaseAction中定义的相关属性其它的就省略了（也可定义在自己的Action，换下访问修饰符即可）
/**
*Action基类
**/
public class BaseAction extends ActionSupport {
protected List<File> upload;
protected List<String> uploadContentType; //文件类型
protected List<String> uploadFileName; //文件名
protected String savePath; //保存路径
}
然后是Action中的一个上传方法，代码如下：
/**
* 8.上传附件
* @param upload
*/
public void uploadAccess(List<File> upload){
try {
if (null != upload) {
for (int i = 0; i < upload.size(); i++) {
String path = getSavePath() + "\\"+ getUploadFileName().get(i);
System.out.println(path);
item.setAccessory(getUploadFileName().get(i));
FileOutputStream fos = new FileOutputStream(path);
FileInputStream fis = new FileInputStream(getUpload().get(i));
byte[] buffer = new byte[1024];
int len = 0;
while ((len = fis.read(buffer)) > 0) {
fos.write(buffer, 0, len);
}
fis.close();
fos.close();
}
}
} catch (Exception e) {
logger.error("上传附件错误。", e);
}
}
接着是我的struts2.xml文件
<action name="itemRDAction_*" method="{1}">
<param name="savePath">e:\upload</param>
<interceptor-ref name="defaultStack">
<param name="fileUpload.allowedTypes">
application/octet-stream,image/pjpeg,image/bmp,image/jpg,image/png,image/gif,image/jpeg,application/msword,application/vnd.openxmlformats-officedocument.wordprocessingml.document,application/vnd.ms-excel
</param>
<param name="fileUpload.maximumSize">8000000</param>
</interceptor-ref>
<result name="show_item_rd_upd"> /WEB-INF/jsp/page_item/updItem_rd.jsp</result>
<result name="show_item_rd_list"> /WEB-INF/jsp/page_item/listItem_rd.jsp</result>
<result name="show_item_rd_push"> /WEB-INF/jsp/page_item/pushItem_rd.jsp</result>
</action>
savePath为保存路径，fileUpload.allowedTypes用来限制上传文件类型 fileUpload.maximumSize 文件大小限制
2.文件下载
首先是页面中的下载链接
<tr>
<td width="20%" align="right" bgcolor="#F5F8F9"><b>附件：</b></td>
<td width="40%" bgcolor="#FFFFFF">
<div style="width:355px; float: left;">${item.accessory}</div>
<c:if test="${!empty item.accessory}">
<div style="float: left;"><a style="color: black; text-decoration: none;" href="download.action?filename=${item.accessory}">下载</a></div>
</c:if>
</td>
<td width="40%" bgcolor="#FFFFFF"> </td>
</tr>
接着是DownloadAction的代码：
/**
* DownloadAction
*
* @author zhaoxz
*
*/
@Controller("downloadAction")
@Scope("prototype")
public class DownloadAction extends BaseAction {
/**
*
*/
private static final long serialVersionUID = -4278687717124480968L;
private static Logger logger = Logger.getLogger(DownloadAction.class);
private String filename;
private InputStream inputStream;
private String savePath;
private String mimeType;
public InputStream getInputStream() {
try {
String path = getSavePath() + "//"+ new String(filename.getBytes("ISO8859-1"), "utf-8");
System.out.println(path);
mimeType = ServletActionContext.getServletContext().getMimeType(path)+ ";charset=UTF-8";
inputStream = new FileInputStream(path);
String agent = request.getHeader("USER-AGENT");
System.out.println(agent);
if (null != agent) {
if (-1 != agent.indexOf("Firefox")) {// Firefox
mimeType = mimeType.replace("UTF-8", "ISO8859-1");
} else {// IE7+ Chrome
System.out.println("IE,Chrome");
filename = new String(filename.getBytes("ISO8859-1"),"utf-8");
filename = java.net.URLEncoder.encode(filename, "UTF-8");
}
}
} catch (Exception e) {
logger.error("下载文件信息出错。", e);
}
if (null == inputStream) {
System.out.println("getResource error");
}
return inputStream;
}
public void setInputStream(InputStream inputStream) {
this.inputStream = inputStream;
}
@Override
public String execute() throws Exception {
return SUCCESS;
}
/*************************** get set ******************************/
public String getSavePath() {
return this.savePath;
}
public void setSavePath(String savePath) {
this.savePath = savePath;
}
public String getFilename() {
return filename;
}
public void setFilename(String filename) {
this.filename = filename;
}
}
«上一页12下一页»
相关评论
|
__label__pos
| 0.982489 |
Path length function?
First of all ... it's amazing to see that Stardog always has a helpful article whatever one googles for ...
I googled "using path queries in subqueries" and landed on "Path queries as subqueries": Stored Query Service
Follow-up question ...
It says at the bottom of that post:
"We added one obvious function — length — which simply returns the length of a path. One can use it to get the average path length"
Where do I find this function? :sweat_smile:
I am trying to get an end result like:
startClass, someSuperClass, length (distance)
a, b, 1
a, c, 2
a, d, 3
a, e, 4
Thank you!
Hi Daniel,
Thanks for the kind word, appreciate it!
There's one example at https://docs.stardog.com/query-stardog/stored-query-service#path-subqueries:
prefix sqs: <tag:stardog:api:sqs:>
prefix stardog: <tag:stardog:api:>
SELECT ?start (avg(stardog:length(?path)) as ?avg_length) {
SERVICE <query://paths> {
[] sqs:vars ?start, ?path
}
} GROUP BY ?start
That groups all paths by the start node and computes their average length.
Best,
Pavel
1 Like
Thank you, that solved it!
And again, outstanding documentation and articles. :star2:
image
@pavel: One more follow up if I may ...
I have two helper functions:
a)
// Stored Query: GetClassPath
PATHS START ?start END ?end VIA {
?start rdfs:subClassOf ?end .
}
b)
// Stored Query: GetClassDistance
prefix sqs: <tag:stardog:api:sqs:>
SELECT ?start ?end (stardog:length(?path) as ?length) {
service <query://GetClassPath> {
[] sqs:vars ?path ; sqs:var:start ?start ; sqs:var:end ?end .
}
}
... and a final query that makes use of the above:
SELECT ?start ?classSubject ?depth {
service <query://GetClassDistance> {
[] sqs:vars ?path ; sqs:var:start ?start ; sqs:var:end ?classSubject ; sqs:var:length ?depth .
}
# Which classes do we want to target?
FILTER(?start IN (:country))
}
My desired response is:
start, classSubject, depth
:country, :country, 0
:country, :place, 1
:country, :object, 2
(edited)
But I don't get :country as a classSubject unless I do ?start rdfs:subClassOf* ?end . in the first helper function (note the star).
But if I do that the length function doesn't work. All lengths are set to 1.
So question:
a) Can I include the starting class in this chain with the length calculation intact?
b) If it's not possible, is there a way of adding it by UNION/Values to an outer query that uses this?
Happy to demo on a 5-min video call.
Thanks!
Hi Daniel,
What's in your data? I just quickly checked on this simplest snippet:
@prefix : <urn:> .
:country rdfs:subClassOf :place .
:place rdfs:subClassOf :object .
and got the expected result. :country cannot be a classSubject because it's not a super class of anything. It's not a classSubject in your desired response either. Am I missing something?
Cheers,
Pavel
Argh, apologies. Typo! My desired response (or any workaround solution) should include :country together with the other ones:
start, classSubject, depth
:country, :country, 0
:country, :place, 1
:country, :object, 2
The idea is to be able to operate over :country, :place, :object when asking for SHACL shape in the next step (outside this example code).
Complete example:
prefix sqs: <tag:stardog:api:sqs:>
SELECT ?start ?classSubject ?depth ?property ?datatype ?minCount ?maxCount ?class {
# Is there any SHACL shape with class as target?
# And does that SHACL container have any Property shape objects?
?nodeShape sh:targetClass ?classSubject;
sh:property ?propertyShape .
# What property is the PropertyShape targeting?
?propertyShape sh:path ?property .
# What specific SHACL constraints are defined for that property?
OPTIONAL { ?propertyShape sh:minCount ?minCount . }
OPTIONAL { ?propertyShape sh:maxCount ?maxCount . }
OPTIONAL { ?propertyShape sh:class ?class . }
# Start of subquery – runs first to narrow down classes of relevance for shape inheritance
{
# What are the parent classes?
# And what is the distance from the classSubject
# to the top of the class chain (:meta_root)
SELECT ?start ?classSubject ?depth {
service <query://GetClassDistance> {
[] sqs:vars ?path ; sqs:var:start ?start ; sqs:var:end ?classSubject ; sqs:var:length ?depth .
}
# Which classes do we want shape for?
FILTER(?start IN (:country))
}
}
# Which properties do we want shape for?
#FILTER(?property in (:p_wikidata_id))
}
Oh I see. Right, you cannot really force a path query to return paths of zero length.
Here's one workaround, modify your GetClassDistance as follows:
prefix sqs: <tag:stardog:api:sqs:>
SELECT ?start ?end ?length {
{
service <query://GetClassPath> {
[] sqs:vars ?path ; sqs:var:start ?start ; sqs:var:end ?end .
}
bind (stardog:length(?path) as ?length)
}
union {
?start a owl:Class
bind(?start as ?end)
bind(0 as ?length)
}
}
and make sure each class has a declaration axiom, eg. :country a owl:Class (they need to be there anyway according to the OWL spec but this workaround relies on them explicitly).
Cheers,
Pavel
1 Like
That worked wonders!
Biggest thanks and have a nice weekend!
This topic was automatically closed 14 days after the last reply. New replies are no longer allowed.
|
__label__pos
| 0.942146 |
How to Calculate the Odds of Winning a Slot Machine
Slot
A major appeal of slot machines is their low-cost nature. Not only do they offer players the chance to win thousands of dollars, but many slot machines have jackpots in the millions. In 2003, a software engineer won the largest ever slot machine jackpot with a wager of just $100. That person then spent the rest of his life living a luxurious lifestyle thanks to the jackpot of 39.7 million dollars. However, not all players are as lucky as this man!
A machine’s payouts are based on the combination of symbols and payouts. The machine’s computer program is programmed to spin thousands of numbers per second. When a player presses the button, the program correlates the results to what they see on the screen. While the odds of winning are similar to other types of games, the exact payout depends on many factors, including the number of coins inserted. Therefore, learning how to calculate the odds of winning is not as straightforward as knowing the odds of playing a specific slot game.
One of the biggest dangers of slot machines is becoming greedy. Unlike table games, slot machines have no real strategy. The goal is to enjoy yourself while playing without stress or tension. It is also a good idea to avoid playing more machines than you can afford. It is not uncommon to see people pushing up their chairs against the machine. This may cause someone to confront them when they return to the game. Remember, slot machines are single-use devices and should be played responsibly.
|
__label__pos
| 0.626896 |
Pen Settings
HTML
CSS
CSS Base
Vendor Prefixing
Add External Stylesheets/Pens
Any URLs added here will be added as <link>s in order, and before the CSS in the editor. You can use the CSS from another Pen by using its URL and the proper URL extension.
+ add another resource
JavaScript
Babel includes JSX processing.
Add External Scripts/Pens
Any URL's added here will be added as <script>s in order, and run before the JavaScript in the editor. You can use the URL of any other Pen and it will include the JavaScript from that Pen.
+ add another resource
Packages
Add Packages
Search for and use JavaScript packages from npm here. By selecting a package, an import statement will be added to the top of the JavaScript editor for this package.
Behavior
Auto Save
If active, Pens will autosave every 30 seconds after being saved once.
Auto-Updating Preview
If enabled, the preview panel updates automatically as you code. If disabled, use the "Run" button to update.
Format on Save
If enabled, your code will be formatted when you actively save your Pen. Note: your code becomes un-folded during formatting.
Editor Settings
Code Indentation
Want to change your Syntax Highlighting theme, Fonts and more?
Visit your global Editor Settings.
HTML
<h1>Sortable Grid Using GreenSock <a href="https://greensock.com/draggable" target="_blank">Draggable</a></h1>
<div class="layout">
<label><input id="mixed" name="layout" type="radio" value="mixed" checked=""/> Mixed Tile Size</label>
<label><input id="fixed" name="layout" type="radio" value="fixed"/> Fixed Tile Size</label>
<label><input id="column" name="layout" type="radio" value="column"/> Single Column</label>
</div>
<button id="add">ADD TILE</button>
<div id="list"></div>
!
CSS
$background: #3A3A4D;
$tile-color: yellowgreen;
$color: #efefef;
html {
font-family: RobotoDraft, 'Helvetica Neue', Helvetica, Arial;
}
body {
background: $background;
padding: 25px;
}
*, *:before, *:after {
box-sizing: border-box;
}
h1 {
color: $color;
font-weight: normal;
font-weight: 300;
margin-bottom: 30px;
a {
color: $color;
border-bottom: 2px solid $color;
text-decoration: none;
padding-bottom: 3px;
&:hover {
color: $tile-color;
border-color: $tile-color;
}
}
}
button {
padding: 12px 30px;
box-shadow: none;
border: none;
color: $color;
cursor: pointer;
background-color: rgba(0, 0, 0, 0.3);
&:hover {
background-color: rgba(0, 0, 0, 0.35);
color: $tile-color;
}
&:focus {
outline: none;
}
}
#list {
background-color: rgba(0, 0, 0, 0.2);
width: 250px;
width: 90%;
width: 0;
position: relative;
display: block;
margin-top: 50px;
}
.tile {
display: block;
position: absolute;
background-color: $tile-color;
color: $background;
padding: 5px;
font-weight: bold;
}
.layout {
color: $color;
margin-bottom: 15px;
label {
margin-right: 15px;
cursor: pointer;
&:hover {
color: $tile-color;
}
}
input {
margin-right: 3px;
}
}
!
JS
// GRID OPTIONS
var rowSize = 100;
var colSize = 100;
var gutter = 7; // Spacing between tiles
var numTiles = 25; // Number of tiles to initially populate the grid with
var fixedSize = false; // When true, each tile's colspan will be fixed to 1
var oneColumn = false; // When true, grid will only have 1 column and tiles have fixed colspan of 1
var threshold = "50%"; // This is amount of overlap between tiles needed to detect a collision
var $add = $("#add");
var $list = $("#list");
var $mode = $("input[name='layout']");
// Live node list of tiles
var tiles = $list[0].getElementsByClassName("tile");
var label = 1;
var zIndex = 1000;
var startWidth = "100%";
var startSize = colSize;
var singleWidth = colSize * 3;
var colCount = null;
var rowCount = null;
var gutterStep = null;
var shadow1 = "0 1px 3px 0 rgba(0, 0, 0, 0.5), 0 1px 2px 0 rgba(0, 0, 0, 0.6)";
var shadow2 = "0 6px 10px 0 rgba(0, 0, 0, 0.3), 0 2px 2px 0 rgba(0, 0, 0, 0.2)";
$(window).resize(resize);
$add.click(createTile);
$mode.change(init);
init();
// ========================================================================
// INIT
// ========================================================================
function init() {
var width = startWidth;
// This value is defined when this function
// is fired by a radio button change event
switch (this.value) {
case "mixed":
fixedSize = false;
oneColumn = false;
colSize = startSize;
break;
case "fixed":
fixedSize = true;
oneColumn = false;
colSize = startSize;
break;
case "column":
fixedSize = false;
oneColumn = true;
width = singleWidth;
colSize = singleWidth;
break;
}
// For images demo
//window.stop();
$(".tile").each(function() {
Draggable.get(this).kill();
$(this).remove();
});
TweenLite.to($list, 0.2, { width: width });
TweenLite.delayedCall(0.25, populateBoard);
function populateBoard() {
label = 1;
resize();
for (var i = 0; i < numTiles; i++) {
createTile();
}
}
}
// ========================================================================
// RESIZE
// ========================================================================
function resize() {
colCount = oneColumn ? 1 : Math.floor($list.outerWidth() / (colSize + gutter));
gutterStep = colCount == 1 ? gutter : (gutter * (colCount - 1) / colCount);
rowCount = 0;
layoutInvalidated();
}
// ========================================================================
// CHANGE POSITION
// ========================================================================
function changePosition(from, to, rowToUpdate) {
var $tiles = $(".tile");
var insert = from > to ? "insertBefore" : "insertAfter";
// Change DOM positions
$tiles.eq(from)[insert]($tiles.eq(to));
layoutInvalidated(rowToUpdate);
}
// ========================================================================
// CREATE TILE
// ========================================================================
function createTile() {
var colspan = fixedSize || oneColumn ? 1 : Math.floor(Math.random() * 2) + 1;
var element = $("<div></div>").addClass("tile").html(label++);
var lastX = 0;
Draggable.create(element, {
onDrag : onDrag,
onPress : onPress,
onRelease : onRelease,
zIndexBoost : false
});
// NOTE: Leave rowspan set to 1 because this demo
// doesn't calculate different row heights
var tile = {
col : null,
colspan : colspan,
height : 0,
inBounds : true,
index : null,
isDragging : false,
lastIndex : null,
newTile : true,
positioned : false,
row : null,
rowspan : 1,
width : 0,
x : 0,
y : 0
};
// Add tile properties to our element for quick lookup
element[0].tile = tile;
$list.append(element);
layoutInvalidated();
function onPress() {
lastX = this.x;
tile.isDragging = true;
tile.lastIndex = tile.index;
TweenLite.to(element, 0.2, {
autoAlpha : 0.75,
boxShadow : shadow2,
scale : 0.95,
zIndex : "+=1000"
});
}
function onDrag() {
// Move to end of list if not in bounds
if (!this.hitTest($list, 0)) {
tile.inBounds = false;
changePosition(tile.index, tiles.length - 1);
return;
}
tile.inBounds = true;
for (var i = 0; i < tiles.length; i++) {
// Row to update is used for a partial layout update
// Shift left/right checks if the tile is being dragged
// towards the the tile it is testing
var testTile = tiles[i].tile;
var onSameRow = (tile.row === testTile.row);
var rowToUpdate = onSameRow ? tile.row : -1;
var shiftLeft = onSameRow ? (this.x < lastX && tile.index > i) : true;
var shiftRight = onSameRow ? (this.x > lastX && tile.index < i) : true;
var validMove = (testTile.positioned && (shiftLeft || shiftRight));
if (this.hitTest(tiles[i], threshold) && validMove) {
changePosition(tile.index, i, rowToUpdate);
break;
}
}
lastX = this.x;
}
function onRelease() {
// Move tile back to last position if released out of bounds
this.hitTest($list, 0)
? layoutInvalidated()
: changePosition(tile.index, tile.lastIndex);
TweenLite.to(element, 0.2, {
autoAlpha : 1,
boxShadow: shadow1,
scale : 1,
x : tile.x,
y : tile.y,
zIndex : ++zIndex
});
tile.isDragging = false;
}
}
// ========================================================================
// LAYOUT INVALIDATED
// ========================================================================
function layoutInvalidated(rowToUpdate) {
var timeline = new TimelineMax();
var partialLayout = (rowToUpdate > -1);
var height = 0;
var col = 0;
var row = 0;
var time = 0.35;
$(".tile").each(function(index, element) {
var tile = this.tile;
var oldRow = tile.row;
var oldCol = tile.col;
var newTile = tile.newTile;
// PARTIAL LAYOUT: This condition can only occur while a tile is being
// dragged. The purpose of this is to only swap positions within a row,
// which will prevent a tile from jumping to another row if a space
// is available. Without this, a large tile in column 0 may appear
// to be stuck if hit by a smaller tile, and if there is space in the
// row above for the smaller tile. When the user stops dragging the
// tile, a full layout update will happen, allowing tiles to move to
// available spaces in rows above them.
if (partialLayout) {
row = tile.row;
if (tile.row !== rowToUpdate) return;
}
// Update trackers when colCount is exceeded
if (col + tile.colspan > colCount) {
col = 0; row++;
}
$.extend(tile, {
col : col,
row : row,
index : index,
x : col * gutterStep + (col * colSize),
y : row * gutterStep + (row * rowSize),
width : tile.colspan * colSize + ((tile.colspan - 1) * gutterStep),
height : tile.rowspan * rowSize
});
col += tile.colspan;
// If the tile being dragged is in bounds, set a new
// last index in case it goes out of bounds
if (tile.isDragging && tile.inBounds) {
tile.lastIndex = index;
}
if (newTile) {
// Clear the new tile flag
tile.newTile = false;
var from = {
autoAlpha : 0,
boxShadow : shadow1,
height : tile.height,
scale : 0,
width : tile.width
};
var to = {
autoAlpha : 1,
scale : 1,
zIndex : zIndex
}
timeline.fromTo(element, time, from, to, "reflow");
}
// Don't animate the tile that is being dragged and
// only animate the tiles that have changes
if (!tile.isDragging && (oldRow !== tile.row || oldCol !== tile.col)) {
var duration = newTile ? 0 : time;
// Boost the z-index for tiles that will travel over
// another tile due to a row change
if (oldRow !== tile.row) {
timeline.set(element, { zIndex: ++zIndex }, "reflow");
}
timeline.to(element, duration, {
x : tile.x,
y : tile.y,
onComplete : function() { tile.positioned = true; },
onStart : function() { tile.positioned = false; }
}, "reflow");
}
});
if (!row) rowCount = 1;
// If the row count has changed, change the height of the container
if (row !== rowCount) {
rowCount = row;
height = rowCount * gutterStep + (++row * rowSize);
timeline.to($list, 0.2, { height: height }, "reflow");
}
}
!
999px
Console
|
__label__pos
| 0.998251 |
Find maximum element from each sub-array of size 'k'| Set 1
Given input as an integer array and an integer 'k', find and print element with maximum value from each sub-array of size 'k'.
For example, for the input array {4,2,12,34,23,35,44,55} and for k = 3, output should be 12,34,34,35,44,55. Observe that 12 is the largest element in the first sub-array {4,2,12}, 34 is the largest element in the second and third sub-arrays - {2,12,34} and {12,34,23} and so on.
Video coming soon!
Subscribe for more updates
Algorithm/Insights
Simple approach: One of the intuitive ways to solve this problem could be to check all the sub-arrays of size 'k' and find out the maximum element from each of these sub-arrays. To check all the sub-arrays of size 'k', we start from a sub-array(window) of the first 'k' elements and then shift this window by one element until we reach the end of the array. There would be total 'n-k+1' such windows of 'k' elements each where 'n' denotes the size of the input array. In the following code snippet, this approach is implemented in function 'simplePrintMaxfromEachSubarray(int[] array, int k)'. Finding maximum element from sub-array of size 'k' takes O(k) time and there are total 'n-k+1' such sub-arrays. Therefore, overall time complexity of this approach is O(k(n-k+1)) which is equivalent to O(nk).
Optimized approach: In this approach, we make use of AVL trees (self-balanced Binary Search Tree) as explained in the following algorithm. We will be referring to AVL tree as a BST in below steps because it would make understanding the intuition of the algorithm easier.
1. Create a BST from first 'k' elements of the input array.
2. Find node with maximum value from the BST created in step #1. Print this node's value. This would represent an element with maximum value from first sub-array of size 'k'.
3. Now starting from i = 0 upto i = n-k-1
a. Search and delete element with value array[i] from the BST.
b. Insert node with value as array[i+k] into the BST. Now this BST represents next sub-array of size 'k'.
c. Find node with maximum value from the BST. Print this node's value.
Handling duplicates: Note that this optimized approach which uses BST won't be able to handle the case when input array has more than one elements with same value. We can handle this case easily by pre-processing the input array such that whenever there are duplicates, their values are modified by adding different decimal offsets. For example, if the given input array is {2,3,6,5,6,5} then it could be modified to {2,3,6.01,5.01,6.02,5.02}. Now this modified input does not have any duplicates and above algorithm should work. Care has to taken to remove the decimal part while printing the value of maximum element in a sub-array.
In the above algorithm, creating BST takes O(logk) time. Search and delete operation, insert operation and finding maximum valued node operation, each of these operations takes O(logk) time and we need to perform each of these operations 'n-k' times. Therefore, overall time complexity of this algorithm is O(nlogk). Extra space taken in the form of BST is O(k).
You can check out function 'printMaxfromEachSubarray(int[] array, int k)' in the following code snippet for implementation details.
Please feel free to add any queries/comments you might have in the below section.
Algorithm Visualization
Code Snippet
package com.ideserve.nilesh.questions;
/**
* <b>IDeserve <br>
* <a href="https://www.youtube.com/c/IDeserve">https://www.youtube.com/c/IDeserve</a>
* In O(nlogk) time, this code finds out maximum element from each sub-array of size 'k'.
* @author Nilesh
*/
public class MaximumfromEachSubarray
{
private class AVLTreeNode
{
int data;
AVLTreeNode left;
AVLTreeNode right;
int height;
AVLTreeNode(int data)
{
this.data = data;
this.height = 1;
}
}
AVLTreeNode root;
public MaximumfromEachSubarray()
{
}
private int getHeight(AVLTreeNode node)
{
if (node == null)
return 0;
return node.height;
}
private void updateHeight(AVLTreeNode node)
{
if (node == null) return;
node.height = Math.max(getHeight(node.left), getHeight(node.right)) + 1;
}
private AVLTreeNode rotateRight(AVLTreeNode node)
{
if (node == null) return node;
AVLTreeNode beta = node.left;
AVLTreeNode t2 = beta.right;
beta.right = node;
node.left = t2;
// we first need to update the height of node because height of beta uses height of node
updateHeight(node);
// now we update height of beta
updateHeight(beta);
return beta;
}
private AVLTreeNode rotateLeft(AVLTreeNode node)
{
if (node == null) return node;
AVLTreeNode beta = node.right;
AVLTreeNode t2 = beta.left;
beta.left = node;
node.right = t2;
// we first need to update the height of node because height of beta uses height of node
updateHeight(node);
// now we update height of beta
updateHeight(beta);
return beta;
}
private int getBalance(AVLTreeNode node)
{
if (node == null)
{
return 0;
}
int balance;
balance = getHeight(node.left) - getHeight(node.right);
return balance;
}
private int getMaxValue(AVLTreeNode node)
{
// though this case should not be hit ever for the usecase this function is employed for
if (node == null) return Integer.MIN_VALUE;
// if this is the left-most node
if (node.right == null) return node.data;
return getMaxValue(node.right);
}
private int getMinValue(AVLTreeNode node)
{
// though this case should not be hit ever for the usecase this function is employed for
if (node == null) return Integer.MIN_VALUE;
// if this is the left-most node
if (node.right == null) return node.data;
return getMaxValue(node.left);
}
private AVLTreeNode delete(AVLTreeNode node, int key)
{
// if empty tree
if (node == null) return null;
if (key < node.data)
{
node.left = delete(node.left, key);
}
else if (key > node.data)
{
node.right = delete(node.right, key);
}
else // key to be deleted is equal to node data
{
// one child/no child case
if (node.left == null)
{
node = node.right;
}
else if (node.right == null)
{
node = node.left;
}
// two children case
// copy value of inorder successor into current node and delete inorder successor
// since right sub-tree would be modified, update node.right
else
{
int inorderSuccessorValue = getMinValue(node.right);
node.data = inorderSuccessorValue;
node.right = delete(node.right, inorderSuccessorValue);
}
}
// if there was only one node in the tree which got deleted above return null
if (node == null)
{
return null;
}
// update the height of the node
updateHeight(node);
// check the balance at this node and perform rotations accordingly
int balance = getBalance(node);
if (balance > 1) // indicates either left-left or left-right case
{
if (getBalance(node.left) >= 0) // confirms left-left case
{
node = rotateRight(node);
}
else // confirms left-right case
{
node.left = rotateLeft(node.left);
node = rotateRight(node);
}
}
else if (balance < -1) // indicates either right-right or right-left case
{
if (getBalance(node.right) <= 0) // confirms right-right case
{
node = rotateLeft(node);
}
else // confirms right-left case
{
node.right = rotateRight(node.right);
node = rotateLeft(node);
}
}
return node;
}
private AVLTreeNode insert(AVLTreeNode node, int key)
{
// do usual BST insertion first
if (node == null)
{
this.root = new AVLTreeNode(key);
return this.root;
}
if (key < node.data)
{
node.left = insert(node.left, key);
}
else if (key > node.data)
{
node.right = insert(node.right, key);
}
else
{
return node;
}
// now update the height of the node
updateHeight(node);
// check the balance at this node and perform rotations accordingly
int balance = getBalance(node);
if (balance > 1) // indicates either left-left or left-right case
{
if (key < node.left.data) // confirms left-left case
{
node = rotateRight(node);
}
else // confirms left-right case
{
node.left = rotateLeft(node.left);
node = rotateRight(node);
}
}
else if (balance < -1) // indicates either right-right or right-left case
{
if (key > node.right.data) // confirms right-right case
{
node = rotateLeft(node);
}
else // confirms right-left case
{
node.right = rotateRight(node.right);
node = rotateLeft(node);
}
}
return node;
}
public void insert(int key)
{
root = insert(this.root, key);
return;
}
public void delete(int key)
{
root = delete(this.root, key);
return;
}
private void printMax(int[] array, int low, int high)
{
int maxValue = Integer.MIN_VALUE;
for (int i = low; i <= high; i++)
{
if (array[i] > maxValue)
{
maxValue = array[i];
}
}
System.out.println(maxValue);
}
public void simplePrintMaxfromEachSubarray(int[] array, int k)
{
// {4,2,12,34,23,35,44,55};
int low = 0, high = low + k - 1;
while (high < array.length)
{
printMax(array, low, high);
low += 1;
high += 1;
}
}
public void printMaxfromEachSubarray(int[] array, int k)
{
// create a balanced BST by inserting first 'k' keys
for (int i = 0; i < k; i++)
{
insert(array[i]);
}
// print the maximum value of first sub-array of 'k' elements
System.out.println(getMaxValue(root));
/*
* Now find maximum element for each remaining contiguous sub-array of size 'k'
* 1. delete 'i'th element from BST
* 2. insert (i+k)th element into BST
* 3. find element with maximum value in this modified BST
*/
for (int i = 0; i < array.length-k; i++)
{
delete(array[i]);
insert(array[i+k]);
System.out.println(getMaxValue(root));
}
}
public static void main(String[] args)
{
MaximumfromEachSubarray solution = new MaximumfromEachSubarray();
int[] array = {4,2,12,34,23,35,44,55};
int k = 3;
System.out.println("Maximum elements from each sub-array of specified size are - ");
solution.printMaxfromEachSubarray(array, k);
}
}
Order of the Algorithm
Time Complexity is O(nlogk), n: size of input array, k: sub-array size
Space Complexity is O(k)
Contribution
• Sincere thanks from IDeserve community to Nilesh More for compiling current post.
Nilesh More
|
__label__pos
| 0.99934 |
Call now! (ID:)+
HomeWeb Hosting ArticlesShared Web Hosting Defined
Shared Web Hosting Defined
The most elemental and frequently availed of sort of web hosting is the hosting solution. It represents a means to host your web portal without having to know much about programming and administrating a web hosting server. In addition, it's also the cheapest type of web hosting and it's quite affordable for everybody. However, what is hosting?
What is hosting?
As the name implies, the hosting service is a sort of service where multiple customers share the resources of the same web hosting server. This means that all server components like CPU, hard disk drives, RAM, network interface cards etc. are divided among the clients whose accounts are on that same server. This is normally made feasible by opening separate accounts for the separate users and appointing some limits and resource usage quotas for each of them. Those restrictions are allocated so as to prevent the clients from meddling with each other's accounts and, of course, to prevent the server from overloading. Usually, hosting customers do not have complete root access to the server's configuration files, which primarily goes to say that they do not have access to anything else on the web hosting server but their very own web hosting account. The website hosting features that each account may use are determined by the hosting supplier that owns the hosting server and by the given website hosting plan. That results in the second essential question:
How are the shared hosting servers shared among the customers?
Hosting suppliers that provide hosting services usually have diverse website hosting plans. Those plans include different amounts of website hosting features and specifications, which actually set the limitations that a website hosting account will include. The customer may choose between the different web hosting plans and sign up for the one that he deems will suit him best. The web hosting plan will then determine what limits the customer's account will have, once created. The costs and the specs of the website hosting packages are set by the particular hosting provider. Depending on the policy of the firm, the hosting solution falls into 2 types - the free hosting service and the classic shared solution, most recently very famous among "cPanel hosting" companies as a cloud web hosting one. It's not possible to affirm, which one is more preferable, since they are very different from each other and they really are determined by the marketing strategy of the particular distributor and, of course, the needs of the specific user.
What is the distinction between the free of charge and the common hosting service?
Of course, the primary difference between the free of charge and the paid service is in the quantity of features that they contain. Free web hosting providers are not able to keep a huge number of web hosting servers, hence, they plainly host more customers on a single web server by reducing the quantity of system resources offered by the accounts. This will be effective only in case the servers are supervised and maintained appropriately, since the large number of accounts may make the server crash regularly. Most of the free web hosting companies, though, overlook the quality of the service and hence, it's quite tough to stumble upon a free of cost website hosting service that's in fact worth the time. The top free hosting vendors commonly offer free technical support even to the free hosting clients, since they want their web pages to get bigger so that they subsequently migrate to a paid web hosting plan, which offers more web hosting features. One such firm, for example, is FreeHostia.com, which is among the biggest and oldest free website hosting corporations worldwide.
On the other hand, established hosting distributors like us, may afford to keep a lot of web hosting servers and hence, we may afford to offer much more feature-rich website hosting plans. Of course, that influences the cost of the website hosting packages. Paying a higher price for a hosting plan, however, does not necessarily mean that this account has a better quality. The most optimal services are the balanced ones, which offer a fee that corresponds to the concrete service which you're getting. In addition, we also provide a free bonus with the hosting package, such as the 1-click applications installer, accompanied by hundreds of cost-free web layouts. As a web hosting provider, we do look after our reputation and that's why if you go with us, you can rest calm that you won't get hoaxed into buying a plan that you cannot in fact utilize.
What should I anticipate from a hosting service?
The hosting service is best for individuals who want to host an average web portal, which is going to generate a small or medium amount of web traffic every month. You cannot anticipate, however, that a hosting account will be sufficient for your needs, since as your business grows bigger, your web page will become more and more resource consuming. Therefore, you will have to eventually migrate to a more feature-rich web hosting solution like a semi-dedicated servers, a VPS (aka a virtual hosting server, or VPS), or even a dedicated server. So, when picking a hosting provider, you should also think about scalability, otherwise you might end up migrating your domain manually to a separate provider, which can bring about site problems and even continuous downtime for your web portal. If you go with as your web hosting supplier, you can rest safe that we can present you with the required domain name and hosting services as you grow, is vital and will spare you lots of hassles in the future.
|
__label__pos
| 0.528875 |
C++ Map Library - swap() Function
Advertisements
Description
The C++ function std::map::swap() exchanges the content of map with contents of map x.
Declaration
Following is the declaration for std::map::swap() function form std::map header.
C++98
template <class Key, class T, class Compare, class Alloc>
void swap (map<Key,T,Compare,Alloc>& first,
map<Key,T,Compare,Alloc>& second);
Parameters
• first − First map object.
• second − Second map object of same type.
Return value
None
Exceptions
This function doesn't throw exception.
Time complexity
Constant i.e. O(1)
Example
The following example shows the usage of std::map::swap() function.
#include <iostream>
#include <map>
using namespace std;
int main(void) {
map<char, int> m1 = {
{'a', 1},
{'b', 2},
{'c', 3},
{'d', 4},
{'e', 5},
};
map<char, int> m2;
swap(m1, m2);
cout << "Map contains following elements" << endl;
for (auto it = m2.begin(); it != m2.end(); ++it)
cout << it->first << " = " << it->second << endl;
return 0;
}
Let us compile and run the above program, this will produce the following result −
Map contains following elements
a = 1
b = 2
c = 3
d = 4
e = 5
map.htm
Advertisements
|
__label__pos
| 0.902873 |
10
I was wondering how I could determine in my ItemWriter, whether Spring Batch was currently in chunk-processing-mode or in the fallback single-item-processing-mode. In the first place I didn't find the information how this fallback mechanism is implemented anyway.
Even if I haven't found the solution to my actual problem yet, I'd like to share my knowledge about the fallback mechanism with you.
Feel free to add answers with additional information if I missed anything ;-)
• can you explain the real world problem which brings you to the question "how do the writer know the current processing-mode"? – Michael Pralow May 16 '13 at 11:20
• Sure :-) I'm storing a business log (beside my technical logging). In this log, messages for each item should only appear once. In case of an exception during processing I will also write error logs for this item in the business log. If an item has been processed but rolled back, I'm not interested in its error logs. I only want to log these error if I'm in single item processing. Otherwise if I'm in chunk mode, I might log errors for items which are fine, just because they are in a bad chunk. – Peter Wippermann May 16 '13 at 14:17
15
The implementation of the skip mechanism can be found in the FaultTolerantChunkProcessor and in the RetryTemplate.
Let's assume you configured skippable exceptions but no retryable exceptions. And there is a failing item in your current chunk causing an exception.
Now, first of all the whole chunk shall be written. In the processor's write() method you can see, that a RetryTemplate is called. It also gets two references to a RetryCallback and a RecoveryCallback.
Switch over to the RetryTemplate. Find the following method:
protected <T> T doExecute(RetryCallback<T> retryCallback, RecoveryCallback<T> recoveryCallback, RetryState state)
There you can see that the RetryTemplate is retried as long as it's not exhausted (i.e. exactly once in our configuration). Such a retry will be caused by a retryable exception. Non-retryable exceptions will immediately abort the retry mechanism here.
After the retries are exhausted or aborted, the RecoveryCallback will be called:
e = handleRetryExhausted(recoveryCallback, context, state);
That's where the single-item-processing mode will kick-in now!
The RecoveryCallback (which was defined in the processor's write() method!) will put a lock on the input chunk (inputs.setBusy(true)) and run its scan() method. There you can see, that a single item is taken from the chunk:
List<O> items = Collections.singletonList(outputIterator.next());
If this single item can be processed by the ItemWriter correctly, than the chunk will be finished and the ChunkOrientedTasklet will run another chunk (for the next single items). This will cause a regular call to the RetryCallback, but since the chunk has been locked by the RecoveryTemplate, the scan() method will be called immediately:
if (!inputs.isBusy()) {
// ...
}
else {
scan(contribution, inputs, outputs, chunkMonitor);
}
So another single item will be processed and this is repeated, until the original chunk has been processed item-by-item:
if (outputs.isEmpty()) {
inputs.setBusy(false);
That's it. I hope you found this helpful. And I even more hope that you could find this easily via a search engine and didn't waste too much time, finding this out by yourself. ;-)
1
A possible approach to my original problem (the ItemWriter would like to know, whether it's in chunk or single-item mode) could be one of the following alternatives:
• Only when the passed chunk is of size one, any further checks have to be done
• When the passed chunk is a java.util.Collections.SingletonList, we would be quite sure, since the FaultTolerantChunkProcessor does the following:
List items = Collections.singletonList(outputIterator.next());
Unfortunately, this class is private and so we can't check it with instanceOf.
• In reverse, if the chunk is an ArrayList we could also be quite sure, since the Spring Batch's Chunk class uses it:
private List items = new ArrayList();
• One blurring left would be buffered items read from the execution context. But I'd expect those to be ArrayLists also.
Anyway, I still find this method too vague. I'd rather like to have this information provided by the framework.
An alternative would be to hook my ItemWriter in the framework execution. Maybe ItemWriteListener.onWriteError() is appropriate.
Update: The onWriteError() method will not be called if you're in single-item mode and throw an exception in the ItemWriter. I think that's a bug a filed it: https://jira.springsource.org/browse/BATCH-2027
So this alternative drops out.
Here's a snippet to do the same without any framework means directly in the writer
private int writeErrorCount = 0;
@Override
public void write(final List<? extends Long> items) throws Exception {
try {
writeWhatever(items);
} catch (final Exception e) {
if (this.writeErrorCount == 0) {
this.writeErrorCount = items.size();
} else {
this.writeErrorCount--;
}
throw e;
}
this.writeErrorCount--;
}
public boolean isWriterInSingleItemMode() {
return writeErrorCount != 0;
}
Attention: One should rather check for the skippable exceptions here and not for Exception in general.
• We can't call instanceof on the singleton list, but this works: Class<?> singletonListClazz = Class.forName( "java.util.Collections$SingletonList" ); boolean retrying = false; if( items.getClass().equals( singletonListClazz ) ){ retrying = true; } – slh777 Sep 3 '15 at 21:41
• @slh777 Thanks for your advice. That would work indeed. However, I'm not happy with this approach since we heavily rely on the implementation details of the framework (i.e. a SingletonList is returned, when in SingleItemMode). I'd rather like to see this encapsulated by the framework. :-( – Peter Wippermann Oct 6 '15 at 12:35
Your Answer
By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy
Not the answer you're looking for? Browse other questions tagged or ask your own question.
|
__label__pos
| 0.585886 |
Fragmentation in SQL Server| Internal and External Fragmentation
Neerajprasadsharma
Posted by in Sql Server category on for Advance level | Points: 250 | Views : 8643 red flag
In this article you will have a clearer understanding about fragmentation in SQL Server, how it occurs, what are the causes and some problem related to it.
Introduction
In this article you will have a clearer understanding about fragmentation in SQL Server, how it occurs, what are the causes and some problem related to it.
This article is dedicated to SQL Server fragmentation only, if you are following the current long running series, then you know, we wrote in the index internals article about basics of fragmentation, but as the series grows, we need to understand everything about fragmentation as it will help understand the complete series.
Background
Like all my articles, this article will be filled with practical based knowledge only. I will start with the same article I linked when I was trying to explain about SQL Server fragmentation so here as well, please read below.
"There are two different types of fragmentation in SQL Server: Internal and External. Internal fragmentation is the result of index pages taking up more space than needed.
It is like having a book where some of the pages are left blank; we do not know what pages are blank until we read the entire book and the same applies to SQL Server, which has to read all the pages in the index wasting extra-time and server resources in the empty pages. External fragmentation occurs when the pages are not contiguous on the index. Following the book analogy,it is like having a book where pages are not ordered in a logical way (page 1, then page 2, then page 3 and so on) causing you to go back and forward to compound the information and make sense of the reading. Heavily used tables, that contains fragmented indexes will impact your database performance. "
Ok, let us try to understand it with an example, suppose in SQL Server a page can hold only 6 pages, for every 7th row Storage Engine has to assign a new page in the storage devices.
The storage engines always try to insert the SQL server index data page in a contiguous sequential manner, for example:
Suppose a user has a unique index on the ID column and insert data like 2,4,6,8,10,12,16,18,20,22,24,26,28,30,32 then most probably you won't see any external fragmentation (for efficient, faster insert Some time storage engine can create a little external fragmentation, but that is negligible) because Storage Engine takes care of that, but still you will find little internal fragmentation because when the some pages are Not full by the data, technically it is internally fragmented right?
Look into the image below:
Look at how the data in the index are nicely continuing physically, but what if user insert 3 in the index?
There would be two options first inserts the row into the index at the right place and rearrange all the index accordingly.
That would be certainly very-very expensive approach, so the better approach would be to allocate a new page and link it to Page 1 and 3, move the data accordingly.
That is what Storage Engine does it allocate a new page for index, move approx half of rows in the new page and link it back.
Look at the image below:
User inserted row 3 in the index, since id 3 should be in the page 1 and there is no space left on the page so storage engine has allocated page id 4 and move half of the data in the page id 4. Now technically the index has fragmented internally and externally.
How internally? Because first page which has a capacity of containing 5 rows now only contains 3 rows in page id 1 and 4, and externally because now the pages are not physically contiguous, so if a user requested rows in the key order, storage engine has to move forward and then come backward to get IDs in order.
Thus, when the index is externally fragmented you will have random IO and random IOs are pretty slower than sequential IOs specially in the conventional rotating disks. I am quoting BOL on that and want readers to visit the link on that
"Accessing data sequentially is much faster than accessing it randomly because of the way in which the disk hardware works. The seek operation, which occurs when the disk head positions itself at the right disk cylinder to access data requested, takes more time than any other part of the I/O process. Because reading randomly involves a higher number of seek operations than does sequential reading, random reads deliver a lower rate of throughput"
Set up the table to test fragmentation in SQL Server
We will try to use the almost same example, what we tried to explain above, but we will create a large table because Storage Engine doesn`t take care of fragmentation in the small indexes for obvious reasons.
As we used the number table to populate the table in this series, again we will use the number table to populate the table, but this time the table will have even rows to match the above example.
Script below:
CREATE TABLE [DBO].Fragmented (
Primarykey int NOT NULL,
Keycol VARCHAR(50) NOT NULL
,SearchCol INT
,SomeData char(20),
Keycol2 VARCHAR(50) NOT NULL
,SearchCol2 INT
,SomeData2 char(20),
Keycol3 VARCHAR(50) NOT NULL
,SearchCol3 INT
,SomeData3 char(20))
GO
INSERT INTO [DBO].Fragmented
SELECT
n,
n,
n
,'Some text..'
,
n,
n
,'Some text..'
,
n,
n
,'Some text..'
FROM Numbers WHERE N/2.0=N/2
GO
CREATE UNIQUE CLUSTERED INDEX CI_Fragmented ON [DBO].Fragmented (Primarykey)
GO
Ok, the table has created with even clustered index IDs so first look at the external fragmentation and internal fragmentation using sys.dm_db_index_physical_stats DMV, in this DMV external fragmentation shows as “avg_fragmentation_in_percent “column and internal fragmentation show as “avg_page_space_used_in_percent”.
Let us run the query against our newly created table and look at the fragmentation.
SELECT OBJECT_NAME(OBJECT_ID) Table_Name, index_id,index_type_desc,index_level,
avg_fragmentation_in_percent fragmentation, avg_page_space_used_in_percent Page_Density, page_count,Record_count
FROM sys.dm_db_index_physical_stats
(db_id('tutorialsqlserver'), object_id ('Fragmented'), null, null,'DETAILED')
Where
Index_Level =0 and Index_Id=1
There is no external fragmentation in the index and even all pages all almost fully occupied nicely.
So, we can start. Earlier we use to paste the DBCC IND command output into the excel find and then we use to filter the results accordingly to the requirement, but since I as passionate in SQL server not in Excel so I decided to create a table with a couple of stored procedures to extract the result of DBCC IND command and insert it in the INDOUTPUT table.
You can get the script from here. So, let us get the DBCC IND command result in the INDOUTPUT table using our stored procedure "REFRESH_INDOUTPUT".
EXEC REFRESH_INDOUTPUT 'TutorialSQLServer', 'Fragmented', 1
Ok, great the table has refreshed, now let us look into the first three data pages in the index.
Select Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
Where IndexLevel = 0 and PrevPagePID = 0
Select Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
Where PagePID=264129
Select Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
Where PagePID=264130
Look into the above result all the first three page id and next to each other this is the same condition as we described in the introduction section.Let us go into one step deeper and look into the index pages using DBCC IND COMMAND.
DBCC PAGE ('TutorialSQLServer',1,264128,3) WITH TABLERESULTS
GO
DBCC PAGE ('TutorialSQLServer',1,264129,3) WITH TABLERESULTS
GO
DBCC PAGE ('TutorialSQLServer',1,264130,3) WITH TABLERESULTS
GO
We can clearly see in the output of DBCC IND command, the first data page/leaf level "264128" contains primary key 2 to 160 with its respective data from, next data page "264129" contains primarykey 162 to 316 and data page "264130" contains primarykey 318 to 472 all the pages are nicely externally and internally defragmented.
So let us fragment it by inserting the primarykey 3 in the table and see what happen.
INSERT INTO [DBO].Fragmented
SELECT
n,
n,
n,
'Some text..',
n,
n,
'Some text..',
n,
n
,'Some text..'
FROM Numbers WHERE N=3
GO
Row with primarykey (key value) 3 has inserted into the table so now as described in introduction section the index should be fragmented,
so let us again look into the output of sys.dm_db_index_physical_stats.
SELECT OBJECT_NAME(OBJECT_ID) Table_Name, index_id,index_type_desc,index_level,
avg_fragmentation_in_percent fragmentation, avg_page_space_used_in_percent Page_Density, page_count,Record_count
FROM sys.dm_db_index_physical_stats
(db_id('tutorialsqlserver'), object_id ('Fragmented'), null, null,'DETAILED')
Where
Index_Level =0 and Index_Id=1
Look at the result of DMV, now the index is slightly fragmented, before inserting the above row DMV output was not showing any fragmentation and now its showing 0.0579206487112656 percent, and there is a change in page density and page count and offcorse the record count as well.
So as per out theory the second page should be out of order and linked to the first page with approx half of rows should be moved to the new page and this newly created page`s next page should be linked to the second page of the last result.
So let us again used our "REFRESH_INDOUTPUT" stored procedure to get the result of IND command.
EXEC REFRESH_INDOUTPUT 'TutorialSQLServer', 'Fragmented', 1
GO
SELECT Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
WHERE IndexLevel = 0 and PrevPagePID = 0
--274014
SELECT Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
WHERE PagePID=274014
--264129
SELECT Table_Name, PrevPagePID, PagePID, NextPagePID from INDOUTPUT
WHERE PagePID=264129
Great, as expected the second page is out of order, then linked back to the ordered next page.
Let us again look inside the all the first three pages using DBCC PAGE command.
DBCC PAGE ('TutorialSQLServer',1,264128,3) WITH TABLERESULTS
GO
DBCC PAGE ('TutorialSQLServer',1,274014,3) WITH TABLERESULTS
GO
DBCC PAGE ('TutorialSQLServer',1,264130,3) WITH TABLERESULTS
GO
Look at the results, now the first page contains the newly inserted row, but because there was no space left to insert a new row in the page,
The Storage Engine has created a new page moved half of rows into this newly created page and then linked it to the first and second page, half of the rows moved to a new page with id "274014" you can see there is no change in data of page id "264129" as it wasn`t required.
Write performance problem with fragmentation (Page Split)
In this section we will look into a fragmentation related performance problem which occurs during the page split.
For that we will insert a row and look into the Statistics IO and transaction log activity using sys.dm_tran_database_transactions.
In last test we have externally fragmented the table by inserting the 3 in the Unique Clustered Key value so we are cleaning the table and index by deleting key key value and alter that we will Rebuild the table, So the Index can be de-Fragmented again and we can play further:)
Delete DBO.Fragmented
Where Primarykey = 3
GO
Alter Table DBO.Fragmented Rebuild
GO
So let us first insert a new row in sequence.
SET STATISTICS IO ON
-- transaction log activity
Begin Tran
INSERT INTO [DBO].Fragmented
SELECT
500001,
500001,
500001,
'Some text..',
500001,
500001,
'Some text..',
500001,
500001,
'Some text..'
SELECT [database_transaction_log_bytes_used] FROM sys.dm_tran_database_transactions
WHERE [database_id] = DB_ID ('tutorialsqlserver');
Commit
Look at the output of statistics IO results, to complete this insert requested SQL Server required 3 logical reads and this transaction has generated 344 bytes for the transaction log.
Now let us again insert the row, but this time we will insert a row in between the clustered key so page split can occur.
In the below test we are inserting 3 in the Clustered Index key column.
-- Transaction log activity
Begin Tran
INSERT INTO [DBO].Fragmented
SELECT
3 ,
3,
3
,'Some text..'
,
3,
3
,'Some text..'
,
3,
3
,'Some text..'
SELECT [database_transaction_log_bytes_used] FROM sys.dm_tran_database_transactions
WHERE [database_id] = DB_ID ('tutorialsqlserver');
Commit
Look into the result of Statistics IO and sys.dm_tran_database_transactions DMV , in this transaction page split occurs and it required 12 pages logical read to complete the transaction and 12148 bytes generated for the transaction log, more transaction logs means more sized transaction log database, and all the transaction log database should be backed up/log shipped/mirrored etc to save the database for disaster recovery. Thus more transaction log means more overhead.
In this particular scenario, this transaction is almost 4 times expensive compare to the last transaction. So where ever page split occurs, it takes more time, more IO and generate more logs, hence become more expensive in all terms, so if the page split can be avoided, then it should be avoided in all teams.
Summary
In todays article we learnt what is fragmentation, how fragmentation occurs why it is considered expensive, how SQL Server deals with it. But this topic hasn`t finished yet, we are coming up with the follow up article which will give you more insights about the fragmentation till then I have a question for readers.
Do you know fragmentation doesn`t matter when...
Complete the sentence and win…
What you think about this subject, write in the comment section.
Quote
"That’s been one of my mantras—focus and simplicity. Simple can be harder than complex; you have to work hard to get your thinking clean to make it simple."—Steve Jobs
Page copy protected against web site content infringement by Copyscape
About the Author
Neerajprasadsharma
Full Name: Neeraj Prasad Sharma
Member Level: Bronze
Member Status: Member
Member Since: 5/13/2016 8:42:37 AM
Country: India
Contact for Free SQL Server Performance Consulting and Training for you or your Organization.
Neeraj Prasad Sharma is a SQL Server developer who started his work as a dot net programmer. He loves SQL Server query optimizer`s capability to process the queries optimally. For the last six years he has been experimenting and testing Query Optimizer default behaviour and if something goes wrong his goal is to identify the reason behind it and fix it. I write technical article here: https://www.sqlshack.com/author/neeraj/ https://www.codeproject.com/script/Articles/MemberArticles.aspx?amid=12524731 https://www.mssqltips.com/sqlserverauthor/243/neeraj-prasad-sharma-/
Login to vote for this post.
Comments or Responses
Posted by: NEERAJPRASADSHARMA on: 2/4/2018 | Points: 25
So you are saying more CPU usage means better and faster query Really ?
Login to post response
Comment using Facebook(Author doesn't get notification)
|
__label__pos
| 0.554361 |
Skip to content
Which of the following shows the graph of the inverse of the
Which of the following shows the graph of the inverse of the function below?y = x2 – 4A.B.C.D.============================If f(-4) = 5, which of the following could not be the inverse of f(x)?g(x) = x – 9g(x) = x2 – 11g(x) = x2 – 5x – 4 g(x) = -x2 + 3x + 6==================Which of the following is the inverse of the function below?————————————–Which of the following is the inverse of the function below?y = -3x – 6y = -x/3 – 2y = x/3 + 2y = 3x + 6y = -x/3 + 2===========================Which of the following is the inverse of the function below? y = -2x – 1y = 2x + 1y = x/2 + ½y = x/2 + ½y = -x/2 – ½==========================Which of the following graphs shows the inverse of the graph below?A.B.C.D.============================Which of the following graphs shows the inverse of the graph below?A.B.C.D.============================If f(2) = 10, which of the following could be the inverse of f(x)?g(x) = 10×2 – 2xg(x) = 2×3 – 10xg(x) = x3 + 2g(x) = x2 – 9x – 8==================Find the inverse of the relation {(6, 11), (-2, 7), (0, 3), (-5, 3)}{(6, 11), (-2, 7), (0, 3), (-5, 3)}{(-6, -11), (2, -7), (0, -3), (5, -3)}{(-11,-6),(-7, 2), (-3, 0), (-3, 5)}{(11,6),(7, -2), (3, 0), (3, -5)}==================Find the inverse of the relation {(-1, -2), (-3, -2), (-1, -4), (0, 6)}{(1, 2), (3, 2), (1, 4), (0, -6)}{(-2, -1), (-2, -3), (-4, -1), (6, 0)}{(2, 1), (2, 3), (4, 1), (-6, 0)}{(-1, -2), (-3, -2), (-1, -4), (0, 6)}========================
Do You Need A Similar Assignment Done For You From Scratch? We Have Qualified Writers To Help You. We Assure You An A+ Quality Paper That Is Free From Plagiarism. Order Now For An Amazing Discount!
Very professional support! Highly recommended.
Amos
Researched complicated topic and delivered report by the requested deadline. The paper is of a high standard reflecting careful research and clear assessments. I recommend iLecturers.com.com
Daria K
It’s a 5 start for me. Excellent research and writing. The paper reflects a careful assessment of scientific information.
Mahone
With Gib’s reflection, they wrote a really specific essay. Very well written, containing all of the languages I required and excellent references. The writer follows my instructions and writes clearly in English.
Mario G
I am so grateful and I am 100% happy with your work.
Julie
A fantastic piece of work 👏. The writer demonstrated full knowledge of the topic and applying the relevant provided material. Well done.
Dieny
|
__label__pos
| 0.92863 |
Mathematical Definition
In the above equation, the values $a$, $b$ and $c$ are constants and are usually chosen as $a=20$, $b=0.2$ and $c=2\pi$.
Plots
Ackley Function
Ackley Function
The contour of the function is as presented below:
Ackley Function
Description and Features
• The function is continuous.
• The function is not convex.
• The function is defined on n-dimensional space.
• The function is multimodal.
Input Domain
The function can be defined on any input domain but it is usually evaluated on $x_i \in [-32, 32]$ for all $i = 1,…,n$.
Global Minima
The function has one global minimum at: $f(\textbf{x}^{\ast})=0$ at $\textbf{x}^{\ast} = (0, …, 0)$.
Implementation
An implementation of the Ackley Function with MATLAB is provided below.
% Computes the value of Ackley benchmark function.
% SCORES = ACKLEYFCN(X) computes the value of the Ackey function at point
% X. ACKLEYFCN accepts a matrix of size M-by-N and returns a vetor SCORES
% of size M-by-1 in which each row contains the function value for each row
% of X.
% For more information please visit:
% https://en.wikipedia.org/wiki/Test_functions_for_optimization
%
% Author: Mazhar Ansari Ardeh
% Please forward any comments or bug reports to mazhar.ansari.ardeh at
% Google's e-mail service or feel free to kindly modify the repository.
function scores = ackleyfcn(x)
n = size(x, 2);
ninverse = 1 / n;
sum1 = sum(x .^ 2, 2);
sum2 = sum(cos(2 * pi * x), 2);
scores = 20 + exp(1) - (20 * exp(-0.2 * sqrt( ninverse * sum1))) - exp( ninverse * sum2);
end
The function can be represented in Latex as follows:
f(\textbf{x}) = f(x_1, ..., x_n)= -a.exp(-b\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2})-exp(\frac{1}{n}\sum_{i=1}^{n}cos(cx_i))+ a + exp(1)
See also:
References:
|
__label__pos
| 0.972804 |
User
Resueltoos - cerrar menú Resueltoos - abrir menú
Inicio
>
Blog
>
Matematicas
>
Puntos de corte con los ejes
La plataforma que te asegura el aprobado en matemáticas
Prueba GRATIS ahora
Resueltoos
Puntos de corte con los ejes
¿Qué son los puntos de corte?
Los puntos de corte son los puntos en los que dos funciones o una función y un eje de coordenadas se intersecan. Estos puntos se caracterizan por tener una abscisa (coordenada (x, 0) ) y una ordenada (coordenada (x, 0)) específicas, que corresponden a las coordenadas del punto de intersección (o punto de corte) en el plano cartesiano.
En general, los puntos de corte son importantes para analizar el comportamiento de las funciones, ya que pueden proporcionar información sobre las raíces de la función, las asíntotas y los puntos críticos, y pueden ser útiles para resolver problemas matemáticos y físicos que involucren el estudio de gráficas de funciones en el plano cartesiano. Por ejemplo, en una gráfica de velocidad frente a tiempo, el punto de corte con el eje (x, 0) corresponde al momento en el que el objeto se detiene (velocidad cero).
¿Cuáles son los tipos de puntos de corte?
En general, podemos identificar tres tipos de puntos de corte:
1. Punto de corte con el eje X: Es el punto en el que la gráfica de una función interseca el eje X, siendo la ordenada y igual a cero en dicho punto, es decir, un punto de la forma (x, 0).
2. Punto de corte con el eje Y: Es el punto en el que la gráfica de una función interseca el eje Y, siendo la abscisa X igual a cero en dicho punto, es decir, un punto de la forma (0, y).
3. Punto de corte entre dos funciones: Es el punto en el que dos funciones se intersecan, que es un punto general (x, y). Este punto es común a ambas funciones, luego cumple las ecuaciones de las dos.
¿Cómo afectan los puntos de corte a la gráfica de una función?
Los puntos de corte son muy útiles para dibujar o representar gráficamente una función, pues indican el punto donde la función cruza al eje correspondiente. Por ende, sirven como referencia para saber por dónde pasa la función, y, sabiendo otros puntos notables como máximos, mínimos y puntos de inflexión, se puede determinar de manera muy precisa la forma de la función.
¿Hay una fórmula para calcular los puntos de corte?
No existe una única fórmula para calcular los puntos de corte de diferentes tipos de funciones, ya que el proceso para encontrar estos puntos depende de la forma específica de cada función. Sin embargo, se pueden utilizar diferentes métodos, entre ellos el análisis algebraico (igualar las ecuaciones de las funciones que se intersecan y resolverlas) y el estudio gráfico de la función (determinar los puntos de intersección visualmente).
En cualquier caso, el proceso para encontrar los puntos de corte puede ser más o menos complejo dependiendo de la complejidad de las funciones involucradas y las herramientas matemáticas disponibles para resolverlas.
¿Cómo se calculan los puntos de corte de una función?
Los puntos de corte en el eje X son los valores de X donde la función se corta con el eje X, es decir, donde y = 0. Así, se deben igualar las funciones a cero (f(x) = 0) y obtener los valores de X que lo cumplen, siendo estos los puntos (x, 0) de la función que cortan al eje X.
Por ejemplo, para la función f(x) = x² - 4, se iguala a cero: x² - 4 = 0. Luego se resuelve para encontrar que:
x² - 4 = 0 ⇒ x² = 4 ⇒ x = +-2
Por lo tanto, los puntos de corte de esta función con el eje X son (-2, 0) y (2,; 0). Los puntos de corte en el eje Y son los valores de y donde la función se corta con el eje Y, es decir, donde x=0. Así, se deben sustituir en las funciones el valor de X por cero (f(0)=y) y el valor obtenido de Y es el punto de corte (0, y). Por ejemplo, para la función f(x)= x² - 4, se sustituye x=0:
f(0) = 0² - 4 = -4
Por lo tanto, el punto de corte de esta función con el eje Y se sustituye (0, -4.).
Es importante tener en cuenta que las funciones pueden tener uno o ningún punto de corte con el eje Y, pero no pueden tener varios. Con el eje X pueden tener tantos puntos de corte como se quiera, pues, por ejemplo: una función constante no tiene puntos de corte en el eje X, pero una función sinusoidal (seno o coseno) tiene infinitos puntos de corte con el eje X.
¿Los puntos de corte de una función mantienen la simetría?
Los puntos de corte preservan la simetría, pues si una curva es simétrica con respecto a un eje, los puntos de corte con cualquier eje perpendicular a este mantendrán dicha simetría, conservando también el tipo (par o impar).
Por ejemplo, la parábola f(x) = x²- 4 es simétrica par con respecto al eje Y, luego sus puntos de corte con el eje X (los puntos (2, 0) y (-2, 0) son también simétricos con respecto al eje Y.
¿Cómo se calculan los puntos de corte entre dos funciones?
Para encontrar los puntos de corte entre dos funciones se deben igualar las dos funciones entre sí (f(x) = g(x)) y resolver para encontrar los valores de X que lo cumplen. Estos valores de X deben sustituirse en cualquiera de las dos funciones para obtener los valores de Y correspondientes a los puntos de corte (x, y) de las dos funciones. Al final, hallar los puntos de corte de dos funciones equivale a resolver la ecuación que se obtiene de igualarlas. Así, si las funciones son muy complicadas puede ser necesario utilizar métodos numéricos o aproximaciones gráficas para encontrar los puntos de corte.
Cabe destacar que los valores de Y obtenidos serán los mismo independientemente de en qué función se sustituya el valor de X, pues los puntos de corte son comunes a ambas funciones.
Es importante tener en cuenta también que las dos funciones pueden no se cortarse en ningún punto, en cuyo caso no existirán puntos de corte (por ejemplo, las funciones f(x) = x² ∙ y g(x) = -x² - 1). También es posible que haya más de un punto de corte, dependiendo de la forma de las funciones.
¿Cuándo dos funciones se intersecan en un solo punto?
Para que dos funciones puedan intersecarse en un solo punto, sus rectas tangentes en dicho punto tendrán que ser distintas. Si no, existirá un pequeño intervalo en el que las funciones se superpongan y no haya un único punto de corte.
¿Cuál es la relación entre resolver una ecuación y hallar los puntos de corte de una función?
Si se considera una función f(x), esta puede expresarse como una ecuación algebraica y=f(x), luego hallar los puntos de corte de una función con el eje X es equivalente a resolver la ecuación asociada para y=f(x) = 0. De esta forma, resolver una ecuación 0 = f(x) o calcular los puntos de corte con el eje X de la función f(x) es equivalente.
Por ejemplo, en el caso de dos funciones que se cortan, el punto de corte puede interpretarse en términos de las soluciones de un sistema de ecuaciones o la intersección de dos curvas en un plano. En el caso de una función y un eje de coordenadas, el punto de corte puede indicar el valor de la función al pasar por el eje (el corte de una curva con una recta), o el valor de la variable correspondiente en dicho eje cuando la función pasa por él.
Para seguir aprendiendo:
1- La inteligencia artificial ya esta aquí y se llama chat GPT
2- ¿Quieres ser una gran científica? Estamos a tu lado
3- Llévate el éxito
4- Recorrido de una función
5- Puntos de corte de otras funciones.
La plataforma que te asegura el aprobado en matemáticas
Prueba GRATIS ahora
Resueltoos
Articulos relacionados
Resueltoos - blog
Dominio de una función
01-06-2023
Resueltoos - blog
Recorrido de una función
02-03-2023
Resueltoos - blog
Puntos de corte de funciones racionales, logarítmicas, trigonométricas
02-03-2023
Resueltoos - blog
Vector unitario
02-03-2023
|
__label__pos
| 0.940021 |
HOME MY TEACHING TOOLS FORUM ABOUT US CONTACT US
Speaking Activities
Self-Study Quizzes
SpokenSkills Lab
Job Market
Home > Self-Study Quizzes > Quiz
GMAT Math practice - (5/36 questions - random)
GMAT practice questions. Once you complete the quiz you will see your score and explanation of the correct answer.
1. Which of the following inequalities have a finite range of values of "x" satisfying them?
2. The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m+2?
3. Set A contains all the even numbers between 2 and 50 inclusive. Set B contains all the even numbers between 102 and 150 inclusive. What is the difference between the sum of elements of set B and that of set A?
4. The average weight of a group of 30 friends increases by 1 kg when the weight of their football coach was added. If average weight of the group after including the weight of the football coach is 31kgs, what is the weight of their football coach in kgs?
5. How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?
Share on Facebook Share on Twitter
STUDENT ACTIVITIES
Speaking Activities
Self-Study Quizzes
WEBTOOLS
SpokenTest
EnglishPresentations
CUSTOMER SERVICE
Privacy Policy
Support
COMPANY INFORMATION
About Us
Contact Us
SpokenSkills News
Vernon, British Columbia, Canada Telephone: 250-307-0541
©SpokenSkills. All Rights Reserved.
|
__label__pos
| 1 |
answersLogoWhite
0
Best Answer
It is: -16
User Avatar
Wiki User
2014-05-19 12:38:00
This answer is:
🙏
0
🤨
0
😮
0
User Avatar
Study guides
Algebra
20 cards
A polynomial of degree zero is a constant term
The grouping method of factoring can still be used when only some of the terms share a common factor A True B False
The sum or difference of p and q is the of the x-term in the trinomial
A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
J's study guide
1 card
What is the name of Steve on minecraft's name
➡️
See all cards
Steel Tip Darts Out Chart
96 cards
170
169
168
167
➡️
See all cards
Add your answer:
Earn +20 pts
Q: What is the answer to this 8 times minus 2?
Write your answer...
Submit
Related questions
What is 8 minus 3 times 2?
2
4 6-2 plus 8 2 -2?
The solution to 4 times 6 minus 2 plus 8 times 2 minus 2 is 36. Archimedes is known as the father of mathematics.
What is 3 minus 7 times a negative 2 minus 8?
The answer is nine.
What times 8 minus 1 equal 2?
8 times -1/8=-8/8, and -8/8 is -1, -1-1=2
What is 20 minus 8 times 2?
20 - (8 x 2) = 4
What is 10 minus 4 times 2?
Ans: 2 10 minus 4 times 2 = 10 - (4*2) = 10-8 = 2
What is 8 to the power of 2 minus 2 times 8 plus 3?
It is: 82-(2*8)+3 = 51
What is 4 multiply minus 2?
2
How much is 2 times 8 minus 7?
9
What is minus 8 minus minus 2?
-8 - -2 = -6
What is 10 to the second power minus 4 times 8 divided by 8 plus 9?
10^2 minus 4 times 8 divided by 8 plus 9 is equal to 105.
What is the answer to 4 times 2 minus 3 divided by 3?
4 times 2 minus 3 divided by 3 is equal to 8 minus 3 divided by 3 which is equal to 5 divided by 3 which gives you 1 and 2 thirds.
What is 6 times 5 to second power minus 5 times 4 plus 8?
Six times 5^2 minus 5 times 4 plus 8 is equal to 138.
What is two minus eight divided by two times four minus one?
(2 - 8) / (2 x 4 - 1) = -6 / (8 - 1) = -6/7
What times what minus what equals 30?
4 x 8 - 2
What is 7 times 8 plus 6 minus 2?
60
What is 5 times 3 minus 8 divided by 2 times 3?
1.6 repeating
9 squared minus 2 times 15 plus 16 minus 8?
Without any parenthesis is: 92 - 2 * 15 + 16 - 8 = 59
What is Negative 8 times 2?
-8 x -8 = +16anything which is minus x minus = plus (- x - = +)
8 minus 13 minus 8 minus 2 multiplied by 2 equals?
-18
What is minus 8 times minus 7 minus 14?
(-8) x (-7) - 14 = 42
What is 2 times 12 divided by 24 times 8 minus 1 times 7 dived by 49?
1
What is the value of minus 6 minus 3 minus 4 divided by minus 8 multiply by 2?
- 6 - 3 - 4 ÷ - 8 x 2 = -8
What is 10 minus 5 times 2 plus 8 divided by 2?
10-5=5x2=10+8=18/2=9
What does minus 6 minus minus 8 equals?
-6 - (-8) = 2
|
__label__pos
| 1 |
Posts Tagged ‘Script’
As you may be knowing, you can use Windows PowerShell to change registry values. In this article I am going to do five things.
I have created few registry entries to use in this example as seen below. In real world you can use whatever entries in your registry. It is always advisable to backup your registry before changing it.
clip_image002[13]
1. Set a registry key value.
To set a value you need to use the “Set-ItemProperty” cmdlet as below.
Set-ItemProperty -Path "HKLM:\Software\Test\Live" -Name "TestValue2" –Value “TestData2”
Above command will put “TestData2” in the registry key “TestValue2” located in HKEY_LOCAL_MACHINE\Software\Test\Live.
2. Read a registry key value.
Reading from the registry can be done by using the cmdlet “Get-ItemProperty”.
Below command will get the value in the “TestValue1” key.
Get-ItemProperty -Path "HKLM:\Software\Test\Live" -Name "TestValue1"
3. Using variables in PowerShell.
Here I am going to read a registry key value and put it to another registry key. This can be done using a variable. First you need to read the value into a variable using the “Get-ItemProperty” cmdlet and that value can be saved using the “Set-ItemProperty” cmdlet.
1. # Check for the existance of the registry key.
2. IF (Get-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “TestValue1” -ea 0)
3. {
4. # Fetching the value from TestValue1.
5. $OldValue = Get-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “TestValue1”
6. }
7. ELSE
8. {
9. # Inserting a blank, if the registry key is not present.
10. $OldValue = “”
11. }
12. # Printing the value in the variable.
13. Write-Host $OldValue.TestValue1
14. # Setting the value to TestValue2.
15. Set-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “TestValue2” -Value $OldValue.TestValue1
4. Working with registry keys with spaces.
In case your registry keys contain spaces, you need to use double quotes in your script as seen below.
1. # Check for the existance of the registry key.
2. IF (Get-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “Test Value 1” -ea 0)
3. {
4. # Fetching the value from Test Value 1.
5. $OldValue = Get-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “Test Value 1”
6. }
7. ELSE
8. {
9. # Inserting a blank, if the registry key is not present.
10. $OldValue = “”
11. }
12. # Printing the value in the variable.
13. Write-Host $OldValue.“Test Value 1”
14. # Setting the value to Test Value 2.
15. Set-ItemProperty -Path “HKLM:\Software\Test\Live” -Name “Test Value 2” -Value $OldValue.“Test Value 1”
5. Saving PowerShell commands as scripts and running them.
Both above can be saved as a PowerShell script by saving it in a file with the extension ps1. For example I did save it as “ChangeReg.ps1” in my C drive inside the folder “new”. Then the script can be run by browsing to the folder and using the command “.\ChangeReg.ps1”.
clip_image002[10]
After the script is run my registry keys looked like this.
clip_image002[3]
In case you need to retrieve values from other registry hives (locations), following table may be helpful.
Registry Hive
Abbreviation
1. HKEY_CLASSES_ROOT HKCR
2. HKEY_CURRENT-USER HKCU
3. HKEY_LOCAL_MACHINE HKLM
4. HKEY_USERS HKU
5. HKEY_CURRENT_CONFIG HKCC
In case you need to read more on “Get-ItemProperty” and “Set-ItemProperty”, use the links to visit official documentation from Microsoft TechNet.
|
__label__pos
| 0.921743 |
[December-2020]Free 350-401 Exam PDF and VCE Offered by Braindump2go[Q91-Q104]
2020/December Latest Braindump2go 350-401 Exam Dumps with PDF and VCE Free Updated Today! Following are some new 350-401 Real Exam Questions!
QUESTION 91
Which access controls list allows only TCP traffic with a destination port range of 22-433, excluding port 80?
A. Deny tcp any any eq 80
Permit tco any any gt 21 it 444
B. Permit tcp any any ne 80
C. Permit tco any any range 22 443
Deny tcp any any eq 80
D. Deny tcp any any ne 80
Permit tcp any any range 22 443
Answer: A
QUESTION 92
Which feature does Cisco TrustSec use to provide scalable, secure communication throughout a network?
A. security group tag ACL assigned to each port on a switch
B. security group tag number assigned to each port on a network
C. security group tag number assigned to each user on a switch
D. security group tag ACL assigned to each router on a network
Answer: B
Explanation:
Cisco TrustSec uses tags to represent logical group privilege. This tag, called a Security Group Tag (SGT), is used in access policies. The SGT is understood and is used to enforce traffic by Cisco switches, routers and firewalls . Cisco TrustSec is defined in three phases: classification, propagation and enforcement.
When users and devices connect to a network, the network assigns a specific security group. This process is called classification. Classification can be based on the results of the authentication or by associating the SGT with an IP, VLAN, or port-profile (-> Answer A and answer C are not correct as they say “assigned … on a switch” only. Answer D is not correct either as it says “assigned to each router”).
QUESTION 93
Which action is the vSmart controller responsible for in an SD-WAN deployment?
A. onboard vEdge nodes into the SD-WAN fabric
B. distribute security information for tunnel establishment between vEdge routers
C. manage, maintain, and gather configuration and status for nodes within the SD-WAN fabric
D. gather telemetry data from vEdge routers
Answer: A
Explanation:
The major components of the vSmart controller are:
+ Control plane connections – Each vSmart controller establishes and maintains a control plane connection with each vEdge router in the overlay network. (In a network with multiple vSmart controllers, a single vSmart controller may have connections only to a subset of the vEdge routers, for load-balancing purposes.) Each connection, which runs as a DTLS tunnel, is established after device authentication succeeds, and it carries the encrypted payload between the vSmart controller and the vEdge router. This payload consists of route information necessary for the vSmart controller to determine the network topology, and then to calculate the best routes to network destinations and distribute this route information to the vEdge routers. The DTLS connection between a vSmart controller and a vEdge router is a permanent connection. The vSmart controller has no direct peering relationships with any devices that a vEdge router is connected to on the service side (so answer C is not correct as vSmart only manages vEdge routers only, not the whole nodes within SD-WAN fabric).
+ OMP (Overlay Management Protocol) – The OMP protocol is a routing protocol similar to BGP that manages the Cisco SD-WAN overlay network. OMP runs inside DTLS control plane connections and carries the routes, next hops, keys, and policy information needed to establish and maintain the overlay network. OMP runs between the vSmart controller and the vEdge routers and carries only control plane information. The vSmart controller processes the routes and advertises reachability information learned from these routes to other vEdge routers in the overlay network.
+ Authentication – The vSmart controller has pre-installed credentials that allow it to authenticate every new vEdge router that comes online (-> Answer A is correct). These credentials ensure that only authenticated devices are allowed access to the network.
+ Key reflection and rekeying – The vSmart controller receives data plane keys from a vEdge router and reflects them to other relevant vEdge routers that need to send data plane traffic.
+ Policy engine – The vSmart controller provides rich inbound and outbound policy constructs to manipulate routing information, access control, segmentation, extranets, and other network needs.
+ Netconf and CLI – Netconf is a standards-based protocol used by the vManage NMS to provision a vSmart controller. In addition, each vSmart controller provides local CLI access and AAA.
Reference: https://www.cisco.com/c/en/us/td/docs/routers/sdwan/configuration/sdwan-xe-gs-book/system-overview.html
QUESTION 94
Refer to the exhibit. Link1 is a copper connection and Link2 is a fiber connection The fiber port must be the primary port for all forwarding. The output of the show spanning-tree command on SW2 shows that the fiber port is blocked by spanning tree. An engineer enters the spanning- tree port-priority 32 command on GO/1 on SW2. but the port remains blocked.
Which command should be entered on the ports that are connected to Lmk2 to resolve the issue?
A. Enter spanning-tree port-priority 32 on SW1.
B. Enter spanning-tree port-priority 224 on SW1.
C. Enter spanning-tree port-priority 4 on SW2.
D. Enter spanning-tree port-priority 64 on SW2.
Answer: A
Explanation:
SW1 needs to block one of its ports to SW2 to avoid a bridging loop between the two switches. Unfortunately, it blocked the fiber port Link2. But how does SW2 select its blocked port? Well, the answer is based on the BPDUs it receives from SW1. A BPDU is superior than another if it has:
1. A lower Root Bridge ID
2. A lower path cost to the Root
3. A lower Sending Bridge ID
4. A lower Sending Port ID
These four parameters are examined in order. In this specific case, all the BPDUs sent by SW1 have the same Root Bridge ID, the same path cost to the Root and the same Sending Bridge ID. The only parameter left to select the best one is the Sending Port ID (Port ID = port priority + port index). And the port index of Gi0/0 is lower than the port index of Gi0/1 so Link 1 has been chosen as the primary link.
Therefore we must change the port priority to change the primary link. The lower numerical value of port priority, the higher priority that port has. In other words, we must change the port-priority on Gi0/1 of SW1 (not on Gi0/1 of SW2) to a lower value than that of Gi0/0.
QUESTION 95
Which requirement for an Ansible-managed node is true?
A. It must be a Linux server or a Cisco device
B. It must have an SSH server running
C. It must support ad hoc commands.
D. It must have an Ansible Tower installed
Answer: A
Explanation:
Ansible can communicate with modern Cisco devices via SSH or HTTPS so it does not require an SSH server -> Answer B is not correct.
An Ansible ad-hoc command uses the /usr/bin/ansible command-line tool to automate a single task on one or more managed nodes. Ad-hoc commands are quick and easy, but they are not reusable -> It is not a requirement either -> Answer C is not correct.
Ansible Tower is a web-based solution that makes Ansible even more easy to use for IT teams of all kinds. But it is not a requirement to run Ansible -> Answer D is not correct.
Therefore only answer A is the best choice left. An Ansible controller (the main component that manages the nodes), is supported on multiple flavors of Linux, but it cannot be installed on Windows.
QUESTION 96
Refer to this output. What is the logging severity level?
R1#Feb 14 37:15:12:429: %LINEPROTO-5-UPDOWN Line protocol on interface GigabitEthernet0/1. Change state to up
A. Notification
B. Alert
C. Critical
D. Emergency
Answer: A
Explanation:
Syslog levels are listed below:
Number “5” in “%LINEPROTO-5- UPDOWN” is the severity level of this message so in this case it is “notification”.
QUESTION 97
Which DNS lookup does an access point perform when attempting CAPWAP discovery?
A. CISCO-DNA-CONTROlLLER.local
B. CAPWAP-CONTROLLER.local
C. CISCO-CONTROLLER.local
D. CISCO-CAPWAP-CONTROLLER.local
Answer: D
Explanation:
The Lightweight AP (LAP) can discover controllers through your domain name server (DNS). For the access point (AP) to do so, you must configure your DNS to return controller IP addresses in response to CISCO-LWAPP-CONTROLLER.localdomain, where localdomain is the AP domain name. When an AP receives an IP address and DNS information from a DHCP server, it contacts the DNS to resolve CISCO-CAPWAP-CONTROLLER.localdomain. When the DNS sends a list of controller IP addresses, the AP sends discovery requests to the controllers.
The AP will attempt to resolve the DNS name CISCO-CAPWAP-CONTROLLER.localdomain. When the AP is able to resolve this name to one or more IP addresses, the AP sends a unicast CAPWAP Discovery Message to the resolved IP address(es). Each WLC that receives the CAPWAP Discovery Request Message replies with a unicast CAPWAP Discovery Response to the AP.
Reference: https://www.cisco.com/c/en/us/support/docs/wireless/4400-series-wireless-lan-controllers/107606-dns-wlc-config.html
QUESTION 98
At which Layer does Cisco DNA Center support REST controls?
A. EEM applets or scripts
B. Session layer
C. YMAL output from responses to API calls
D. Northbound APIs
Answer: D
QUESTION 99
Which two statements about IP SLA are true? (Choose two)
A. SNMP access is not supported
B. It uses active traffic monitoring
C. It is Layer 2 transport-independent
D. The IP SLA responder is a component in the source Cisco device
E. It can measure MOS
F. It uses NetFlow for passive traffic monitoring
Answer: BC
Explanation:
IP SLAs allows Cisco customers to analyze IP service levels for IP applications and services, to increase productivity, to lower operational costs, and to reduce the frequency of network outages. IP SLAs uses active traffic monitoring–the generation of traffic in a continuous, reliable, and predictable manner–for measuring network performance.
Being Layer-2 transport independent, IP SLAs can be configured end-to-end over disparate networks to best reflect the metrics that an end-user is likely to experience.
Reference: https://www.cisco.com/c/en/us/td/docs/ios-xml/ios/ipsla/configuration/15-mt/sla-15-mt-book/sla_overview.html
QUESTION 100
Which two statements about Cisco Express Forwarding load balancing are true?
A. Cisco Express Forwarding can load-balance over a maximum of two destinations
B. It combines the source IP address subnet mask to create a hash for each destination
C. Each hash maps directly to a single entry in the RIB
D. Each hash maps directly to a single entry in the adjacency table
E. It combines the source and destination IP addresses to create a hash for each destination
Answer: DE
Explanation:
Cisco IOS software basically supports two modes of CEF load balancing: On per-destination or per-packet basis.
For per destination load balancing a hash is computed out of the source and destination IP address (-> Answer E is correct). This hash points to exactly one of the adjacency entries in the adjacency table (-> Answer D is correct), providing that the same path is used for all packets with this source/destination address pair. If per packet load balancing is used the packets are distributed round robin over the available paths. In either case the information in the FIB and adjacency tables provide all the necessary forwarding information, just like for non-load balancing operation.
The number of paths used is limited by the number of entries the routing protocol puts in the routing table, the default in IOS is 4 entries for most IP routing protocols with the exception of BGP, where it is one entry. The maximum number that can be configured is 6 different paths -> Answer A is not correct.
Reference: https://www.cisco.com/en/US/products/hw/modules/ps2033/prod_technical_reference09186a00800afeb7.html
QUESTION 101
What is the main function of VRF-lite?
A. To allow devices to use labels to make Layer 2 Path decisions
B. To segregate multiple routing tables on a single device
C. To connect different autonomous systems together to share routes
D. To route IPv6 traffic across an IPv4 backbone
Answer: B
QUESTION 102
Which two steps are required for a complete Cisco DNA Center upgrade? (Choose two.)
A. golden image selection
B. automation backup
C. proxy configuration
D. application updates
E. system update
Answer: DE
QUESTION 103
Based on this interface configuration, what is the expected state of OSPF adjacency?
A. Full on both routers
B. not established
C. 2WAY/DROTHER on both routers
D. FULL/BDR on R1 and FULL/BDR on R2
Answer: B
Explanation:
On Ethernet interfaces the OSPF hello intervl is 10 second by default so in this case there would be a Hello interval mismatch -> the OSPF adjacency would not be established.
QUESTION 104
Which statement about TLS is true when using RESTCONF to write configurations on network devices?
A. It is provided using NGINX acting as a proxy web server.
B. It is no supported on Cisco devices.
C. It required certificates for authentication.
D. It is used for HTTP and HTTPs requests.
Answer: C
Explanation:
The https-based protocol-RESTCONF (RFC 8040), which is a stateless protocol, uses secure HTTP methods to provide CREATE, READ, UPDATE and DELETE (CRUD) operations on a conceptual datastore containing YANG-defined data -> RESTCONF only uses HTTPs.
RESTCONF servers MUST present an X.509v3-based certificate when establishing a TLS connection with a RESTCONF client. The use of X.509v3-based certificates is consistent with NETCONF over TLS -> Answer C is correct.
Reference: https://tools.ietf.org/html/rfc8040
QUESTION 105
Which controller is the single plane of management for Cisco SD-WAN?
A. vBond
B. vEdge
C. vSmart
D. vManange
Answer: D
Explanation:
The primary components for the Cisco SD-WAN solution consist of the vManage network management system (management plane), the vSmart controller (control plane), the vBond orchestrator (orchestration plane), and the vEdge router (data plane).
+ vManage – This centralized network management system provides a GUI interface to easily monitor, configure, and maintain all Cisco SD-WAN devices and links in the underlay and overlay network.
+ vSmart controller – This software-based component is responsible for the centralized control plane of the SD-WAN network. It establishes a secure connection to each vEdge router and distributes routes and policy information via the Overlay Management Protocol (OMP), acting as a route reflector. It also orchestrates the secure data plane connectivity between the vEdge routers by distributing crypto key information, allowing for a very scalable, IKE-less architecture.
+ vBond orchestrator – This software-based component performs the initial authentication of vEdge devices and orchestrates vSmart and vEdge connectivity. It also has an important role in enabling the communication of devices that sit behind Network Address Translation (NAT).
+ vEdge router – This device, available as either a hardware appliance or software-based router, sits at a physical site or in the cloud and provides secure data plane connectivity among the sites over one or more WAN transports. It is responsible for traffic forwarding, security, encryption, Quality of Service (QoS), routing protocols such as Border Gateway Protocol (BGP) and Open Shortest Path First (OSPF), and more.
Reference: https://www.cisco.com/c/dam/en/us/td/docs/solutions/CVD/SDWAN/CVD-SD-WAN-Design-2018OCT.pdf
Resources From:
1.2020 Latest Braindump2go 350-401 Exam Dumps (PDF & VCE) Free Share:
https://www.braindump2go.com/350-401.html
2.2020 Latest Braindump2go 350-401 PDF and 350-401 VCE Dumps Free Share:
https://drive.google.com/drive/folders/1EIsykNTrKvqjDVs9JMySv052qbrCpe8V?usp=sharing
3.2020 Free Braindump2go 350-401 PDF Download:
https://www.braindump2go.com/free-online-pdf/350-401-Dumps(131-145).pdf
https://www.braindump2go.com/free-online-pdf/350-401-PDF(117-130).pdf
https://www.braindump2go.com/free-online-pdf/350-401-PDF-Dumps(91-103).pdf
https://www.braindump2go.com/free-online-pdf/350-401-VCE(104-116).pdf
https://www.braindump2go.com/free-online-pdf/350-401-VCE-Dumps(146-160).pdf
Free Resources from Braindump2go,We Devoted to Helping You 100% Pass All Exams!
Comments are closed.
|
__label__pos
| 0.890366 |
The dining philosophers problem was invented by E.W. Dijkstra, a concurrency pioneer, to clarify...
The dining philosophers problem was invented by E.W. Dijkstra, a concurrency pioneer, to clarify the notions of deadlock- and starvation-freedom. Imagine five philosophers who spend their lives just thinking and feasting on rice. They sit around a circular table. However, there are only five chopsticks (forks, in the original formulation). Each philosopher thinks. When he gets hungry, he picks up the two chopsticks closest to him. If he can pick up both chopsticks, he can eat for a while. After a philosopher finishes eating, he puts down the chopsticks and again starts to think. 1. Write a program to simulate the behavior of the philosophers, where each philosopher is a thread and the chopsticks are shared objects. Note that you must prevent a situation where two philosophers hold the same chopstick at the same time. 2. Amend your program so that it never reaches a state where philosophers are deadlocked, that is, it is never the case that every philosopher holds one chopstick and is stuck waiting for another to get the second chopstick. 3. Amend your program so that no philosopher ever starves. 4. Write a program to provide a starvation-free solution for n philosophers for any natural number n.
Oct 15 2021| 08:15 AM |
Earl Stokes
Earl Stokes Verified Expert
9 Votes
8464 Answers
This is a sample answer. Please purchase a subscription to get our verified Expert's Answer. Our rich database has textbook solutions for every discipline. Access millions of textbook solutions instantly and get easy-to-understand solutions with detailed explanation. You can cancel anytime!
Related Questions
c++ program: This problem needs to be solved by using recursion: Write a recursive function...
c++ program: This problem needs to be solved by using recursion: Write a recursive function PrintPattern1 in the form void PrintPattern1(int n, int k, char ch1, char ch2); to print pattern that takes two integer arguments n and k. n is the starting number while k is the ending limit and ch1 is the first symbol and ch2 is the seacond symbol Example: void PrintPattern1(int n, int k, char ch1, char...
Oct 15 2021
Prove that the following 3 statements are all EQUIVALENT. 1)x is a rational number 2) x/3 is a...
Prove that the following 3 statements are all EQUIVALENT. 1)x is a rational number 2) x/3 is a rational number 3) 4x + 1 is a rational number.
Oct 15 2021
Given the following JS code snippet, rewrite the entire statement using a template literal. You...
Given the following JS code snippet, rewrite the entire statement using a template literal. You can assume that all the variables used in the concatenated expression have been declared and are ready to use. You must include a closing semicolon. const fulldate = dayName + ", " + d.getDate() + ", "+ monthName + " " + year; (update: That is the entire snippet given)
Oct 15 2021
Create a simple python program for the following inheritance concept. • Create a Vehicle class...
Create a simple python program for the following inheritance concept. • Create a Vehicle class without any variables and methods. • Create a Vehicle class with 3 instance attributes as name, max_speed and mileage. • Create child class Car that will inherit all of the variables and methods of the Vehicle class. • Create a Car object that will inherit all of the variables and methods of the Vehicle...
Oct 15 2021
What is the single greatest advantage of having a checksum cover only the datagram header and not...
What is the single greatest advantage of having a checksum cover only the datagram header and not the payload? What is the disadvantage? Do you expect a high-speed local area network to have larger or smaller MTU than a wide area network? Why?
Oct 14 2021
Join Quesbinge Community
5 Million Students and Industry experts
• Need Career counselling?
• Have doubts with course subjects?
• Need any last minute study tips?
Interact with your Peers for FREE!
|
__label__pos
| 0.665466 |
latest
Configuration Reference
DISTINCTCOUNTTHETASKETCH
This section contains reference documentation for the DISTINCTCOUNTTHETASKETCH function.
The Theta Sketch framework enables set operations over a stream of data, and can also be used for cardinality estimation. Pinot leverages the Sketch Class and its extensions from the library org.apache.datasketches:datasketches-java:1.2.0-incubating to perform distinct counting as well as evaluating set operations.
Signature
DistinctCountThetaSketch(<thetaSketchColumn>, <thetaSketchParams>, predicate1, predicate2..., postAggregationExpressionToEvaluate) -> Long
• thetaSketchColumn (required): Name of the column to aggregate on.
• thetaSketchParams (required): Parameters for constructing the intermediate theta-sketches.
• Currently, the only supported parameter is nominalEntries (defaults to 4096).
• predicates (optional)_: _ These are individual predicates of form lhs <op> rhs which are applied on rows selected by the where clause. During intermediate sketch aggregation, sketches from the thetaSketchColumn that satisfies these predicates are unionized individually. For example, all filtered rows that match country=USA are unionized into a single sketch. Complex predicates that are created by combining (AND/OR) of individual predicates is supported.
• postAggregationExpressionToEvaluate (required): The set operation to perform on the individual intermediate sketches for each of the predicates. Currently supported operations are SET_DIFF, SET_UNION, SET_INTERSECT , where DIFF requires two arguments and the UNION/INTERSECT allow more than two arguments.
Usage Examples
These examples are based on the Batch Quick Start.
1
select distinctCountThetaSketch(teamID) AS value
2
from baseballStats
Copied!
value
149
1
select distinctCountThetaSketch(teamID, 'nominalEntries=10') AS value
2
from baseballStats
Copied!
value
146
We can also provide predicates and a post aggregation expression to compute more complicated cardinalities. For example, we could can find the intersection of the following queries:
1
select yearID
2
from baseballStats
3
where teamID = 'SFN' AND numberOfGames = 28 AND homeRuns = 1
Copied!
yearID
1986
1985
1
select yearID
2
from baseballStats
3
where teamID = 'CHN' AND numberOfGames = 28 AND homeRuns = 1
Copied!
yearID
1937
2003
1979
1900
1986
1978
2012
(the yearId 1986 is the only one in common)
By running the following query:
1
select distinctCountThetaSketch(
2
yearID,
3
'nominalEntries=4096',
4
'teamID = ''SFN'' AND numberOfGames=28 AND homeRuns=1',
5
'teamID = ''CHN'' AND numberOfGames=28 AND homeRuns=1',
6
'SET_INTERSECT($1, $2)'
7
) AS value
8
from baseballStats
Copied!
value
1
Copy link
|
__label__pos
| 0.842088 |
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!
Homework Help: Integral of 1/(x*ln(x)) converging or diverging?
1. May 4, 2006 #1
In order to find the integral of 1/(x*ln(x)) dx, I tried using the substitution method where
u = 1/x and dv = 1/ln(x) dx .
Then du = ln(x) dx.
However this is where I got stuck.
What would v equal? Or is their another way I should be approaching this integral in order to find if it is diverging or converging and if converging, to what value?
2. jcsd
3. May 4, 2006 #2
siddharth
User Avatar
Homework Helper
Gold Member
First of all, is it a definite integral? What are the limits of integration?
Then, if u=1/x then du is not ln(x) dx. Why don't you try the substitution u=ln(x) and see what happens?
4. May 4, 2006 #3
My mistake and sorry for not mentioning the limits. Its an improper integral with lower limit 2 and upper limit infinity.
I think I found the solution...
u=ln(x)
du=(1/x)
int(1/u) du
And then solve for the integral, with the lower limit being ln(2) and the upper limit being infinity.
It ends up diverging, correct?
5. May 5, 2006 #4
siddharth
User Avatar
Homework Helper
Gold Member
Yes, I think you're right.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
|
__label__pos
| 0.999503 |
Add Layers (Google Maps) - Samsung DROID Charge SCH-I510
Discussion in 'Droid Charge FAQ' started by danDroid, Jun 9, 2011.
1. danDroid
danDroid DF Administrator Staff Member
Joined:
Oct 20, 2009
Messages:
2,404
Likes Received:
154
Trophy Points:
128
Ratings:
+157
1. From the Home screen, select Applications.
[IMG]
2. Select Maps.
[IMG]
3. Select Layers.
[IMG]
4. Select the layer. Traffic Highways will be displayed in green, yellow or red based on real-time traffic data. Satellite Maps will be displayed from a satellite view. Latitude Enables Latitude. Map will show GPS locations. [IMG]
5. If desired, select More layers.
[IMG]
6. Select the layer. My Maps Displays custom maps from the Google account Wikipedia Displays location links from Wikipedia. Transit Lines Displays subway, bus and train schedules. [IMG]
Search tags for this page
add layer to google navigation
,
adding layers to google navigation
,
android google maps custom layers
,
android how to add navigation layers
,
google maps android layers
,
google navigation how to add layers
,
how to add layer in android google map
,
how to insert layer in google earth android
,
how to show layers in google maps android
,
samsung s4 google maps layers
|
__label__pos
| 0.889711 |
Question
(a) Determine the critical value for a right-tailed test of a population mean at the α = 0.01 level of significance with 15 degrees of freedom.
(b) Determine the critical value for a left-tailed test of a population mean at the α = 0.05 level of significance based on a sample size of n = 20.
(c) Determine the critical values for a two-tailed test of a population mean at the α = 0.05 level of significance based on a sample size of n = 13.
Sales0
Views193
Comments0
• CreatedApril 28, 2015
• Files Included
Post your question
5000
|
__label__pos
| 0.667779 |
Add event handler (ExecutionListener or TaskListener) at runtime
According to Camunda’s doc (https://docs.camunda.org/manual/latest/user-guide/process-applications/process-application-event-listeners/) a “global” event handler (ExecutionListener or TaskListener) can be added to the ProcessApplication.
Nonetheless, I have not been able to find a way to add a similar (“global”) event handler at runtime. This feature was present in Activiti using the method addEventListener of the engine’s RuntimeService (https://www.activiti.org/javadocs/org/activiti/engine/RuntimeService.html#addEventListener-org.activiti.engine.delegate.event.ActivitiEventListener-) but is no longer present in Camunda’s RuntimeService.
How could I add a “global” event handler at runtime?
Note: The ProcessApplication to which the event handlers will be added can not be modified since I want to add the handlers from a different library.
Thank you all,
So, you want to add global listeners by simply adding libraries to the classpath?
Assuming that your listeners can implement a custom marker interface like e.g. GlobalTaskListener and that Spring is available, you could enhance your ProcessApplication to have two autowired fields that collect all global task/execution listeners and return these lists in the respective methods.
Spring will then then collect and inject the listeners during startup.
@Autowired
List<GlobalTaskListeners> taskListeners;
public TaskListener getTaskListener() {
return taskListeners;
}
Thanks for your answer.
But how do I “enhance” my process application? The way I would like to add the listeners is via an annotation for example, this means that I cannot modify the process application.
Is my reasoning correct?
You mentioned the ProcessApplication in your original post. That gave me the impression that you already created your own ProcessApplication subclass by extending one of the provided ones, just like it is shown in the docs you linked.
By enhancing I meant to add the mentioned fields and methods to your ProcessApplication class.
For sake of clarity, my original idea in more detail:
@ProcessApplication
public class MyProcessApplication extends ServletProcessApplication {
@Autowired
List<GlobalTaskListeners> taskListeners;
public TaskListener getTaskListener() {
return taskListeners;
}
}
Please note that the above code is just a concept and is not guaranteed to work as-is.
Oh, I see. In fact I have no access whatsoever to the ProcessApplication.
More generally, would you happen to know any way in Camunda to handle any event (e.g. transitions, task achieved, etc.) and that can be added in runtime (or at least in any way to a third-party / deployed application)?
Hi @underscorenico,
There is no such feature. I also don’t think an API similar to Activiti’s API would be a good idea. In a shared engine scenario, this can easily result in memory leaks when you register a listener that is an instance of a class that belongs to a web application. When the web application is undeployed, the listener is still registered with the engine.
Please expand on your use case, i.e. why you can’t have this code as part of a process application.
Cheers,
Thorben
Does the camunda-bpm-reactor-spring-starter extension handle just this sort of thing> It allows you to tap into all sorts of life-cycle events using a set of selectors. For example I can listen to all user task creation events in my application like so:
@Component
@CamundaSelector(type = "userTask", event = TaskListener.EVENTNAME_CREATE)
public class UserTaskListener extends ReactorTaskListener {
...
}
You can find it here: https://github.com/camunda/camunda-bpm-reactor
|
__label__pos
| 0.558072 |
Cannot extend a vec of trait objects
I have a Vec containing boxed trait objects which I’d like to extend by the result of an iterator. Performing an loop-and-push approach works fine but using the extend-method fails with:
error[E0271]: type mismatch resolving `<std::iter::Map<std::ops::Range<usize>, [closure@src/main.rs:7:47: 7:72]> as std::iter::IntoIterator>::Item == std::boxed::Box<dyn MyTrait>`
--> src/main.rs:13:9
|
13 | vec.extend(new_elements);
| ^^^^^^ expected struct `MyStruct`, found trait MyTrait
|
= note: expected type `std::boxed::Box<MyStruct>`
found type `std::boxed::Box<dyn MyTrait>`
This is my code: https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=fa8d7cd89a62760c681bc2e2aa5f6f2e
struct MyStruct {}
trait MyTrait {}
impl MyTrait for MyStruct {}
fn main() {
let mut vec: Vec<Box<MyTrait>> = vec![];
let mut new_elements = (0..16_usize).map( |_| Box::new(MyStruct {}));
// works
vec.push(new_elements.next().unwrap());
// doesn't work
vec.extend(new_elements);
// works (std implementation of Vec::extend)
while let Some(element) = new_elements.next() {
let len = vec.len();
if len == vec.capacity() {
let (lower, _) = new_elements.size_hint();
vec.reserve(lower.saturating_add(1));
}
unsafe {
std::ptr::write(vec.get_unchecked_mut(len), element);
// NB can't overflow since we would have had to alloc the address space
vec.set_len(len + 1);
}
}
}
I find the error message confusing - I expected the expected and found types to be swapped.
Do I have to provide some type hints? Can I add a map to cast the type into a trait object?
Like you mention map is one fix. .map(|b| b as _)
Alternate is to start with the trait object coming from the iterator. Box::new(MyStruct {}) as _
Type inference works out the _.
Box<Struct> and Box<dyn Trait> are different types (even of different size) in most cases you have to be explicit in converting.
2 Likes
The problem comes from the closure |_| Box::new(My Struct {}) having a signature of type _ -> Box<MyStruct>.
You can either add the as _ in the closure body as @jonh suggested (the return type then becomes something that can be coerced into from a Box<MyStruct> (e.g., a Box<dyn MyTrait>).
Or you can annotate the signature of the closure:
let mut new_elements =
(0 .. 16)
.map(|_| -> Box<dyn MyTrait> {
Box::new(MyStruct {}) // as Box<dyn MyTrait> /* implicit coercion if MyStruct : MyTrait */
})
;
Aside
Using trait objects without the dyn keyword is ill-advised; I suggest you add the
#![deny(bare_trait_objects)]
lint at the beginning of your code.
5 Likes
Thank you both for all of your input!
I initially tried to cast the inner type which failed because of the missing sizedness
let mut new_elements = (0..16).map(|_| Box::new(MyStruct {} as MyTrait));
@Yandros Thanks for the dyn hint - I expected the compiler to be picky by default. Will add a --deny bare_trait_objects to the command line in my IDE to make this a global default.
RUSTFLAGS="-D bare_trait_objects" environment variable should do it.
3 Likes
This topic was automatically closed 90 days after the last reply. New replies are no longer allowed.
|
__label__pos
| 0.924393 |
3Com WX2200 3CRWX220095A Switch User Manual
Managing the Management Services 119
The command lists the TCP port number on which the switch listens for
HTTPS connections. The command also lists the last 10 devices to
establish HTTPS connections with the switch and when the connections
were established.
If a browser connects to a WX switch from behind a proxy, then only the
proxy IP address is shown. If multiple browsers connect using the same
proxy, the proxy address appears only once in the output.
Changing the Idle
Timeout for CLI
Management
Sessions
By default, MSS automatically terminates a console or Telnet session that
is idle for more than one hour. To change the idle timeout for CLI
management sessions, use the following command:
set system idle-timeout seconds
You can specify from 0 to 86400 seconds (one day). The default is 3600
(one hour). If you specify 0, the idle timeout is disabled. The timeout
interval is in 30-second increments. For example, the interval can be 0, or
30 seconds, or 60 seconds, or 90 seconds, and so on. If you enter an
interval that is not divisible by 30, the CLI rounds up to the next 30-second
increment. For example, if you enter 31, the CLI rounds up to 60.
This command applies to all types of CLI management sessions: console,
Telnet, and SSH. The timeout change applies to new sessions only.
The following command sets the idle timeout to 1800 seconds (one half
hour):
WX1200# set system idle-timeout 1800
success: change accepted.
To reset the idle timeout to its default value, use the following command:
clear system idle-timeout
To display the current setting (if the timeout has been changed from the
default), use the display config area system command. If you are not
certain whether the timeout has been changed, use the display config
all command.
|
__label__pos
| 0.588969 |
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free.
I have a multiple line EditText that does not permit line returns. Right now I am replacing returns with some spaces as soon as they click save. Is there any way I can replace the on screen enter button with a Done button? (like it is for single line EditText)
I am aware that I should still strip out returns (\r\n|\r|\n) because the on screen keyboard is not the only way to add them.
Here is my current XML
<EditText android:layout_width="fill_parent" android:layout_height="wrap_content"
android:minLines="3" android:gravity="left|top"
android:inputType="textMultiLine|textAutoCorrect|textCapSentences"
android:imeOptions="actionDone" />
share|improve this question
stackoverflow.com/a/5037488/185022 – AZ_ Dec 31 '14 at 3:49
4 Answers 4
up vote 13 down vote accepted
android:inputType="textEmailAddress|textEmailSubject"
You need to set the input type as email address or email subject. Either one will give you your desired result. shouldAdvanceFocusOnEnter() is a private method in TextView which determines whether to enter a new line or move focus to next field.
share|improve this answer
3
does this actually work for multiline text? i can only get the done button if i don't pipe in textMultiLine. – tote Sep 7 '12 at 14:13
I suggest to read this article
http://savagelook.com/blog/android/android-quick-tip-edittext-with-done-button-that-closes-the-keyboard
really good example
XML:
<EditText
android:id="@+id/edittext_done"
android:layout_width="fill_parent"
android:layout_height="wrap_content"
android:hint="Enter some text"
android:imeOptions="actionDone"
/>
Custom Action Class:
class DoneOnEditorActionListener implements OnEditorActionListener {
@Override
public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
if (actionId == EditorInfo.IME_ACTION_DONE) {
InputMethodManager imm = (InputMethodManager)v.getContext().getSystemService(Context.INPUT_METHOD_SERVICE);
imm.hideSoftInputFromWindow(v.getWindowToken(), 0);
return true;
}
return false;
}
}
Activity Class:
public class SampleActivity extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.sample_activity_layout); // sample_activity_layout contains our target EditText, target_edittext
EditText targetEditText = (EditText)findViewById(R.id.target_edittext);
targetEditText.setOnEditorActionListener(new DoneOnEditorActionListener());
// The rest of the onCreate() code
}
}
share|improve this answer
Since link-only answers are discouraged, I went ahead and edited in the relevant code. Thanks! – George Bailey Sep 12 '11 at 20:31
@GeorgeBailey why does the example focus on single line edit text when you automatically get the "next" or "done" functionality by default with a single line. as soon as you pipe in inputType textMultiLine, setting imeOption actionDone will not turn your enter key into the done button as far as i can tell. – tote Sep 7 '12 at 14:18
@towpse, I don't know. I didn't use this solution. Perhaps you can formulate a question and get a better answer from someone who does know. If I was to guess I would say that imeOption actionDone only works if you properly combine that with the two sections of Java Class source code. Or perhaps you should read the linked article for further explanation. – George Bailey Sep 7 '12 at 17:59
@GeorgeBailey were you successfully able to modify the enter/action key for a multi line text box? i can only get the done button if i don't pipe in textMultiLine. – tote Sep 7 '12 at 19:33
@towpse, I don't remember. It's been 6 months. I suggest you ask a new question, with the example code you have tried, and the current results, and a description of your desired results. – George Bailey Sep 10 '12 at 14:48
If you're using android:inputType="textMultiLine|..." in your XML, or using the corresponding Java code:
editField.setInputType( InputType.TYPE_CLASS_TEXT | InputType.TYPE_TEXT_FLAG_MULTI_LINE );
then the only solution to show a "Done" or "Search" button is to follow the answer here:
http://stackoverflow.com/a/5037488/1071942
or see
Multiline EditText with Done SoftInput Action Label on 2.3
This is because whenever you enable the "textMultiLine" option, it ignores any setting of android:imeOptions="actionDone" or android:imeActionLabel="actionDone", which is very strange and confusing.
share|improve this answer
I do this for multiline texts with an actionLabel:
editText.setSingleLine(true);
editText.setLines(10);
editText.setHorizontallyScrolling(false);
editText.setImeActionLabel(getString(R.string.ready), 0);
share|improve this answer
Your Answer
discard
By posting your answer, you agree to the privacy policy and terms of service.
Not the answer you're looking for? Browse other questions tagged or ask your own question.
|
__label__pos
| 0.701746 |
#1
1. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
Text indenting in drop down menu
I am learning to do drop down menus and so far, so good. What I cannot figure out is that in the drop downs the text seems to be indented. For example, if the text said "New information", the "New" part starts what appears to be a few spaces to the right. I did, most recently, set the text to be centered but it still does this. The effect is that if the text wraps to a second line, then the second line is further to the left. Looks odd to me but I can't figure out how to fix it.
Thanks for any help.
Richard
2. #2
3. Thanks Johnny Hart (BC) R.I.P.
Join Date
May 2003
Location
Dallas
Posts
5,262
Rep Power
1960
Without seeing your actual markup and css, we can't have a clue what's going on. Please either post the code, or give us a link to your test case.
cheers,
gary
There are those who manage to build a web site without knowing what they're doing; thereby proving to themselves they do, indeed, know what they're doing.
My html and css workshop, demos and tutorials.
Ask a better question, get a better answer.
4. #3
5. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
My CSS code
Here is the CSS code I am using for the navigation, drop down menus. I am new to this forum but don't see a way to post a picture or I would show you what it looks like at the web page menus.
_____________________________
body { color: black; text-align: left; font: 1em verdana, sans-serif, arial; padding: 0 0 0 12px; direction: ltr;
background-color: #FDFFF6;
}
div#listmenu { width: 750px; float: left;
font: .8em tahoma, arial, helvetica, sans-serif;
}
div#listmenu ul {
margin: 5px 0 0 10px;
}
div#listmenu li { float: left; background-color: #F8C099; color: black; list-style-type: none; border-right: 1px solid #9c3200; position: relative; padding: 4px 5px !important;
border-top: 1px solid #9c3200;
border-bottom: 1px solid #9c3200;
padding-top: 10px;
text-align: center;
}
div#listmenu a {
text-decoration: none;
color: maroon;
padding: 0 10px;
}
div#listmenu a:hover {
color: #F33;
}
div#listmenu li:first-child { border-left: 1px solid #9c3200;
}
div#listmenu li:hover {
background-color: #FFF2C4;
}
* html div#listmenu ul {
float: left;
border-left: 1px solid #9c3200;
margin-left: 15px;
}
* html a {
display: block;
}
div#listmenu ul li ul { margin: 4px; width: 12em; position: absolute; left: -5px; padding: 0; }
div#listmenu ul li ul li {
width: 100%;
border-right: 1px solid #9c3200;
border-left: 1px solid #9c3200;
border-bottom: 1px solid #9c3200;
border-top-style: none;
}
div#listmenu ul li ul li:first-child {
border-top: 1px solid #9c3200;
}
body div#listmenu ul li ul {
display: none;
}
div#listmenu ul li:hover ul, div#listmenu ul li:hover ul li ul:hover {
display: block;
}
body div#listmenu ul li ul li ul {
visibility: hidden;
top: -3px;
left: 12.6em;
}
div#listmenu ul li ul li:hover ul {
visibility: visible;
}
6. #4
7. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
HTML code
Hmm, I am thinking when you say "actual markup" you mean the html part, right? If so, I tried to post the code but get a message I cannot post anything with url's included.
Here is a little piece of it, edited, that is representative, all the rest of the listings in the same format.
_____________________________
<!--Start Navigations-->
<Div Id="Listmenu">
<Ul>
<Li><A Href="#l" Id="Home" Name="Home" Title="">Home</A></Li>
<Li><A Href="#" Id="Classes" Name="Classes" Title="Classes">Classes</A>
<Ul > <!-- Drop Down Menus -->
<Li><A Href="#" Id="Pccourse" Name="Pccourse" Title="">Veterinary Homeopathy Training</A></Li>
<Li><A Href="#" Id="Webinar" Name="Webinar" Title="">Free Webinars</A></Li>
</Ul>
</Li>
_____________________________
Then next is another <li> and so on, finishing the section with closing </ul>
8. #5
9. Thanks Johnny Hart (BC) R.I.P.
Join Date
May 2003
Location
Dallas
Posts
5,262
Rep Power
1960
Oh, wow! What a mess.
Don't take that personally; it's a common affliction in newcomers to css. You're over thinking.
First, to your problem: Lists are normally indented. Browsers use 40px padding on the ul. Older versions of IE and Opera use margin. To kill the indent:
Code:
ul {
margin: 1em 0;
padding: 0;
}
The 1 em top/bottom margin is the default. If you don't need it, or want a different value set the whole margin to zero, or set the top/bottom to what you want.
You have a div as a container for the menu. You don't need it. The ul is already a container for the list.
For the sake of your favorite deity, do not format your rulesets in a single line. Put each property on its own line, and indent it 2 or 3 spaces; do not tab. For example, not this,
Code:
body { color: black; text-align: left; font: 1em verdana, sans-serif, arial; padding: 0 0 0 12px; direction: ltr; background-color: #FDFFF6;}
but this
Code:
body {
color: black;
text-align: left; /*usually redundant*/
font: 1em verdana, sans-serif, arial; /*wrong order*/
padding: 0 0 0 12px;
direction: ltr; /*why? is the site's language Hebrew? (rtl)*/
background-color: #FDFFF6;
}
It is good practice to use all lowercase for the markup and css.
cheers,
gary
There are those who manage to build a web site without knowing what they're doing; thereby proving to themselves they do, indeed, know what they're doing.
My html and css workshop, demos and tutorials.
Ask a better question, get a better answer.
10. #6
11. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
Thanks for the reply. Sorry to ask likely simple or not-smart questions but when you say "what a mess" is it because I have too much coding in? Or something else?
I will correct the ul code. Thanks for identifying it for me.
Regarding rulesets in a line I am using CSSEdit software (Mac) which does it this way. I don't see a way to change settings and would have to edit manually. Is this important?
Regarding the uppercase of the html code, in my page and program it is all lowercase. I don't know why it looks this way in the posting but when I pasted it in must have changed it.
In the body code you have some comments.
When you say font in wrong order I don't know what that means (well, I understand order but not why it is wrong). I am using CSSEdit software (Mac) and this is what it puts in when I select a font family. Does it matter?
For the direction of text, ltr, I don't know if needed or not. Sounds like not but how do I know which of these things should be put in? Mind you I am an amateur at this, my main job being something else entirely.
Thanks for your help and, I presume, patience.
Richard
12. #7
13. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
Also I meant to add, regarding the use of div, that I took this code from the book "Stylin' With CSS: A Designer's Guide" by Charles Wyke-Smith. Seems a good book, clear, and it was suggested I do it this way. I don't know enough to do all this on my own.
14. #8
15. Thanks Johnny Hart (BC) R.I.P.
Join Date
May 2003
Location
Dallas
Posts
5,262
Rep Power
1960
Originally Posted by rpitcairn
Thanks for the reply. Sorry to ask likely simple or not-smart questions but when you say "what a mess" is it because I have too much coding in? Or something else?
Yes, both. There are unnecessary elements in the markup and unnecessary properties in the css. Additionally, the formatting is difficult to read.
<snip>
Regarding rulesets in a line I am using CSSEdit software (Mac) which does it this way. I don't see a way to change settings and would have to edit manually. Is this important?
Yes, in my opinion it is very important. The eye scans down a column considerably better than across a row. It doesn't matter to the computer, but you or someone else down the road will have to debug or maintain these files. If they're formatted poorly, the debug/maintenance costs are increased and the maintainer becomes highly irritated. Code as if the maintainer will be a violent psychotic who knows where you live. I speak from experience on this; my primary job has been to fix pages that aren't working as desired. I have to read the client's disorganized, badly formatted excuse for a stylesheet or web page. Very often, once I've rewritten the files, the solution is obvious.
If you can't configure CSSEdit to format properly, dump it. Use a plain old text editor.
<snip>
In the body code you have some comments.
When you say font in wrong order I don't know what that means (well, I understand order but not why it is wrong). I am using CSSEdit software (Mac) and this is what it puts in when I select a font family. Does it matter?
Very much so. Those font families are a list of options that the browser go down until reaching one it can supply. For this,
Code:
{font: 1em verdana, sans-serif, arial;}
the order is to use verdana if possible. If not, try sans-serif. But there is always a generic available, so the browser never gets to arial.
Again, junk the CSSEdit if it is screwing up, but in this case, I suspect user error .
For the direction of text, ltr, I don't know if needed or not. Sounds like not but how do I know which of these things should be put in? Mind you I am an amateur at this, my main job being something else entirely.
A good rule of thumb is to use only what you know you need, and nothing more. Browsers have built in style sheets which define the default presentation, so there's always a reasonable fallback. To see Firefox's, go to wherever you keep Firefox and in the res directory, view html.css. (In my Debian Gnu/Linux install, it's at file:///usr/lib/iceweasel/xulrunner/res/html.css.
cheers,
gary
There are those who manage to build a web site without knowing what they're doing; thereby proving to themselves they do, indeed, know what they're doing.
My html and css workshop, demos and tutorials.
Ask a better question, get a better answer.
16. #9
17. Thanks Johnny Hart (BC) R.I.P.
Join Date
May 2003
Location
Dallas
Posts
5,262
Rep Power
1960
Originally Posted by rpitcairn
Also I meant to add, regarding the use of div, that I took this code from the book "Stylin' With CSS: A Designer's Guide" by Charles Wyke-Smith. Seems a good book, clear, and it was suggested I do it this way. I don't know enough to do all this on my own.
I haven't read that book, so can't comment on its overall worth. I will say that even the most respected and knowledgeable web developers can have stupid moments. Web designers have a lot more of them.
As with css properties, if you can't find a compelling reason to use an element, don't. This applies especially to the div and span elements. The other elements specifically define what their content is. That's semantic value. The div has no semantic value and is used to aggregate other block elements into a contextually defined structure. The span is another semantically neutral element and its purpose is to segregate inline content for special rendering.
cheers,
gary
There are those who manage to build a web site without knowing what they're doing; thereby proving to themselves they do, indeed, know what they're doing.
My html and css workshop, demos and tutorials.
Ask a better question, get a better answer.
18. #10
19. No Profile Picture
Registered User
Devshed Newbie (0 - 499 posts)
Join Date
Jun 2010
Posts
6
Rep Power
0
Thanks for the guidance. I manually adjusted all the code and sent a message to the developer asking if it could be made default. I like the program because I can see the changes immediately as I make them. Good way for me to learn what the functions are.
IMN logo majestic logo threadwatch logo seochat tools logo
|
__label__pos
| 0.582896 |
tencent cloud
Feedback
DescribeStatisticData
Last updated: 2022-05-05 16:02:21
1. API Description
Domain name for API request: monitor.tencentcloudapi.com.
This API is used to query monitoring data by dimension conditions.
A maximum of 20 requests can be initiated per second for this API.
We recommend you to use API Explorer
Try it
API Explorer provides a range of capabilities, including online call, signature authentication, SDK code generation, and API quick search. It enables you to view the request, response, and auto-generated examples.
2. Input Parameters
The following request parameter list only provides API request parameters and some common parameters. For the complete common parameter list, see Common Request Parameters.
Parameter Name Required Type Description
Action Yes String Common Params. The value used for this API: DescribeStatisticData.
Version Yes String Common Params. The value used for this API: 2018-07-24.
Region Yes String Common Params. For more information, please see the list of regions supported by the product.
Module Yes String Module, whose value is fixed at monitor
Namespace Yes String Namespace. Valid values: QCE/TKE
MetricNames.N Yes Array of String Metric name list
Conditions.N No Array of MidQueryCondition Dimension condition. The = and in operators are supported
Period No Integer Statistical period in seconds. Default value: 300. Optional values: 60, 300, 3,600, and 86,400.
Due to the storage period limit, the statistical period is subject to the time range of statistics:
60s: The time range is less than 12 hours, and the timespan between StartTime and the current time cannot exceed 15 days.
300s: The time range is less than three days, and the timespan between StartTime and the current time cannot exceed 31 days.
3,600s: The time range is less than 30 days, and the timespan between StartTime and the current time cannot exceed 93 days.
86,400s: The time range is less than 186 days, and the timespan between StartTime and the current time cannot exceed 186 days.
StartTime No String Start time, which is the current time by default, such as 2020-12-08T19:51:23+08:00
EndTime No String End time, which is the current time by default, such as 2020-12-08T19:51:23+08:00
GroupBys.N No Array of String groupBy by the specified dimension
3. Output Parameters
Parameter Name Type Description
Period Integer Statistical period
StartTime String Start time
EndTime String End time
Data Array of MetricData Monitoring data
RequestId String The unique request ID, which is returned for each request. RequestId is required for locating a problem.
4. Example
Example1 Querying monitoring data by dimension condition
Input Example
POST / HTTP/1.1
Host: monitor.tencentcloudapi.com
Content-Type: application/json
X-TC-Action: DescribeStatisticData
<Common request parameters>
{
"Module": "monitor",
"Namespace": "QCE/TKE",
"MetricNames": [
"K8sPodCpuCoreUsed"
],
"Period": 300,
"Conditions": [
{
"Key": "tke_cluster_instance_id",
"Operator": "=",
"Value": [
"cls-mw2w40s7"
]
}
],
"StartTime": "2020-11-24T15:15:50+08:00",
"EndTime": "2020-11-24T15:25:50+08:00"
}
Output Example
{
"Response": {
"Data": [
{
"MetricName": "K8sPodCpuCoreUsed",
"Points": [
{
"Dimensions": [
{
"Name": "tke_cluster_instance_id",
"Value": "cls-mw2w40s7"
},
{
"Name": "node",
"Value": "cls-gn7vut9a-virtual-kubelet-subnet-node-2"
},
{
"Name": "pod_name",
"Value": "proxy-edge-j7j95"
},
{
"Name": "workload_name",
"Value": "init-test-nfs02"
},
{
"Name": "namespace",
"Value": "default-test"
},
{
"Name": "node_role",
"Value": "node"
},
{
"Name": "un_instance_id",
"Value": "cls-gn7vut9a-virtual-kubelet-subnet-nmpi5ecw-1"
},
{
"Name": "workload_kind",
"Value": "Deployment"
}
],
"Values": [
{
"Timestamp": 1606202100,
"Value": 41.066
},
{
"Timestamp": 1606202400,
"Value": 38.666
},
{
"Timestamp": 1606202700,
"Value": 37.866
}
]
},
{
"Dimensions": [
{
"Name": "namespace",
"Value": "default-test"
},
{
"Name": "node",
"Value": "cls-gn7vut9a-virtual-kubelet-subnet-node-3"
},
{
"Name": "node_role",
"Value": "node"
},
{
"Name": "pod_name",
"Value": "network-tools-b94d6dd89-flf7n"
},
{
"Name": "un_instance_id",
"Value": "cls-gn7vut9a-virtual-kubelet-subnet-nmpi5ecw-1"
},
{
"Name": "workload_kind",
"Value": "Deployment"
},
{
"Name": "workload_name",
"Value": "init-test-nfs00"
},
{
"Name": "tke_cluster_instance_id",
"Value": "cls-mw2w40s7"
}
],
"Values": [
{
"Timestamp": 1606202100,
"Value": 42.666
},
{
"Timestamp": 1606202400,
"Value": 33.6
},
{
"Timestamp": 1606202700,
"Value": 40.266
}
]
}
]
}
],
"EndTime": "2020-11-24 15:25:00",
"Period": 300,
"RequestId": "0d25d659-71ee-4c93-b8e1-f3992c61ff46",
"StartTime": "2020-11-24 15:15:00"
}
}
5. Developer Resources
SDK
TencentCloud API 3.0 integrates SDKs that support various programming languages to make it easier for you to call APIs.
Command Line Interface
6. Error Code
The following only lists the error codes related to the API business logic. For other error codes, see Common Error Codes.
Error Code Description
AuthFailure Error with CAM signature/authentication.
AuthFailure.UnauthorizedOperation The request is not authorized. For more information on the authentication, see the CAM documentation.
FailedOperation Operation failed.
FailedOperation.DataColumnNotFound The data table field doesn't exist.
FailedOperation.DataQueryFailed Failed to query the data.
FailedOperation.DataTableNotFound The data table doesn't exist.
FailedOperation.DbQueryFailed Failed to query the database.
FailedOperation.DbRecordCreateFailed Failed to create the database record.
FailedOperation.DbRecordDeleteFailed Failed to delete the database record.
FailedOperation.DbRecordUpdateFailed Failed to update the database record.
FailedOperation.DbTransactionBeginFailed Failed to start the database transaction.
FailedOperation.DbTransactionCommitFailed Failed to submit the database transaction.
FailedOperation.DimQueryRequestFailed Failed to query the service in the request dimension.
FailedOperation.DivisionByZero The dividend is zero.
InternalError Internal error.
InternalError.CallbackFail Error with the callback.
InternalError.DependsApi Error with another dependent API.
InternalError.DependsDb Error with the dependent db.
InternalError.DependsMq Error with the dependent mq.
InternalError.ExeTimeout Execution timed out.
InternalError.System System error.
InternalError.TaskResultFormat An error occurred while parsing the task result.
InvalidParameter Invalid parameter.
InvalidParameter.DupTask The task has already been submitted.
InvalidParameter.InvalidParameter Invalid parameter.
InvalidParameter.InvalidParameterParam Invalid parameter.
InvalidParameter.MissAKSK The platform configuration is missing.
InvalidParameter.ParamError Incorrect parameter.
InvalidParameter.SecretIdOrSecretKeyError Error with the platform configuration.
InvalidParameter.UnsupportedProduct This product doesn't support scan.
InvalidParameterValue The parameter value is incorrect.
InvalidParameterValue.DashboardNameExists The dashboard name already exists.
InvalidParameterValue.VersionMismatch The version does not match.
LimitExceeded Quota limit is reached.
LimitExceeded.MetricQuotaExceeded Quota limit on metrics is reached. Requests containing unregistered metrics are prohibited.
MissingParameter Missing parameter.
OperationDenied Operation denied.
RequestLimitExceeded The number of requests exceeds the frequency limit.
ResourceInUse The resource is in use.
ResourceInsufficient Insufficient resources.
ResourceNotFound The resource is not found.
ResourceNotFound.NotExistTask The task does not exist.
UnauthorizedOperation Unauthorized operation.
UnknownParameter Unknown parameter.
UnsupportedOperation Unsupported operation.
Contact Us
Contact our sales team or business advisors to help your business.
Technical Support
Open a ticket if you're looking for further assistance. Our Ticket is 7x24 avaliable.
7x24 Phone Support
|
__label__pos
| 0.876037 |
12
\$\begingroup\$
I am working my way through "Cracking the Coding Interview" and I came up the question(3.6) to design a data structure to manage an Animal shelter with 2 animals, Cats, and Dogs such that when we dequeue the oldest animal should be removed first (FIFO)
E.g. if the animals were taken in order C,C,D,C,C,D (C- Cat, D- Dog) and someone wants to adopt an animal there could be 3 ways either adopt D, adopt C, or adopt any(I have just implemented for any as other two are trivial). If the person chooses to adopt D the first D to come in would be chosen from the queue similarly with C. If the person has no preference either of C, D who came first can be chosen.
I implemented it as below. What all improvements can be done here? One possible way is to improve printQueue() function using 'this'.
#include <iostream>
#include <queue>
using namespace std;
class Animal {
public:
virtual string getClassName() = 0;
inline int getOrder() {
return _order;
}
void setOrder(int order) {
_order = order;
}
int setType(string type) {
_type = type;
}
inline string getType() {
return _type;
}
bool Compare(Animal* animal) {
if (this->_order > animal->_order)
return true;
return false;
}
private:
int _order;
string _type;
};
class Cat : public Animal {
public:
Cat(string name) : _name(name) {
}
inline string getClassName() {
return "Cat";
}
inline string getName() {
return _name;
}
private:
string _name;
};
class Dog : public Animal {
public:
Dog(string name) : _name(name) {
}
inline string getClassName() {
return "Dog";
}
inline string getName() {
return _name;
}
private:
string _name;
};
class AnimalQueue {
public:
void enqueue(Animal* animal) {
if (animal->getClassName() == "Cat") {
Cat* d = dynamic_cast<Cat*>(animal);
if (!d) {
cout << "\nCasting Error";
}
else {
cout << "\nEnqueued Cat";
d->setOrder(++queueOrder);
d->setType(animal->getClassName());
catQueue.push(d);
}
}
else{
Dog* d = dynamic_cast<Dog*>(animal);
if (!d) {
cout << "\nCasting Error";
}
else {
cout << "\nEnqueued Dog";
d->setOrder(++queueOrder);
d->setType(animal->getClassName());
dogQueue.push(d);
}
}
}
void dequeue() {
if (catQueue.empty()) {
dogQueue.pop();
}
if (dogQueue.empty())
{
catQueue.pop();
}
//Pop with smaller timestamp
if (catQueue.front()->Compare(dogQueue.front())) {
dogQueue.pop();
}
else {
catQueue.pop();
}
}
void printQueue() {
queue<Cat*> cQueue = this->catQueue;
queue<Dog*> dQueue = this->dogQueue;
cout << "\nCat Queue\n";
while (!cQueue.empty()) {
cout << cQueue.front()->getName() << " ";
cout << cQueue.front()->getOrder();
cout << endl;
cQueue.pop();
}
cout << "\nDog Queue\n";
while (!dQueue.empty()) {
cout << dQueue.front()->getName() << " ";
cout << dQueue.front()->getOrder();
cout << endl;
dQueue.pop();
}
}
queue<Dog*> getDogQueue() {
return dogQueue;
}
queue<Cat*> getCatQueue() {
return catQueue;
}
private:
queue<Cat*> catQueue;
queue<Dog*> dogQueue;
int queueOrder = -1;
};
int main()
{
Animal* d1 = new Dog("Max"),*d2 = new Dog ("Shaun"), *d3 = new Dog("Tiger");
Animal* c1 = new Cat("Trace"), *c2 = new Cat("Han"), *c3 = new Cat("Meow");
AnimalQueue queue;
queue.enqueue(d1);
queue.enqueue(c1);
queue.enqueue(c2);
queue.enqueue(d2);
queue.enqueue(d3);
queue.enqueue(c3);
cout << endl;
queue.printQueue();
cout << endl;
queue.dequeue();
queue.printQueue();
}
\$\endgroup\$
3
• \$\begingroup\$ You also need something else for this to compile ... string is an undeclared identifier ... \$\endgroup\$
– L. F.
Jul 15, 2019 at 3:53
• \$\begingroup\$ That's why I don't use namespaces, don't know why I did it here @L.F. \$\endgroup\$
– CocoCrisp
Jul 15, 2019 at 4:05
• \$\begingroup\$ Just a note: you could solve this in C++ with templates and some constexpr with much less code. \$\endgroup\$
– mfnx
Oct 5, 2020 at 13:33
6 Answers 6
13
\$\begingroup\$
To supplement the other reviews, here are some other things you might improve.
Use override where appropriate
When a virtual function is being overridden, it should be marked override to allow catching errors at compile time. See C.128.
Make sure all paths return a value
The setType routine claims it returns an int but it does not. That's an error that should be addressed.
Don't use std::endl if you don't really need it
The difference betweeen std::endl and '\n' is that '\n' just emits a newline character, while std::endl actually flushes the stream. This can be time-consuming in a program with a lot of I/O and is rarely actually needed. It's best to only use std::endl when you have some good reason to flush the stream and it's not very often needed for simple programs such as this one. Avoiding the habit of using std::endl when '\n' will do will pay dividends in the future as you write more complex programs with more I/O and where performance needs to be maximized.
Think carefully about object ownership
The traditional role of a shelter is to take in animals and then give them to a new owner on adoption. This shelter seems to only do bookkeeping of the location of animals (by handling only pointers) rather than actually taking ownership of them. What is actually a more appropriate way to express this is by the use of a std::unique_ptr. See R.20
Think carefully about the domain and range of numbers
The _queueOrder increases without bound and is used to assign the _order of each animal. What happens when that number wraps around?
Use polymorphism effectively
Whenever you find yourself writing code like this:
if (animal->getClassName() == "Cat") {
Cat* d = dynamic_cast<Cat*>(animal);
stop and question whether this is really needed. By using animal->getClassName() as a sort of home-grown polymorphism, the code is made much more brittle and hard to maintain. Here's how I'd write that using a std::unique_ptr instead:
void enqueue(std::unique_ptr<Animal> &&animal) {
animal->setOrder(++queueOrder);
if (typeid(*animal) == typeid(Cat)) {
catQueue.push_back(std::move(animal));
} else if (typeid(*animal) == typeid(Dog)) {
dogQueue.push_back(std::move(animal));
} else {
throw std::runtime_error("This animal is not suitable for the shelter");
}
}
Note that this uses true RTTI, built into the language, instead of inventing a poor imitation. It also throws an error if the passed animal is neither a cat nor a dog. This could be handy if someone attempted to drop off a pet rhinocerous.
Don't expose class internals
It seems to me that getDogQueue and getCatQueue are both ill-advised and unneeded. I'd simply omit them both.
Base destructors should be virtual
The destructor of a base class, including a pure virtual one like Animal, should be virtual. Otherwise, deleting the object could lead to undefined behavior and probably memory leaks.
Consolidate common items into a base class
Since all of the derived classes have _name, why not move that functionality into the base class?
Use const where practical
The getName() function does not alter the underlying class because it returns a copy of the name. Similarly, the getClassName() function does not alter the class. Both should be declared const.
Use standard operators
Rather than the vaguely named Compare, better would be to simply use the standard operator<. Here's how I'd write it as a member function of Animal:
bool operator<(const Animal& b) const {
return _order < b._order;
}
Use better names
Most of the names are not bad, but rather than AnimalQueue and enqueue and dequeue, I'd suggest giving them more usage-oriented names rather than describing the internal structure. So perhaps AnimalShelter, dropoff and adopt would be more suitable.
Think carefully about data types
If you use a std::deque instead of a std::queue, you gain access to iterators which are useful for printing as shown in the next suggestion.
Use an ostream &operator<< instead of display
The current code has printQueue() function but what would make more sense and be more general purpose would be to overload an ostream operator<< instead. This renders the resulting function much smaller and easier to understand:
friend std::ostream& operator<<(std::ostream& out, const AnimalShelter& as) {
out << "\nCat Queue\n";
for (const auto& critter : as.catQueue) {
out << *critter << '\n';
}
out << "\nDog Queue\n";
for (const auto& critter : as.dogQueue) {
out << *critter << '\n';
}
return out;
}
I also modified the base class to include _name as mentioned above and wrote this friend function of Animal:
friend std::ostream& operator<<(std::ostream& out, const Animal& a) {
return out << a.getClassName() << ' ' << a._name << ' ' << a._order;
}
Implement the problem specification completely
The description of the problem mentions that one might be able to adopt either a cat or a dog or the first available, but only the latter function has been implemented. Here's how I wrote all three:
std::unique_ptr<Animal> adopt() {
std::unique_ptr<Animal> adoptee{nullptr};
if (catQueue.empty() && dogQueue.empty())
return adoptee;
if (catQueue.empty()) {
std::swap(adoptee, dogQueue.front());
dogQueue.pop_front();
} else if (dogQueue.empty() || (catQueue.front() < dogQueue.front())) {
std::swap(adoptee, catQueue.front());
catQueue.pop_front();
} else {
std::swap(adoptee, dogQueue.front());
dogQueue.pop_front();
}
return adoptee;
}
std::unique_ptr<Animal> adoptCat() {
std::unique_ptr<Animal> adoptee{nullptr};
if (!catQueue.empty()) {
std::swap(adoptee, catQueue.front());
catQueue.pop_front();
}
return adoptee;
}
std::unique_ptr<Animal> adoptDog() {
std::unique_ptr<Animal> adoptee{nullptr};
if (!dogQueue.empty()) {
std::swap(adoptee, dogQueue.front());
dogQueue.pop_front();
}
return adoptee;
}
Results
Here is the modified main to exercise the revised code:
int main()
{
AnimalShelter shelter;
shelter.dropoff(std::unique_ptr<Animal>(new Dog{"Max"}));
shelter.dropoff(std::unique_ptr<Animal>(new Cat{"Trace"}));
shelter.dropoff(std::unique_ptr<Animal>(new Cat{"Han"}));
shelter.dropoff(std::unique_ptr<Animal>(new Dog{"Shaun"}));
shelter.dropoff(std::unique_ptr<Animal>(new Dog{"Tiger"}));
shelter.dropoff(std::unique_ptr<Animal>(new Cat{"Meow"}));
try {
shelter.dropoff(std::unique_ptr<Animal>(new Rhino{"Buster"}));
} catch (std::runtime_error &err) {
std::cout << err.what() << '\n';
}
std::cout << shelter << '\n';
for (int i = 0; i < 2; ++i) {
auto pet = shelter.adoptDog();
if (pet) {
std::cout << "You have adopted " << *pet << "\n";
} else {
std::cout << "sorry, there are no more pets\n";
}
std::cout << shelter << '\n';
}
for (int i = 0; i < 6; ++i) { // adopt any
auto pet = shelter.adopt();
if (pet) {
std::cout << "You have adopted " << *pet << "\n";
} else {
std::cout << "sorry, there are no more pets\n";
}
std::cout << shelter << '\n';
}
std::cout << "Final: \n" << shelter << '\n';
}
\$\endgroup\$
8
• \$\begingroup\$ Thankyou for briefly explaining! I didn't know much about std::endl as I have just used it in college to break lines but this explanation is really nice! Also, I did not think of the range of int, good point! Virtual destructor, how did I miss that! \$\endgroup\$
– CocoCrisp
Jul 15, 2019 at 21:36
• 2
\$\begingroup\$ I would suggest to use std::make_unique instead of std::unique_ptr(new …) if you have C++14 or higher. Not 100% sure that it will work. \$\endgroup\$ Jul 16, 2019 at 13:10
• \$\begingroup\$ @val: You're right about std::make_unique. An earlier iteration of my code didn't work with it but this one does. One can use std::make_unique<Cat>("Whiskers") for a bit more succinct code. \$\endgroup\$
– Edward
Jul 16, 2019 at 13:36
• \$\begingroup\$ Why is the exception not const. Main doesn't return a value. It's a better interface if AnimalShelter had separate dropoffDog() and dropoffCat() functions, as then you wouldn't need any RTTI nonsense and the use of unique_ptr's would be hidden from the user. This is essentially Java. \$\endgroup\$
– James
Jul 16, 2019 at 14:47
• 1
\$\begingroup\$ @nowox I believe you are misinterpreting what it says. While they may not return the same instance, they are guaranteed to return true from their operator==(). Otherwise typeid would be rather useless. \$\endgroup\$
– Edward
Dec 9, 2020 at 11:37
16
\$\begingroup\$
In addition to the suggestions given by 1201ProgramAlarm's answer, I have the following suggestions.
First, your (over)use of RTTI seems overkill here. Even dynamic polymorphism is not necessary. A simple class is enough:
enum class Species { cat, dog };
struct Animal {
Species species;
std::string name;
};
(You are missing #include <string>)
Second, you maintain two queues. Since this is an animal shelter, why not merge them? You can use vector to accommodate for adopters that prefer a specified species.
std::vector<Animal> animals;
Third, dequeue returns void. This is abandonment, not adoption. It should return the Animal. (An animal protectionist can even go further and mark the function [[nodiscard]].)
Also, let print accept an ostream&, so you can print to any stream.
Here's the revised version:
#include <algorithm>
#include <iostream>
#include <stdexcept>
#include <string>
#include <utility>
#include <vector>
enum class Species { cat, dog };
std::string to_string(Species species)
{
switch (species) {
case Species::cat:
return "cat";
case Species::dog:
return "dog";
default:
throw std::invalid_argument{"Unrecognized species"};
}
}
struct Animal {
Species species;
std::string name;
};
class Shelter {
public:
void enqueue(Animal animal)
{
animals.push_back(animal);
}
// adopt a specific species
Animal dequeue(Species species)
{
auto it = std::find_if(animals.begin(), animals.end(),
[=](const Animal& animal) {
return animal.species == species;
});
if (it == animals.end())
throw std::logic_error{"No animal to adopt"};
Animal animal = std::move(*it);
animals.erase(it);
return animal;
}
// adopt any animal
Animal dequeue()
{
if (animals.empty())
throw std::logic_error{"No animal to adopt"};
Animal animal = std::move(animals.front());
animals.erase(animals.begin());
return animal;
}
void print(std::ostream& os)
{
for (const auto& animal : animals) {
os << animal.name << " (" << to_string(animal.species) << ")\n";
}
}
private:
std::vector<Animal> animals;
};
int main()
{
Shelter shelter;
shelter.enqueue({Species::dog, "Max"});
shelter.enqueue({Species::cat, "Trace"});
shelter.enqueue({Species::cat, "Han"});
shelter.enqueue({Species::dog, "Shaun"});
shelter.enqueue({Species::dog, "Tiger"});
shelter.enqueue({Species::cat, "Meow"});
shelter.print(std::cout);
std::cout << "\n";
auto new_pet = shelter.dequeue();
shelter.print(std::cout);
std::cout << "\n";
auto new_dog = shelter.dequeue(Species::dog);
shelter.print(std::cout);
}
Output:
Max (dog)
Trace (cat)
Han (cat)
Shaun (dog)
Tiger (dog)
Meow (cat)
Trace (cat)
Han (cat)
Shaun (dog)
Tiger (dog)
Meow (cat)
Trace (cat)
Han (cat)
Tiger (dog)
Meow (cat)
\$\endgroup\$
7
• 5
\$\begingroup\$ I would note that algorithmic-wise, maintaining differentiating queues has a different complexity trade-off than maintaining a single queue. Popping a Cat off a single-queue is an O(number-animals) operation, while popping a Cat off differentiated queues is O(1). \$\endgroup\$ Jul 15, 2019 at 12:12
• 1
\$\begingroup\$ @MatthieuM. Good point. It seems unlikely that OP would be dealing with a lot of animals in this case, anyway :) \$\endgroup\$
– L. F.
Jul 15, 2019 at 14:12
• 2
\$\begingroup\$ My first thought was "oh another 'let's use inheritance for animals' type problem, let's see how unnecessary complex it can get" and I was not disappointed. Would give +2 if I could, given that other answers already started with virtual destructors and Knuth knows what. It' also easy to maintain different queues for each species with this approach as well if that's really necessary. \$\endgroup\$
– Voo
Jul 16, 2019 at 17:30
• \$\begingroup\$ @MatthieuM. If you maintain two different queues, how would you account for the case where the species doesn't matter and you just want to dequeue an animal? \$\endgroup\$ Jul 17, 2019 at 12:07
• 2
\$\begingroup\$ @AleksandrH: Similarly to what the OP is doing, you need an index (they call "queueOrder"). I would not however store it in the Animal itself, as this is metadata, and instead would store std::pair<Order, Dog> and std::pair<Order, Cat> respectively. \$\endgroup\$ Jul 18, 2019 at 6:27
12
\$\begingroup\$
First off, you should avoid using namespace ::std.
In Animal, getOrder and getType should be const (int getOrder() const). Since they're defined within the class definition, you don't need to include the inline keyword. Similarly, getters in the derived classes should be const.
It is rarely ever necessary to use the this keyword, and the places where you use this-> to access a member variable can do without it.
Compare should also be const (and take its parameter as a const Animal *, and can be simplified to return _order > animal->_order;.
You never delete any of the memory you allocate with new.
You should declare a virtual destructor for Animal since it is used as a base class. It isn't a direct problem here, since you never delete any of the memory you allocate, but if you'd delete Animals you dequeue, you'd still leak memory as the strings in the derived classes would not be freed.
In AnimalQueue you can have separate enqueue methods that take Cat * and Dog * parameters. This would avoid needing to have all that code to check the type, and would work better if you have further derived classes (like "Tiger" derived from "Cat" or "Wolf" derived from "Dog"). The generic enqueue(Animal *) can call the proper overloaded enqueue after checking the result of the dynamic cast. Something like
if (auto cat = dynamic_cast<Cat*>(animal))
enqueue(cat);
else if (auto dog = dynamic_cast<Dog*>(animal))
enqueue(dog);
else /* error handling */;
assuming you're using a compiler that supports the variable declarations in if conditional expressions.
dequeue has several bugs. It will attempt to remove elements from empty containers, and could also try to access elements from empty containers.
printQueue should be const, and should use iterators to print the contents of the queues. Since displaying the two queues has a bunch of duplicated code, make a function that will display one queue and call that for each queue you want to display.
getDogQueue and getCatQueue should be const and return const references.
\$\endgroup\$
0
4
\$\begingroup\$
In addition to other answers: you're using inline a lot. Do not.
Modern compilers (almost all starting from year 2000) don't use it for optimization anyways: they are already clever enough to do so without hints when possible. These days inline have another meaning in some contexts (when function is defined in many translation units), but in your situation it's a simple no-op.
\$\endgroup\$
1
• 1
\$\begingroup\$ Welcome to Code Review! It would be nice if you could add some external resources or a online compiler example to back your claim. \$\endgroup\$
– AlexV
Jul 15, 2019 at 14:05
2
\$\begingroup\$
I'll assume in 2019 that C++17 is available.
All other answers seem to be using two queues, one for each type of animal, but I think it kinda defeats the purpose that the FIFO behaviour has to be on the whole set of animals. With two queues, the FIFO has to be implemented by an _order stored with the animal, which mixes the data with the algorithm. Once the animal is adopted, out of the shelter, the _orderhas no meaning, but is still part of the structure.
Using a queue, I would use a single queue of animals, which implements FIFO. (But the deque allows you to remove from the middle of it)
std::deque<Animal> _animals;
Now, the animal being either a cat or a dog, I would just say so
using Animal = std::variant<Dog,Cat>;
And implement each species as its own class.
class Cat { /* implementation*/ };
class Dog { /* implementation*/ };
Then, borrowing terminology from Edward's answer, I would simply implement:
void dropoff(Animal a) { _animals.emplace_back(std::move(a)) };
std::optional<Animal> adoptAny() {
if(_animals.empty()) return std::nullopt;
auto adoptee = std::move(_animals.front());
_animals.pop_front();
return adoptee; // NRVO
}
template<typename T>
auto adoptFirstOfType() -> std::optional<T> {
// Find first animal of given type
const auto adoptee_it = std::find_if(
begin(_animals),
end(_animals),
[](const Animal& a) { return a.holds_alternative<T>(); };
// If not found, return empty optional.
if(adoptee_it == end(_animals)) {
return std::nullopt;
}
// If found, steal the right alternative, remove from queue and return
auto adoptee = std::get<T>(std::move(*adoptee_it));
_animals.erase(adoptee_it);
return adoptee; //NRVO
}
auto adoptDog() { // type deduced consistently from returned expression
return adoptFirstOfType<Dog>();
}
auto adoptCat() { // type deduced consistently from returned expression
return adoptFirstOfType<Cat>();
}
Edward's main function should work fine as is, because optional has the same "container access" interface as unique_ptr.
This allows to simply drop the _order and operator< hacks from his solution. (because yes, implementing operator< is a hack - it makes no sense that a Dog be more than a Cat because it arrived first in the shelter)
\$\endgroup\$
6
• \$\begingroup\$ Good answer! I also considered the use of std::optional for this. I also like to think of C++17 as everywhere by now but just this morning I was trying to compile a program on a Raspberry Pi and found that alas, the compiler was only up to C++14. As for operator< being a hack, I disagree. The only ordering among animals mentioned in the problem statement is time-in-shelter, so it makes sense that a standard operator would be used to express that. And by the way as I implemented it, it would be Cat < Dog if the cat had arrived first. :) \$\endgroup\$
– Edward
Jul 16, 2019 at 18:59
• \$\begingroup\$ @L.F.: You seem to be right. That would take a deque instead of a queue. Correcting this. \$\endgroup\$ Jul 17, 2019 at 6:12
• \$\begingroup\$ @Edward: You're right, the logic is inverted in my answer. Correcting this \$\endgroup\$ Jul 17, 2019 at 6:16
• \$\begingroup\$ @Edward : thinking again, I think the problem is not in the operator<, but in the fact that the _order should not belong to the Animal class, because once the animal is out of the shelter, the _order matters no more. Maybe the queue(s) should instead store a std::pair<size_t, Animal or Dog or Cat>, and then you could define a cleaner operator< like template<typename T, typename U> bool operator<(const std::pair<size_t,T>&, const std::pair<size_t,U>&) \$\endgroup\$ Jul 17, 2019 at 6:31
• \$\begingroup\$ My single queue is bothering me, come to think of it. Adopting a specific animal implies a lookup, and I suppose adoptAny is not the dominating case. As such, one queue per animal type seem to be constant-time in either case, so would be the proper way to handle the problem. \$\endgroup\$ Jul 17, 2019 at 7:01
1
\$\begingroup\$
Design
Is a single queue acceptable, or does each animal need its own?
Multiple queues need more bookkeeping, as well as slight overhead when enqueueing an animal, or dequeueing the senior resident regardless of species.
While one can put the entrance-order in the animal-class, it certainly does not belong.
A single queue potentially require looking at every single animal to find one of a specific species. On the flip-side, adding additional species comes naturally.
Implementation
1. ::std is not designed for wholesale inclusion. It contains myriad symbols, there is no comprehensive list, and it is always evolving. Your code might not "work" today, but tomorrow it might fail to compile, or worse silently change ist meaning.
Read "Why is “using namespace std” considered bad practice?" for more detail.
2. You are flat-out leaking all your animals.
As the program terminates directly afterwards, and deleteing them has no observable effect you depend on, that could be a good idea.
That is, if it was intentional, well-explained, and not demo-code, but neither is the case.
Consider an appropriate use of smart-pointers, specifically std::unique_ptr, as manual resource-management is error-prone.
3. Generally, polymorphic types should have a virtual dtor, so they can be deleted polymorphically.
4. The string type is set to the classname when enqueueing? Rip that out at the roots, you can use the source directly.
5. Unless you need a friendly name for presistence or display, consider fully relying on builtin RTTI instead of adding your own home-grown variant. Use typeid if you want exact type-matching, or dynamic_cast if subtypes are fair game.
6. All member-functions defined inline are automatically inline.
7. Avoid needlessly copying strings, that's inefficient. Prefer std::string_view where applicable.
8. Mark functions noexcept if you can.
9. Mark overrides override for documentation and to allow the compiler to catch mistakes.
10. printQueue() clears it? That violates all expectations. And anyway, it should be std::ostream& operator<<(std::ostream&, AnimalQueue const&).
You probably should move from std::queue to directly using std::deque when fixing that, even though you can safely access the underlying container.
Using a single shared queue (also live on coliru):
#include <algorithm>
#include <iostream>
#include <memory>
#include <string>
#include <string_view>
#include <utility>
#include <vector>
class Animal {
public:
virtual std::string_view friendlyName() noexcept = 0;
virtual ~Animal() = default;
};
class Cat : public Animal {
std::string _name;
public:
Cat(std::string name) : _name(std::move(name)) {}
std::string_view friendlyName() noexcept override { return "Cat"; }
};
class Dog : public Animal {
std::string _name;
public:
Dog(std::string name) : _name(std::move(name)) {}
std::string_view friendlyName() noexcept override { return "Dog"; }
};
class Shelter {
std::vector<std::unique_ptr<Animal>> _queue;
public:
template <class T = Animal>
std::unique_ptr<T> get() noexcept {
auto pos = std::find_if(_queue.begin(), _queue.end(), [](auto&& x){ return dynamic_cast<T*>(x.get()); });
if (pos == _queue.end())
return {};
std::unique_ptr<T> r(dynamic_cast<T*>(pos->release()));
_queue.erase(pos);
return r;
}
void put(std::unique_ptr<Animal>&& a) {
if (a)
_queue.push_back(std::move(a));
}
friend std::ostream& operator<<(std::ostream& os, Shelter const& s) {
for (auto&& x : s._queue)
os << x->friendlyName() << ' ';
os << '\n';
return os;
}
};
int main()
{
auto d1 = std::make_unique<Dog>("Max");
auto d2 = std::make_unique<Dog>("Shaun");
auto d3 = std::make_unique<Dog>("Tiger");
auto c1 = std::make_unique<Cat>("Trace");
auto c2 = std::make_unique<Cat>("Han");
auto c3 = std::make_unique<Cat>("Meow");
Shelter shelter;
shelter.put(std::move(d1));
shelter.put(std::move(c1));
shelter.put(std::move(c2));
shelter.put(std::move(d2));
shelter.put(std::move(d3));
shelter.put(std::move(c3));
std::cout << shelter;
shelter.get();
std::cout << shelter;
}
\$\endgroup\$
1
• 1
\$\begingroup\$ I also proposed a solution with a single queue, but it has the inconvenient of requiring a linear search for the common case :adopting an animal of a specific species. The OP's solution of multiple queues with an everincreasing 64-bit order of entry gives constant time search in every case. You're making good points otherwise. \$\endgroup\$ Jul 18, 2019 at 14:28
Your Answer
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Not the answer you're looking for? Browse other questions tagged or ask your own question.
|
__label__pos
| 0.984293 |
144...44
144 = 1 2 2 1444 = 3 8 2 \begin{aligned} 144 &= 12^2 \\ 1444 &= 38^2 \end{aligned}
Are there any other numbers 144 44 144 \ldots 44 (starting with 1 and followed by all 4s) that are perfect squares?
Yes No
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.
8 solutions
Henry U
Oct 26, 2018
Every number 1 44 44 n digits 1 \underbrace{44 \ldots 44}_{\text{n digits}} can be written as 1 0 n + 44 44 n digits = 1 0 n + 4 ( 11 11 n digits ) = 1 0 n + 4 ( 1 0 n 1 ) 9 10^n + \underbrace{44 \ldots 44}_{\text{n digits}} = 10^n + 4 \cdot ( \underbrace{11 \ldots 11}_{\text{n digits}} ) = 10^n + 4 \cdot \frac {\left( 10^n-1 \right)} 9 .
For n 2 n \geq 2 , we can factor out 2 2 2^2 , since we only want to prove that the expression is not a perfect square.
1 0 n + 4 ( 1 0 n 1 ) 9 = 4 ( 25 1 0 n 2 + ( 1 0 n 1 ) 9 ) 10^n + 4 \cdot \frac {\left( 10^n-1 \right)} 9 = 4 \cdot \left( 25 \cdot 10^{n-2} + \frac {\left( 10^n-1 \right)} 9 \right)
Let's look at the remainder when dividing by 4 4 . The remainder of a square number divided by 4 4 is always 0 0 or 1 1 .
The remainder of a sum is the sum of the remainders, so the remainder of 25 1 0 n 2 + ( 1 0 n 1 ) 9 25 \cdot 10^{n-2} + \frac {\left( 10^n-1 \right)} 9 is the remainder of 25 1 0 n 2 25 \cdot 10^{n-2} , which is 0 0 for n 4 n \geq 4 , plus the remainder of ( 1 0 n 1 ) 9 = 11 11 n digits \frac {\left( 10^n-1 \right)} 9 = \underbrace{11 \ldots 11}_{\text{n digits}}
To test for divisibility by 4 4 , we look at the two last digits, 11 11 . They give a remainder of 3 3 when divided by 4 4 . But since the remainder of a perfect square is either 0 0 or 1 1 , this number with remainder 3 3 cannot be a perfect square, so there are No other such numbers.
Why can you factor out 16 16 ? 144 44 144\dots44 isn't divisible by 16 16 for n 3 n\ge3 .
Abraham Zhang - 2 years, 7 months ago
Log in to reply
Oh, I'm sorry, that's a mistake. I meant to say 4 4 .
Henry U - 2 years, 7 months ago
12012^2 = 144288144, which starts with 144 and ends on 44. So I'm not wrong :/. The answer is 'Yes'.
Carlo Wood - 2 years, 7 months ago
Log in to reply
The dots were meant to say that there are a bunch of 4 4 s in between. I'm sorry for the confusion.
Henry U - 2 years, 7 months ago
Log in to reply
But it literaly says: "starting with 144... and ending in ...44" thats what I understand then. I also got confused.
Pau Cantos - 2 years, 7 months ago
Log in to reply
@Pau Cantos Yeah, same for me :(
Bk Lim - 2 years, 7 months ago
@Henry U , you could say "(an integer that exists of a 1 followed by four or more 4's)"
Carlo Wood - 2 years, 7 months ago
Log in to reply
@Carlo Wood That's a better wording, thank you. Somehow, I can't edit the problem, so @Calvin Lin, could you replace the part in brackets by "a 1 followed by at least 4 4's"?. Thanks!
Henry U - 2 years, 7 months ago
the answer 'no' is clearly wrong. the mistake in this problem is not in the question, but in the answer. the meaning of the question is unmistakable - starts w 144, ends in 44. says nothing about the middle, which means 'whatever'. any number that starts with 12 pr 38 has enough zeroes in between and ends with 12 or 38 will fit the bill. so, the only correct answer is yes
sabit khakimzhanov - 2 years, 7 months ago
Log in to reply
Then we have a horse of a different colour. In setting questions detail, accuracy, and lack of ambiguity are everything!!
Phil Greene - 2 years, 7 months ago
What should the number of digits of 144...44 be in the solution?
Manan Rastogi - 2 years, 7 months ago
Log in to reply
The question is just whether there is a number with any number of 4's that is a perfect square.
Henry U - 2 years, 7 months ago
So what about 12012^? Starts with 144 and ends with 44? This is a solution or have I missed something here
Phil Greene - 2 years, 7 months ago
Log in to reply
Sorry Carlo just noticed you came to same solution! There another one 38038 !! Starts with 144 and ends with 44!!
Phil Greene - 2 years, 7 months ago
Well done finding that number, but the problem asks about numbers which have only the initial 1 followed by 4’s. Your number is 144288144, which is close, but the question is interested in numbers like 144444444.
Jason Carrier - 2 years, 7 months ago
Log in to reply
He commemted that before I had changed the wording because it was very ambigous before that.
Henry U - 2 years, 7 months ago
if you can factor a 4 out of the sum the remainder modulo 4 should be zero
Jonas Steiner - 2 years, 7 months ago
Log in to reply
The remainder of the number 1 444 444 n 4 ’s 1\underbrace{444\ldots444}_{n \text{ } 4\text{'s}} is indeed 0 0 because it ends in 44 44 which is divisible by 4 4 . At the end, where I mentioned the remainder ( m o d 4 ) \pmod 4 , I was referring to the number 1 444 444 n 4 ’s ÷ 4 = 36 111 111 ( n 2 ) 1 ’s 1 \underbrace {444\ldots444}_{n \text{ } 4\text{'s}} \div 4 = 36\underbrace{111\ldots111}_{(n-2) \text{ } 1\text{'s}} . which leaves a remainder 3 ( m o d 4 ) 3 \pmod 4 .
Henry U - 2 years, 7 months ago
Perfect solution, btw can you explain by using theory that the remainder of a square number divided by 4 is always 0 or 1
Ghally Arrahman - 2 years, 7 months ago
Log in to reply
Every integer n n can be written as either n = 2 k , k Z n = 2k, k \in \mathbb{Z} or n = 2 k + 1 , k Z n = 2k+1, k \in \mathbb{Z} (even and odd numbers).
If we square both forms, we get
( 2 k ) 2 = 4 k 2 = 4 a + 0 ( 2k )^2 = 4k^2 = 4a+0 for some integer a = k 2 a = k^2
( 2 k + 1 ) 2 = 4 k 2 + 4 k + 1 = 4 ( k 2 + k ) + 1 = 4 b + 1 ( 2k+1 )^2 = 4k^2+4k+1 = 4 \left( k^2+k \right) +1 = 4b+1 for some integer b = k 2 + k b = k^2+k
So the square of an even number is always divisible by 4 4 and the square of an odd number is always one more than a multiple of 4 4 .
Henry U - 2 years, 7 months ago
You have a typo in the soln you have written 2 in place of 4
The Undescribed - 2 years, 7 months ago
Log in to reply
Thanks for noting that, I think I have corrected the place you meant
Henry U - 2 years, 7 months ago
Sorry, if this is obvious, but why did you divide by four? If every square number mod 4 is either 0 or 1, evenly dividing by 4 means that the number has a chance to be true square number. But you factored out a 4 and then divided by 4 again? I'm a little confused as to why you needed to do this...
Wilson Guo - 2 years, 7 months ago
Log in to reply
A square number can be written as, if even, (2c)^2 and odd as (2c+1)^2. Divinding by 4 we get: (2c)^2/4 = c^2 => zero remainder (2c+1)^2/4 = (4c^2+4c+1)/4 => we can see that in this case we obtain 1 as remainder.
filipe santos - 2 years, 7 months ago
But what about sqrt(1444...44)? [Ps. It's 1 following by 77 digits of 4]
** If my calc working properly, It's 2^(380058475033045993045349675188918159692)
Log in to reply
That's a nice near-miss you've found there. According to Wolphram Alpha , it's only 0.00177 away from the next square number. But it's not a perfect square.
Henry U - 2 years, 6 months ago
Alon Amit
Nov 4, 2018
The numbers of this form (starting at 144) are all divisible by 4, but after dividing them by 4, the quotient mod 8 quickly stabilizes: it is 4, 1, 3, 7, 7, 7... as can easily be checked. Neither 3 nor 7 are quadratic residues mod 8, so the answer is No.
Matthew Feig
Nov 7, 2018
Consider the remainders of such numbers after dividing by 16. A quick check tells us that perfect squares are always congruent to 0 \color{#20A900} 0 , 1 \color{#20A900} 1 , 4 \color{#20A900} 4 , or 9 \color{#20A900} 9 modulo 16.
14 14 mod 16 144 0 1 , 444 4 14 , 444 12 144 , 444 12 1 , 444 , 444 12 \begin{aligned} 14 &\equiv \color{#D61F06}14 \ \ \color{#333333} \text{mod} \ 16 \\ 144 &\equiv \ \ \color{#20A900}0 \\ 1,444 &\equiv \ \ \color{#20A900}4 \\ 14,444 &\equiv \color{#D61F06}12 \\ 144,444 &\equiv \color{#D61F06}12 \\ 1,444,444 &\equiv \color{#D61F06}12 \end{aligned}
Note that 10 , 000 = 16 × 625 10,000 = 16 \times 625 , so every number from 14 , 444 14,444 on is congruent to 4 , 444 12 4,444 \equiv \color{#D61F06}12 mod 16, and thus not a perfect square.
[This train of thought was inspired by Henry U factoring out 2 2 = 4 2^2 = 4 and then considering divisibility by 4.]
How does one check what perfect squares are congruent to modulo 16? Or any number, for that matter?
Madhur Agrawal - 2 years, 7 months ago
Log in to reply
Look up modular arithmetic . There are only 16 possible remainders when you divide by 16: 0, 1, 2, ..., 14, 15. Just look at the squares of those numbers (so the perfect squares from 0 to 225) and see what the remainders are when dividing by 16. It doesn't take too long, and you see a lot of repeated values.
0 2 = 0 = 16 × 0 + 0 8 2 = 64 = 16 × 4 + 0 0^2 = \ \ 0 = 16 \times 0 + \color{#20A900} 0 \ \ \ \ \ \ \ \ \color{#333333} \ \ 8^2 = \ \ 64 = 16 \times \ 4 + \color{#20A900} 0
1 2 = 1 = 16 × 0 + 1 9 2 = 81 = 16 × 5 + 1 1^2 = \ \ 1 = 16 \times 0 + \color{#20A900} 1 \ \ \ \ \ \ \ \ \color{#333333} \ \ 9^2 = \ \ 81 = 16 \times \ 5 + \color{#20A900} 1
2 2 = 4 = 16 × 0 + 4 1 0 2 = 100 = 16 × 6 + 4 2^2 = \ \ 4 = 16 \times 0 + \color{#20A900} 4 \ \ \ \ \ \ \ \ \color{#333333} 10^2 = 100 = 16 \times \ 6 + \color{#20A900} 4
3 2 = 9 = 16 × 0 + 9 1 1 2 = 121 = 16 × 7 + 9 3^2 = \ \ 9 = 16 \times 0 + \color{#20A900} 9 \ \ \ \ \ \ \ \ \color{#333333} 11^2 = 121 = 16 \times \ 7 + \color{#20A900} 9
4 2 = 16 = 16 × 1 + 0 1 2 2 = 144 = 16 × 9 + 0 4^2 = 16 = 16 \times 1 + \color{#20A900} 0 \ \ \ \ \ \ \ \ \color{#333333} 12^2 = 144 = 16 \times \ 9 + \color{#20A900} 0
5 2 = 25 = 16 × 1 + 9 1 3 2 = 169 = 16 × 10 + 9 5^2 = 25 = 16 \times 1 + \color{#20A900} 9 \ \ \ \ \ \ \ \ \color{#333333} 13^2 = 169 = 16 \times 10 + \color{#20A900} 9
6 2 = 36 = 16 × 2 + 4 1 4 2 = 196 = 16 × 12 + 4 6^2 = 36 = 16 \times 2 + \color{#20A900} 4 \ \ \ \ \ \ \ \ \color{#333333} 14^2 = 196 = 16 \times 12 + \color{#20A900} 4
7 2 = 49 = 16 × 3 + 1 1 5 2 = 225 = 16 × 14 + 1 7^2 = 49 = 16 \times 3 + \color{#20A900} 1 \ \ \ \ \ \ \ \ \color{#333333} 15^2 = 225 = 16 \times 14 + \color{#20A900} 1
Matthew Feig - 2 years, 7 months ago
Log in to reply
Oh interesting, I had the same question! But how does one think of “Hmm.. lemme check the residue mod 16? lol just asking
Golden Boy - 2 years, 5 months ago
Antonio O
Nov 8, 2018
Since any number 1444 444 1444\ldots 444 is even, it must be the square of an even number, i.e. 1444 444 = ( 2 k ) 2 = 4 k 2 1444\ldots 444 = (2k)^2 = 4k^2 .
Thus any 1444 444 1444\ldots 444 is a multiple of 4 4 .
In our case 144 = 1 2 2 = ( 2 6 ) 2 = 4 36 144 = 12^2 = (2\cdot 6)^2 = 4\cdot 36
and 1444 = 3 8 2 = ( 2 19 ) 2 = 4 361 1444 = 38^2 = (2\cdot 19)^2 = 4\cdot 361 .
Now let's take a look at 1444 444 1444\ldots 444 numbers with more than two 4 4 digits: by dividing any such number by 4 4 we get 361 111 361\ldots 111 , in fact
144 4 444 4 = 1440 000 4 + 4 444 4 = 360 000 + 1 111 = 36 1 111 \frac{{\color{#0000ff}{144}}4\ldots 444}{4} = \frac{{\color{#0000ff}{1440\ldots 000}}}{4} + \frac{4\ldots 444}{4} = {\color{#0000ff}{360\ldots 000}} + 1\ldots 111 = {\color{#0000ff}{36}} 1\ldots 111 .
Any 361 111 361\ldots 111 is an odd number, thus it must be the square of an odd number, i.e. 361 111 = ( 2 l + 1 ) 2 = 4 l 2 + 4 l + 1 361\ldots 111 = (2l+1)^2 = 4l^2 + 4l +1 .
By rearranging we see that 361 110 = 4 l 2 + 4 l = 4 ( l 2 + l ) 361\ldots 110 = 4l^2 + 4l = 4(l^2 + l)
i.e. any number on the left must be a multiple of 4 4 ; but any such number must be divisible by 4 4 , thus it must have the last two digits divisible by 4 4 , which in our case is never true for 10 10 is not divisible by 4 4 .
Summing up: we have just proven that any 361 111 361\ldots 111 (with two or more 1 1 digits) can't be the square of an integer number k k (see first line), so the answer is that there are no other 1444 444 1444\ldots 444 that are perfect squares.
Aditya Ghosh
Nov 10, 2018
Let 144...4 (with n fours) be a perfect square. Since its even, we may let 4k^2 = 144...4=10^n +44...4. This implies, k^2 = 25×10^(n-2) + 11...1. Now, if n is greater than 3, then 25×10^(n-2) is a multiple of 4 and the remaining 111....1=11...100+11, which is of the form 4a+3. So the whole RHS is of the form 4b+3, which can never be a perfect square.
K T
Nov 9, 2018
If x 2 = 14...4 x^2=14...4 then x 2 4 = ( x + 2 ) ( x 2 ) x^2-4=(x+2)(x-2) is a multiple of 10, so x is even and x = + / 2 ( m o d 5 ) x=+/-2(mod 5) : combined the last digit of x must be 2 or 8. Below, a, b and c are nonnegative integers.
1) Suppose x = 10 a + 2 x=10a+2 . Then x 2 = 100 a 2 + 40 a + 4 x^2 = 100a^2+40a+4 \implies a must end on 1 or 6 \implies x must end on 12 or 62
1.1) Suppose x = 100 b + 12 x =100b+12 . Then x 2 = 10000 b 2 + 2400 b + 144 x^2 = 10000b^2+2400b+144 , implying that the 3rd digit from the right in x 2 x^2 is odd. The only possibility satisfying the pattern is x 2 = 144 x^2 = 144 .
1.2) Suppose x = 100 b + 62 x =100b+62 . Then x 2 = 10000 b 2 + 12400 b + 3844 x^2 =10000b^2+12400b+3844 . Now 4b+8 must end on 4, so b must end on a 4 or a 9, so x must end on 462 or 962.
1.2.1) Suppose x = 1000 c + 462 x=1000c+462 . Then x 2 = 1000000 c 2 + 2 × 1000 × 462 × c + 213444 x^2=1000000c^2+2×1000×462×c+213444
1.2.2) Suppose x = 1000 c + 962 x=1000c+962 . Then x 2 = 1000000 c 2 + 2 × 1000 × 962 × c + 925444 x^2=1000000c^2+2×1000×962×c+925444
For both these cases there is no solution because the digit at position 4 from the right is odd (and it cannot be the leading 1 since x 2 > 1444 x^2 > 1444 ).
Summarizing: The only solution for possibility 1) is x = 12 x=12 .
2) Suppose x ends on an 8, so x = 10 a + 8 x=10a+8 . Then x 2 = 100 a 2 + 160 a + 64 x^2 = 100a^2+160a+64 \implies 6a+6 must end on a 4 \implies 6a ends on an 8 \implies a ends on a 3 or 8. \implies x ends on 38 or 88
2.1) suppose that x ends on 38. x 2 = ( 100 b + 38 ) 2 = 10000 b 2 + 7600 b + 1444 x^2=(100b+38)^2=10000b^2+7600b+1444 . Possibilities are: b = 0 b=0 or b = 5 ( m o d 10 ) b=5 (mod 10) \implies x ends on 038 or 538
2.1.1) if x ends on 038, the 4th position from the right is odd so b = 0 b=0 \implies x = 38 x=38 is the only solution.
2.1.2) if x ends on 538, x 2 = ( 1000 c + 538 ) 2 = 1000000 c 2 + 1076 c + 289444 x^2=(1000c+538)^2=1000000c^2+1076c+289444 . Here too, position 4 from the right is odd and x 2 > 1444 x^2 >1444 .
2.2) suppose that x ends on 88. x 2 = ( 100 b + 88 ) 2 = 10000 b 2 + 17600 b + 7744 x^2=(100b+88)^2=10000b^2+17600b+7744 . Position 3 from the right is odd and x 2 > 144 x^2 > 144 .
Summarizing: The only solution for possibility 2) is x=38, and 12 and 38 are the only solutions for x.
William Kennedy
Nov 10, 2018
I was lazy and wrote a computer program. I tested it to 100,000,000 numbers and decided that I was pretty sure the answer was no.
function checkNumber (n) {
var arg = n.toString();
var arr = arg.split("");
if (arr[0] === "1") {
var counter = 0;
for (var i = 1; i < arr.length; i++) {
if (arr[i] === "4") {
counter++;
}
}
if (counter === arr.length - 1) {
return true
} else {
return false
}
} else {
return false;
}
}
var check = 0;
for (var i = 4; i <= 100000000; i++) {
if (checkNumber(i*i)) {
check++;
}
}
console.log(check)
Roneel V.
Nov 10, 2018
python for i in range(10000): for j in range(200): if i*i==int('1'+'4'*j): print(i,flush=1) I used brute force.
I don't know much about python and runtime, but I think you could also have used
1
floor(sqrt(int('1'+'4'*j)))=sqrt(int('1'+'4'*j))
That way, you generate the number, take its square root and check if it is an integer.
Henry U - 2 years, 7 months ago
0 pending reports
×
Problem Loading...
Note Loading...
Set Loading...
|
__label__pos
| 0.995545 |
Skip to main content
Sample pairs of particles according to a discrete Gaussian distrbution
Project description
# Sample pairs of particles according to a discrete Gaussian Python code to sample pairs of a given set of particles in n dims, where the probability for each pair is Gaussian
<img src=”examples/figures/sample_2d_counts.png” width=”500”>
## Requirements
Python 3 & Numpy.
## Installation and usage
Use pip: ` pip install samplePairsGaussian ` or manually: ` python setup.py install ` and in your code: ` from samplePairsGaussian import * ` See also the [examples](examples) folder.
## Idea
Given a set of n particles with positions in d-dimensional space denoted by x_i for i=0,1,…,n.
We want to sample a pair of particles i,j where i =/= j, where the probability for sampling this pair is given by: ` p(i,j) ~ exp( - |x_i - x_j|^2 / 2 sigma^2 ) ` where we use |x| to denote the L_2 norm, and sigma is some chosen standard deviation.
This problem is easy to write down, but difficult to implement for large numbers of particles since it requires computing N^2 distances.
A further problem is that we may want to:
1. Add a particle.
2. Remove a particle.
3. Move a particle.
In this case, not all distances are affected - these operations should be of order N. However, if we sample the discrete distribution by forming the CDF, we will need to recalculate it, which is expensive. Alternatively, if we use rejection sampling, we must have a good candidate (envelope) distribution such that the acceptance ratio is high.
This library attempts to come up with the most efficient way to perform these operations in Python.
A key way this library reduces computational cost is by introducing a cutoff for particle distances, where pairs of particles separated by a distance greater than the cutoff are not considered for sampling. It is natural to let this be some chosen multiple of the std. dev., i.e. m*sigma for some m. If we use rejection sampling where candidates are drawn from a uniform distribution, the acceptance ratio should be approximately ( sqrt(2 * pi) * sigma ) / ( 2 * m * sigma ) = 1.253 / m. (in the first equation: the area of the Gaussian is 1, divided by the area of the uniform distribution of width 2 * m * sigma and height 1 / (sqrt(2 * pi) * sigma )).
In general, we avoid all use of for loops, and rely extensively on array operations using numpy.
### Multiple species
Multiple species are also supported, where we have multiple species but want to draw two particles of the same species (two particles of any species can be done by simply ignoring the species labels).
Specifically, the classes ProbCalculatorMultiSpecies and SamplerMultiSpecies implement this.
## Examples
See the [examples](examples) folder.
Project details
Download files
Download the file for your platform. If you're not sure which to choose, learn more about installing packages.
Source Distribution
SamplePairsGaussian-1.1.tar.gz (12.1 kB view hashes)
Uploaded Source
Built Distribution
SamplePairsGaussian-1.1-py3.7.egg (45.8 kB view hashes)
Uploaded Source
Supported by
AWS AWS Cloud computing and Security Sponsor Datadog Datadog Monitoring Fastly Fastly CDN Google Google Download Analytics Microsoft Microsoft PSF Sponsor Pingdom Pingdom Monitoring Sentry Sentry Error logging StatusPage StatusPage Status page
|
__label__pos
| 0.989523 |
The Houseware Platform
Learn about the modular components in Houseware, and how they enable a platform-first approach for the Houseware experience!
Overview
Houseware is a no-code, warehouse-native platform designed to streamline product analytics workflows.
Houseware eliminates the need for complex data transfers and modeling by working directly within your data warehouse. The platform-first approach ensures a modular, scalable, and secure solution that can be tailored to various analytical needs. This architecture allows for easy integration with multiple data sources, providing flexibility and a promise of accuracy to the product teams while maintaining strong governance and observability on the data warehouse for data teams.
By leveraging the center of gravity of data in your organization, Houseware offers real-time insights and comprehensive analytics capabilities. The platform's modular components enable customization and extensibility, making it suitable for diverse use cases. With a strong emphasis on security and compliance, Houseware ensures that your data is protected while delivering actionable insights.
Building Blocks
Here are the key building blocks that power the Houseware Platform
• Events: Represent unique user actions on your product.
• Event Properties: Attributes that describe events.
• Users: Unique individuals performing events on your product.
• User Properties: Attributes that describe users.
• Cohorts: Specific groups of users segmented by properties or behaviors.
• Entities: Fundamental components representing distinct objects or items within your product ecosystem.
• Event Schema: The structured format required to make events and user data available for analysis in Houseware.
For detailed information on each building block, please refer to our comprehensive documentation here.
Composability
Houseware's platform is designed for composability, allowing you to build a flexible and scalable analytics environment that meets your specific needs. This modular approach allows individual components to be integrated, extended, and replaced without affecting the overall system, ensuring seamless compatibility with your existing data infrastructure. This approach ensures that Houseware can integrate seamlessly into your existing data infrastructure and grow with your business.
We follow the principle of "Configuration over Convention", allowing maximum flexibility to the end users to best design their use cases and solve their specific challenges:
• Choice of Warehouse: Houseware supports integration with multiple data warehouses, providing flexibility and scalability for your analytics needs. Whether you are using Snowflake, BigQuery, or another data warehouse solution, Houseware ensures a consistent product experience.
• Flexible Schema: Houseware’s schema is designed to be customizable to most event streams and flexible to any frequency, whether batch or real-time. This ensures that your data is always ready for analysis regardless of how it is ingested.
• Warehouse-Native: Leverage the center of gravity of data in your organization by working natively within your existing data warehouse. This enables product teams to derive cross-functional insights by merging disparate data sources like clickstream events, transactional data, and SaaS datasets, all on the trusted data warehouse.
• Customizable: Houseware allows users to build custom events, cohorts, and other analytics components tailored to their specific needs. This customization ensures that the analytics setup aligns perfectly with the business requirements.
• Integrations: The composable platform approach enables inbound integrations(with CDPs/transaction DBs) and writeback syncs with a wide range of marketing, sales, and business tools, allowing for a best-of-breed approach to building your stack. These integrations can be managed programatically, allowing developers to build unique workflows.
• Access Controls: Access controls in Houseware are defined once and applied everywhere up the stack, ensuring consistent and secure data access across the platform. This centralized approach simplifies the management of permissions and enhances security, allowing organizational admins to build their own access policies.
Data Platform
Houseware's data platform is designed to provide a robust, scalable, and secure environment for product analytics. A warehouse-native architecture demanded a unique perspective at building multiple workloads, that worked agnostic of the storage layer or the query engines.
Listed below are some of the workloads supported by the Data Platform
Data Schema
Houseware employs an activity schema-esque data schema that organizes events into a set of mandatory and optional attributes. This enables:
• Scalability: Efficiently handles large volumes (up to tens of billions) of event data at runtime. The schema supports high-throughput data ingestion and processing, maintaining robust performance as data scales.
• Flexibility: Accommodates new event types and attributes without requiring schema changes. This allows for seamless integration of new data points as the product evolves, avoiding extensive redesigns.
• Simplicity: The schema is user-friendly and adaptable to various scenarios, making it easy to model and query data. It minimizes the learning curve for users and simplifies data management.
• Efficiency: Optimizes query performance by structuring data for fast retrieval and analysis. The schema supports efficient indexing, clustering and partitioning, which speeds up query execution times.
• Consistency: Ensures uniform data formatting and structuring across the organization. This standardization simplifies data integration, governance, and compliance.
• Reduced Complexity: Minimizes the complexity of data transformations by using a straightforward schema that maps easily to various data sources. This simplification accelerates the ETL process and reduces the potential for errors.
Metadata:
Metadata in Houseware refers to data about events/entities. It serves several critical functions:
• Query Optimization: Enhances query performance by providing statistics, indexing information, and other query-related metadata. This allows the query engine to make informed decisions about query execution plans, improving efficiency and reducing execution times.
• Incremental Processing: Tracks changes and updates in data, enabling efficient incremental query processing. By identifying and processing only the data that has changed since the last update, it minimizes data reprocessing and reduces load on the data warehouse.
• Enhanced User Experience: Metadata allows users to know specific details about their event names, event dimensions, user properties, date ranges, hence informing the users about the available options for analysis.
Key Metadata Components
• Dimensions: Metadata about event attributes that are first-class columns in the events table. This includes information such as the dimension name, possible values, data type, and cardinality.
• Event Names and Properties: Metadata about event names and their associated properties, including property values, data types, and cardinality. This is stored in a dedicated table to facilitate efficient querying and analysis.
• Event Time Range: Metadata about the earliest and latest timestamps of events, allowing for time-based filtering and analysis. This is essential for temporal queries and understanding data recency.
• First-Time Event Mapping: Tracks the first occurrence of specific events for each user, enabling analysis of first-time user actions. This mapping is crucial for understanding user onboarding and initial interactions.
• ID Mapping: Maps device identifiers to user identifiers, resolving multiple occurrences of events from different devices to a single user. This ensures accurate user tracking and analysis.
• Custom Enrichment: Allows users to add custom metadata such as event tags, descriptions, and dimension types, enhancing the analytical capabilities and contextual understanding of the data.
Real-Time Analytics
Realtime analytics in Houseware enables immediate data processing and analysis, providing up-to-date insights as data is generated. This capability is essential for applications requiring timely decision-making based on the latest events.
Realtime analytics is achieved through continuous data ingestion, processing, and querying. Data is streamed from various sources into the data warehouse, where it is immediately available for analysis.
Architecture
The reference architecture for realtime analytics in Houseware, supporting both BigQuery and Snowflake, is as follows:
• Data Ingestion: Data is continuously ingested from various sources, such as user interactions, IoT devices, or application logs. This data is sent to a cloud storage service like Google Cloud Storage (GCS) or AWS S3.
• Data Processing Pipeline:
• GCP DataFlow: For environments using BigQuery, GCP DataFlow processes the ingested data in real-time, applying necessary transformations such as parsing, filtering, and enrichment.
• AWS Kinesis / Apache Kafka: For environments using Snowflake, streaming data is processed using AWS Kinesis or Apache Kafka with AWS Lambda or Apache Flink for transformations.
• Metadata Update: As new data arrives, metadata in the warehouse is updated in real-time. This includes updating dimension tables, event names, and property values, ensuring that the latest data is available for querying.
Caching
Houseware employs multiple caching strategies to enhance performance for end users of the platform:
• Application Cache: Caches visualization queries and metadata requests to reduce load on the warehouse and improve response times. Visualization queries are cached for 365 days, while metadata queries are cached for 1 hour.
• Warehouse Cache: Utilizes native warehouse caching mechanisms (e.g., Snowflake's result cache, local disk cache, and remote disk cache) to speed up queries.
Deployment models for "Query Compute"
Houseware offers flexible deployment models to suit different organizational needs. While the customer connects their own data warehouse, they still have the flexibility to choose between the following models:
Bring-your-own-Compute: In the Bring-your-own-Compute model, customers utilize their existing data warehouse compute resources for query execution. This model provides several advantages:
• Control: Customers maintain full control over their compute resources, allowing them to manage performance, scaling, and cost directly.
• Optimization: Enables customers to optimize query execution based on their specific workload and performance requirements.
• Integration: Seamlessly integrates with existing data governance and security policies within the customer's environment.
Houseware-Compute: In the Houseware-Compute model, Houseware leverages "Snowflake Secure Data Sharing" or "BigQuery Data Sharing" to simplify the compute process for customers. In this model, Houseware absorbs the compute costs and abstracts it away from the customer. Key benefits include:
• Simplified Management: Customers do not need to manage or optimize compute resources, as Houseware handles all aspects of query execution.
• Cost Efficiency: Houseware absorbs the compute costs, providing a predictable pricing model and reducing the customer's operational overhead.
Identity Resolution
Houseware provides advanced identity resolution capabilities to unify user identities across different data sources, helpful for digital applications where non-loggedin user behavior is critical to their product flows.
• Device and User ID Mapping: Houseware's algorithms map device identifiers to user identifiers to track user activities across multiple devices. This mapping enables accurate tracking and analysis of user behavior, even when users switch between devices, ensuring accurate user insights.
• First-Time Event Mapping: Houseware identifies the first occurrence of specific events for each user. This mapping helps in understanding user onboarding and initial interactions with the product, which is crucial for optimizing user experience and engagement strategies.
Cost Management (Budgeting, Quotas, and Cost Observability)
Houseware offers robust tools to manage and observe costs associated with data processing and analytics. These capabilities ensure that customers can maintain fine-grained control over their data warehouse expenditures while optimizing query performance.
• Cost Observability: Costs are broken down by specific query patterns, allowing customers to see which queries or analytical operations are consuming the most resources. The platform continuously monitors compute costs and detects anomalies. If there is an unusual spike or unexpected pattern in costs, alerts are sent to the designated Slack channels, enabling prompt investigation and resolution.
• Budgeting & Quotas: Customers can define usage quotas and budgets directly on their Data Warehouse(Snowflake/BigQuery). Budgets can be defined for overall data warehouse usage as well as for specific query types or runtimes. This helps in tracking and controlling spending, ensuring that costs remain within the planned limits.
Query Observability
The Houseware Data Platform has internal query observability to monitor and optimize query performance. These capabilities ensure that queries run efficiently, resource usage is optimized, and any issues are quickly identified and resolved.
Query Logging Houseware tracks and logs all query executions, providing a detailed record of each query's lifecycle:
• Execution Tracking: Logs the start time, end time, and duration of each query, along with the query text and execution context.
• Error Logging: Captures any errors or exceptions that occur during query execution, facilitating troubleshooting and root cause analysis.
• User and Session Information: Logs metadata about the user and session that initiated the query, helping to correlate queries with specific users or applications.
These logs provide a comprehensive history of query activity, which is essential for both performance analysis and security auditing.
Performance Metrics: Houseware collects detailed performance metrics for each query, enabling the identification and resolution of performance bottlenecks:
• Resource Utilization: Tracks metrics such as CPU usage, memory consumption, and I/O operations for each query.
• Execution Plan Analysis: Analyzes the execution plan of each query to identify inefficiencies, such as suboptimal joins or missing indexes.
• Latency Metrics: Measures the response times and latency of queries to ensure they meet performance expectations.
API-first Platform
Houseware is built with a "platform-first" approach, offering a robust and flexible foundation that can be extended to accommodate a wide range of use cases. The APIs are designed to support the Product teams' use cases via the Houseware UI, but are also extensible to build custom analytics interfaces, integrate with, or develop entirely new applications.
All Use Cases, available via APIs: Houseware provides a wide array of API endpoints that cover all essential product analytics use cases:
• Event Tracking: Endpoints to ingest and query user events in real-time, allowing for immediate visibility into user interactions.
• Funnel Analysis: APIs to create and analyze conversion funnels, helping teams understand user drop-off points and optimize conversion rates.
• Retention Analysis: Endpoints for building retention cohorts and tracking user engagement over time, aiding in the measurement of user stickiness and long-term engagement.
• Cohort Analysis: APIs to define, manage, and sync user cohorts based on behaviors and properties, enabling targeted analysis and personalized user experiences.
• Advanced Use Cases: Aside from the aforementioned use cases like "funnels", "retention", "cohorts", there are furthermore use cases that are available:
• Aggregations on Properties: Perform advanced aggregations on event and user properties to derive detailed insights.
• First-time User Analysis: Identify and analyze the first-time occurrence of specific events for users.
• User Activity Lookup: Retrieve detailed records of user activities for a specific user, searchable by a custom identifier
• Joins : Execute complex queries that join events and entities for cross-functional product analysis.
Ease of Integration: Houseware APIs are designed to be easily integrated with your existing systems and applications:
• Authentication: Supports both API Key and JWT (JSON Web Token) authentication methods to secure access and ensure that only authorized users and applications can interact with the API.
• Documentation: Comprehensive and clear API documentation is provided to guide developers through integration, including example requests and responses, parameter descriptions, and use case scenarios.
To get started with the Houseware Platform API, refer to our detailed guide here.
Security
Houseware prioritizes security to provide a safe platform for viewing, exploring, and building your product analytics workflows. We adhere to global industry standards to ensure the highest level of consumer data protection.
• Bring-Your-Own-Warehouse: Houseware operates on a bring-your-own-warehouse model, where only user and application configuration data is stored by Houseware. All clickstream events and datasets remain within the customer's data warehouse, ensuring that sensitive data never leaves your controlled environment.
• Governance: Houseware implements a minimum-access policy, granting users only the permissions necessary to perform their tasks. Our multi-tenant architecture ensures complete segregation of data and execution environments between clients, providing robust data isolation and security.
• SOC 2 Type II Compliance: Houseware is SOC 2 Type II compliant, with annual renewals and rigorous security practices.
Read here to know more: Houseware's Security and Compliance Principles
🚀 By understanding these building blocks and leveraging Houseware’s API's flexibility, you can drive insightful and secure product analytics that adapt to your evolving business needs.
|
__label__pos
| 0.646197 |
Cloudflare Docs
KV
Visit KV on GitHub
Set theme to dark (⇧+D)
Read key-value pairs
To get the value for a given key, call the get() method on any KV namespace you have bound to your Worker code:
NAMESPACE.get(key);
The get() method returns a promise you can await on to get the value. If the key is not found, the promise will resolve with the literal value null.
The get() method may return stale values. If a given key has recently been read in a given location, changes to the key made in other locations may take up to 60 seconds to display.
An example of reading a key from within a Worker:
export default {
async fetch(request, env, ctx) {
const value = await env.NAMESPACE.get("first-key");
if (value === null) {
return new Response("Value not found", { status: 404 });
}
return new Response(value);
},
};
You can read key-value pairs from the command line with Wrangler and from the API.
Types
You can pass in an options object with a type parameter to the get() method:
NAMESPACE.get(key, { type: "text" });
The type parameter can be any of the following:
For simple values, use the default text type which provides you with your value as a string. For convenience, a json type is also specified which will convert a JSON value into an object before returning the object to you. For large values, use stream to request a ReadableStream. For binary values, use arrayBuffer to request an ArrayBuffer.
For large values, the choice of type can have a noticeable effect on latency and CPU usage. For reference, the type can be ordered from fastest to slowest as stream, arrayBuffer, text, and json.
CacheTTL parameter
cacheTTL is a parameter that defines the length of time in seconds that a KV result is cached in the global network location it is accessed from.
Defining the length of time in seconds is useful for reducing cold read latency on keys that are read relatively infrequently. cacheTTL is useful if your data is write-once or write-rarely.
cacheTTL is not recommended if your data is updated often and you need to see updates shortly after they are written, because writes that happen from other global network locations will not be visible until the cached value expires.
The get() options object also accepts a cacheTTL parameter:
NAMESPACE.get(key, { cacheTTL: 3600 });
The cacheTTL parameter must be an integer greater than or equal to 60, which is the default.
The effective cacheTTL of an already cached item can be reduced by getting it again with a lower cacheTTL. For example, if you did NAMESPACE.get(key, {cacheTTL: 86400}) but later realized that caching for 24 hours was too long, you could NAMESPACE.get(key, {cacheTTL: 300}) or even NAMESPACE.get(key) and it would check for newer data to respect the provided cacheTTL, which defaults to 60 seconds.
Metadata
A metadata is a serializable value you append to each KV entry.
Get the metadata associated with a key-value pair alongside its value by calling the getWithMetadata() method on a KV namespace you have bound in your Worker code:
const { value, metadata } = await NAMESPACE.getWithMetadata(key);
If there is no metadata associated with the requested key-value pair, null will be returned for metadata.
You can pass an options object with type and/or cacheTTL parameters to the getWithMetadata() method, similar to get().
|
__label__pos
| 0.736525 |
创世智能标题-滚动隐藏但向上滚动显示
在WordPress Genesis中实现智能粘性页眉的步骤
粘性页眉是一个很好的方式，让访问者在页面或文章的中间浏览您的网站。它帮助读者进一步浏览页眉菜单。
但是有一个问题。
粘性页眉在滚动时始终显示在屏幕上。这没问题，但是如果在小设备上浏览，则会影响可读性。
那么解决方案是什么呢？
一个智能页眉 – 向下滚动时隐藏，但在尝试向上滚动时立即显示。听起来很酷吧？
我已经在Authority Pro主题中测试过了。备份现有文件，以便在出现问题时可以恢复。
functions.php文件中添加以下内容
//* 智能页眉功能
add_action('wp_footer','yaoweibin_header_sticky_script');
function yaoweibin_header_sticky_script()
{
?>
<?php
}
并在style.css文件中添加以下内容
/* 智能页眉 */
header.site-header {
position: fixed;
top: 0;
transition: top 0.3s ease-in-out;
width: 100%;
z-index: 9;
left: 0;
right: 0;
}
header.site-header.shadow {
-webkit-box-shadow: 0 0 50px rgba(0,0,0,.15);
box-shadow: 0 0 50px rgba(0,0,0,.15);
}
body.admin-bar header.site-header{
top: 32px;
}
@media only screen and (max-width: 780px)
{
body.admin-bar header.site-header{
top: 46px;
}
}
刷新页面以查看结果。如果有缓存，请不要忘记清除缓存。
你喜欢这个小优化吗？
类似文章
|
__label__pos
| 0.993603 |
Overview
Writing proper tests can assist in writing better software. If you set up proper test cases you can eliminate most functional bugs and better maintain your software.
Integrating PHPUnit with Phalcon
If you don’t already have phpunit installed, you can do it by using the following composer command:
composer require phpunit/phpunit:^5.0
or by manually adding it to composer.json:
{
"require-dev": {
"phpunit/phpunit": "^5.0"
}
}
Once PHPUnit is installed create a directory called tests in project root directory:
app/
public/
tests/
Next, we need a ‘helper’ file to bootstrap the application for unit testing.
The PHPUnit helper file
A helper file is required to bootstrap the application for running the tests. We have prepared a sample file. Put the file in your tests/ directory as TestHelper.php.
<?php
use Phalcon\Di;
use Phalcon\Di\FactoryDefault;
use Phalcon\Loader;
ini_set("display_errors", 1);
error_reporting(E_ALL);
define("ROOT_PATH", __DIR__);
set_include_path(
ROOT_PATH . PATH_SEPARATOR . get_include_path()
);
// Required for phalcon/incubator
include __DIR__ . "/../vendor/autoload.php";
// Use the application autoloader to autoload the classes
// Autoload the dependencies found in composer
$loader = new Loader();
$loader->registerDirs(
[
ROOT_PATH,
]
);
$loader->register();
$di = new FactoryDefault();
Di::reset();
// Add any needed services to the DI here
Di::setDefault($di);
Should you need to test any components from your own library, add them to the autoloader or use the autoloader from your main application.
To help you build the Unit Tests, we made a few abstract classes you can use to bootstrap the Unit Tests themselves. These files exist in the Phalcon Incubator.
You can use the Incubator library by adding it as a dependency:
composer require phalcon/incubator
or by manually adding it to composer.json:
{
"require": {
"phalcon/incubator": "^3.0"
}
}
You can also clone the repository using the repo link above.
The phpunit.xml file
Now, create a phpunit.xml file as follows:
<?xml version="1.0" encoding="UTF-8"?>
<phpunit bootstrap="./TestHelper.php"
backupGlobals="false"
backupStaticAttributes="false"
verbose="true"
colors="false"
convertErrorsToExceptions="true"
convertNoticesToExceptions="true"
convertWarningsToExceptions="true"
processIsolation="false"
stopOnFailure="false"
syntaxCheck="true">
<testsuite name="Phalcon - Testsuite">
<directory>./</directory>
</testsuite>
</phpunit>
Modify the phpunit.xml to fit your needs and save it in tests. This will run any tests under the tests directory.
Sample Unit Test
To run any Unit Tests you need to define them. The autoloader will make sure the proper files are loaded so all you need to do is create the files and phpunit will run the tests for you.
This example does not contain a config file, most test cases however, do need one. You can add it to the DI to get the UnitTestCase file.
First create a base Unit Test called UnitTestCase.php in your tests directory:
<?php
use Phalcon\Di;
use Phalcon\Test\UnitTestCase as PhalconTestCase;
abstract class UnitTestCase extends PhalconTestCase
{
/**
* @var bool
*/
private $_loaded = false;
public function setUp()
{
parent::setUp();
// Load any additional services that might be required during testing
$di = Di::getDefault();
// Get any DI components here. If you have a config, be sure to pass it to the parent
$this->setDi($di);
$this->_loaded = true;
}
/**
* Check if the test case is setup properly
*
* @throws \PHPUnit_Framework_IncompleteTestError;
*/
public function __destruct()
{
if (!$this->_loaded) {
throw new \PHPUnit_Framework_IncompleteTestError(
"Please run parent::setUp()."
);
}
}
}
It’s always a good idea to separate your Unit Tests in namespaces. For this test we will create the namespace ‘Test’. So create a file called tests\Test\UnitTest.php:
<?php
namespace Test;
/**
* Class UnitTest
*/
class UnitTest extends \UnitTestCase
{
public function testTestCase()
{
$this->assertEquals(
"works",
"works",
"This is OK"
);
$this->assertEquals(
"works",
"works1",
"This will fail"
);
}
}
Now when you execute phpunit in your command-line from the tests directory you will get the following output:
$ phpunit
PHPUnit 3.7.23 by Sebastian Bergmann.
Configuration read from /var/www/tests/phpunit.xml
Time: 3 ms, Memory: 3.35Mb
There was 1 failure:
1) Test\UnitTest::testTestCase
This will fail
Failed asserting that two strings are equal.
--- Expected
+++ Actual
@@ @@
-'works'
+'works1'
/var/www/tests/Test/UnitTest.php:25
FAILURES!
Tests: 1, Assertions: 2, Failures: 1.
Now you can start building your Unit Tests. You can view a good guide here. We also recommend reading the PHPUnit documentation if you’re not familiar with PHPUnit.
|
__label__pos
| 0.539324 |
next up previous contents index
Next: The Misc Config Button: Up: Extraction System: Setup and Previous: Device Templates Contents Index
Format Library File
Xic provides a mechanism for user-specified formatting of physical and electrical netlist output. Such formatting is generated by scripts found in a file named ``xic_format_lib''. This file need not exist if user formatting is not required.
An example xic_format_lib file is included in the distributions. This provides two examples each, for physical and electrical output. In either case, the first example is the Cadence Design Exchange Format (DEF), which is an industry-standard ASCII netlist format. The second format in each case is a simple example, not a ``real'' format. The example library is found in the startup directory, and can be used as-is or as a starting point for customization. The example format scripts include instructive comments.
The xic_format_lib file is searched for in the library search path, and the first such file found will be used.
There are three types of script that can appear in the file: those for generating netlists from physical data, those that generate netlists from electrical data, and those that format the output of LVS runs (this is not supported yet).
Blank lines, and lines that start with the `#' character, are ignored. There are four keywords (outside of the scripts) that are recognized:
PhysFormat name
ElecFormat name
LvsFormat name
EndScript
One of the first three of these keywords and its argument should appear on its own line ahead of a script, and ``EndScript'' should appear on its own line following a script. The name is the name of the format, which will appear on command or menu buttons or is given to script functions to indicate that the following script is to be used for formatting. This should be a short alpha-numeric word or phrase, and must be unique among keywords of a given type. If the name contains white space, it should be double-quoted.
The script lines can contain any of the script library functions and operators. All local variables are static. The script can call functions that have been previously defined in a regular library file.
When the script is executed:
When the script is executing, the following predefined variables are available for use in the script.
Name Type Description
_cellname string name of the cell being output
_viewname string ``physical'' or ``electrical''
_techname string TechnologyName value from technology file
_num_nets integer number of wire nets in cell
_mode integer 0 if physical, 1 if electrical
_list_all integer 1 if list all cells active, 0 otherwise
_bottom_up integer 1 if list bottom-up active, 0 otherwise
_show_geom integer 1 if include geometry active, 0 otherwise
_show_wire_cap integer 1 if show wire cap active, 0 otherwise
_ignore_labels integer 1 if ignore labels active, 0 otherwise
The script will use functions that iterate through the cell and print the desired information in an order and format desired. The function library is being expanded to provide flexibility.
next up previous contents index
Next: The Misc Config Button: Up: Extraction System: Setup and Previous: Device Templates Contents Index
Stephen R. Whiteley 2018-04-19
|
__label__pos
| 0.8739 |
Calculating Probability with a Probability Tree (Probability Tree is a kind of Tree Diagram)
Calculating Probability with a Tree Diagram
Q1. You roll a die twice and add up the dots to get a score. Draw a tree diagram to represent this experiment. What is the probability that your score is a multiple of 5?
solution:
The tree diagram for the experiment is shown above. To save space, probabilities were not indicated on the branches of the tree, but every branch has a probability of ⅙. The multiples of 5 are underlined. Since the probability of each of the underlined outcomes in ⅙×⅙=1/36 and since there are 7 outcomes that are multiples of 5, the probability of getting a multiple of 5 is 7/36.
Probability Trees
Sometimes, at 2+ level you may see complex probability problems that include conditions or restrictions. For such problems it could be helpful to draw a probability tree that include all possible outcomes and their probabilities.
Q2. Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?
solution:
Let’s draw all possible outcomes and calculate all probabilities.
Now, It is pretty obvious that the probability of Julia’s win is:
Q3. What is the probability of throwing at least one five in four rolls of a regular 6-sided die? Hint: do not show all possible outcomes of each roll of the die. We are interested in whether the outcome is 5 or not 5 only.
solution:
probability tree
The outcomes that lead to at least one 5 in four rolls of the die are marked on the tree diagram above. Summing the probabilities along all the branches gives
The Following Probability Tree below Represents the Probabilities of Some Without-Replacement Events
Q4. A bag contains 10 orange balls and 7 black balls. You draw 3 balls from the bag without replacement. What is the probability that you will end up with exactly 2 orange balls? Represent this experiment using a tree diagram.
Solution:
The tree diagram for the experiment is shown below. Since balls are drawn without replacement, the total number of balls (which appears in the denominators of the fractions) decreases by 1 at each step. Depending on whether the ball drawn was orange or black, the numerator either decreases by 1 or stays the same. The outcomes containing exactly 2 orange balls are underlined.
P(two Orange balls)=
Let’s read the post Calculating Probability with a Tree Diagram
Leave a Reply
Your email address will not be published. Required fields are marked *
|
__label__pos
| 0.995742 |
What is a GitHub pull request?
Pull requests let you tell others about changes you’ve pushed to a branch in a repository on GitHub. Once a pull request is opened, you can discuss and review the potential changes with collaborators and add follow-up commits before your changes are merged into the base branch.
What is pull request in Sourcetree?
Pull requests are a feature that makes it easier for developers to collaborate using Bitbucket. They provide a user-friendly web interface for discussing proposed changes before integrating them into the official project.
How do I use pull request GitHub?
In summary, if you want to contribute to a project, the simplest way is to:
1. Find a project you want to contribute to.
2. Fork it.
3. Clone it to your local system.
4. Make a new branch.
5. Make your changes.
6. Push it back to your repo.
7. Click the Compare & pull request button.
8. Click Create pull request to open a new pull request.
What is the difference between pull and pull request?
If you use git pull , you pull the changes from the remote repository into yours. If you send a pull request to another repository, you ask their maintainers to pull your changes into theirs (you more or less ask them to use a git pull from your repository).
What is pull request in Azure Devops?
Pull requests (PRs) are a way to change, review, and merge code in a Git repository on Azure Repos. PRs can come from branches within the same repository or from branches in forks of the repository. Teams use PRs to review code and give feedback on changes before merging the code into the main branch.
Is pull request same as push?
A “pull request” is you requesting the target repository to please grab your changes. A “push request” would be the target repository requesting you to push your changes. When you send a pull request, you’re asking (requesting) the official repo owner to pull some changes from your own repo.
How does pull request work?
A pull request works by allowing developers to create new features or squash bugs without affecting the main project code or what the users are seeing. This way, they are able to write and test code changes locally without having to worry about breaking the overall product.
Who are responsible for the pull request?
The author(s) of the pull request are the one or more people who have made the changes to the project being proposed. The reviewer(s) are usually teammates or co-collaborators with the author(s) and are responsible for reviewing the proposed changes.
What is a pull request in agile?
The pull request holds the merge in a state that allows other developers to see what is being changed. If a developer agrees with the changes, she can proceed with accepting the pull request and executing the corresponding merge and then delete the supporting branch if needed.
How do I install Sourcetree on Windows 10?
– Install Sourcetree. Step-by-step instructions for installation. – Connect your Bitbucket or Github account. If you want to add remote repositories, you need to connect to your hosting service. – Clone a remote repository. – Create a local repository. – Add an existing local repository.
How do I revert a push in Sourcetree?
– create a new local branch called for example cleaning having master as parent (that is behind staging) – pull the remote staging into cleaning. – use git revert {last good commit hash in staging} – now cleaning should be in the good commit and in the same state of remote staging before my bad push, isn’t it?
How to create pull request from command line?
On the top bar,select Menu > Projects and find your project.
• On the left menu,select Repository > Branches .
• Type a branch name and select New branch .
• Above the file list,on the right side,select Create merge request . A merge request is created. The default branch is the target.
• Fill out the fields and select Create merge request .
• How to create a pull request using TortoiseHg?
Create a pull request. To create a pull request, you need to have made your code changes on a separate branch or forked repository. From the open repository, click + in the global sidebar and select Create a pull request under Get to work. Fill out the rest of the pull request form. See the screenshot below for a description of each field.
|
__label__pos
| 0.999861 |
Explore BrainMass
Share
Validity of Statistic Questions
1. True or False? The normal distribution is the most important discrete probability distribution.
2. True or False? Every normal distribution can be transformed to a standard normal
distribution.
3. For a standard normal distribution, what is the value of the standard deviation and the mean?
4. What is the area under a standard normal curve between z = 2:21 and z = 1:27?
5. What is the area under a standard normal curve to the left of z = 1:95?
6. What is the area under a standard normal curve between z = 2 and z = 2?
7. The probability that z lies between a and b is denoted P(a < z < b). Using a standard normal
curve, what is P(1:09 < z < 2:35)?
8. SAT Math Scores: Suppose the population of SAT math scores follows a normal distribution.
Suppose = 574 and = 156. What is P(600 < x < 685)?
Use the Information below to answer questions 9-11.
Beagles: The weights of a group of adult male beagles are normally distributed, with a mean
of 22:5 pounds and a standard deviation of 3:7 pounds. A beagle is randomly selected.
9. Find the probability that the beagle's weight is less than 20 pounds.
10. Find the probability that the beagle's weight is between 18 and 23 pounds.
11. Find the probability that the beagle's weight is more than 25 pounds.
The following directions apply to problems 12 and 13.
Use the Standard Normal Distribution Table to find the z-score that corresponds to the
given cumulative area. If the area is not in the table, use the entry closest to the area.
If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores.
12. Area = 0:9678
13. Area = 0:1220
14. Find the z-score that has 69:5% of the distribution's area to its right.
Use the following information to answer questions 15 and 16.
A population with n = 744 has a mean = 279 and a standard deviation = 53.
15. Find the mean of the sampling distribution of the sample means.
16. Find the standard deviation of the sampling distribution of the sample means.
17. True or False? As the standard deviation increases, the standard deviation of the distribution
of sample means increases.
Use the following information to answer questions 18 and 19.
Finding Probabilities For a sample with n = 49, = 132 and = 35.
18. What is P(x < 125)?
19. Would it be considered unusual if x < 125?
20. Salaries: The population mean annual salary for employees at a software ?rm is $77; 000. A
random sample of 45 employees is selected from this population. What is the probability that
the mean annual salary of the sample is greater than $79; 000. Assume = $6300.
21. Would it be unusual for a randomly selected employee to have a salary greater than $79,000?
Use the following information to answer question 1 - 4.
Lengths of Fish You are performing a study about the lengths of ?sh in a mountain lake. A previous
study found the lengths to be normally distributed, with a mean of 20 inches and a standard deviation
of 4.0 inches. You randomly sample 30 ?sh and ?nd their lengths (in inches) are as follows.
19 17 23 20 20 13 17 19 22 18
14 18 19 15 23 21 26 19 16 19
21 16 10 13 25 25 12 21 24 28
1. Draw a frequency histogram to display the data using 7 classes
2. Is it reasonable to assume the heights are normally distributed? Why?
3. Find the mean and standard deviation of your sample.
4. Compare the mean and standard deviation of your sample with those in the previous study.
5. Using the standard normal distribution, ?nd the probability. P(z > 1:05).
6. Using the standard normal distribution, ?nd the probability. P(1:45 < z < 1:17).
Section 5.2
Use the following information to answer questions 7-9. Please see note about rounding
above.
Health Club Schedule The time per workout an athlete uses a stair climber is normally distributed, with a mean of 25 minutes and a standard deviation of 6 minutes. An athlete is
randomly selected.
7. Find the probability that the athlete uses a stair climber for less than 17 minutes.
8. Find the probability that the athlete uses a stair climber between 20 and 28 minutes.
9. Find the probability that the athlete uses a stair climber for more than 30 minutes.
Use the following information to answer questions 10-11.
Utility Bills Utility are normally distributed with a mean equal to $130 and a standard deviation of $15.
10. What percent of the utility bills are more than $140?
11. If 300 utility bills are randomly selected, about how many would you expect to be less than
$115?
Use the following information to answer questions 12-13.
Peanuts: Assume the mean annual consumption of peanuts is normally distributed, with a
mean of 4.8 pounds per person and a standard deviation of 1.6 pounds per person.
12. What percent of people annually consume less than 3.5 pounds of peanuts person?
13. Would it be unusual for a person to consume less than 3.5 pounds of peanuts in a year?
Explain.
14. Find the z-score for which 94% of the standard normal distribution's area lies between z and
z.
Use the following information to answer questions 15-16.
Heights of Men: In a study of men at a college (ages 20-29), the mean height was 70.9 inches
with a standard deviation of 3.5 inches.
15. What height represents the 90th percentile? (Note: Percentiles were discussed in the text in
the same section as quartiles.)
16. What height represents the third quartile?
Use the following information to answer questions 17-19.
Brake Pads: A brake pad manufacturer claims its brake pads will last for an average of 40,000
miles. You work for a consumer protection agency and you are testing the manufacturer's
brake pads. Assume the life spans of the brake pads are normally distributed. You randomly
select 50 brake pads. In your tests, the mean life for the brake pads is 38,850 miles. Assume
= 900 miles.
17. Assuming the manufacturer's claim is correct, what is the probability the mean of the sample
is 38,850 or less?
18. Using your answer from question 17, what does the probability indicate about the manufacturer's claim?
19. Would it be unusual to have an individual brake pad last for 38,850 miles?
© BrainMass Inc. brainmass.com July 16, 2018, 12:25 am ad1c9bdddf
$2.19
|
__label__pos
| 0.990602 |
2018年3月1日木曜日
開発環境
Head First C ―頭とからだで覚えるCの基本 (David Griffiths (著)、Dawn Griffiths (著)、中田 秀基 (監修)、木下 哲也 (翻訳)、オライリージャパン)の11章(プロセス間通信 - お話は楽しい)、長いエクササイズ(p. 480)を取り組んでみる。
長いエクササイズ(p. 480)
//
// main.c
// sample1
//
// Created by kamimura on 2018/02/28.
// Copyright © 2018 kamimura. All rights reserved.
//
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <sys/socket.h>
#include <arpa/inet.h>
#include <stdbool.h>
#include <unistd.h>
#include <signal.h>
int catch_signal(int sig, void (*handler)(int)) {
struct sigaction action;
action.sa_handler = handler;
sigemptyset(&action.sa_mask);
action.sa_flags = 0;
return sigaction(sig, &action, NULL);
}
int read_in(int socket, char *buf, int len) {
char *s = buf;
int slen = len;
int c = (int)recv(socket, s, slen, 0);
while (c > 0 && s[c - 1] != '\n') {
s += c;
slen -= c;
c = (int)recv(socket, s, slen, 0);
}
if (c < 0) {
return c;
} else if (c == 0) {
buf[0] = '\0';
} else {
s[c - 1] = '\0';
}
return len - slen;
}
int listener_d;
void handle_shutdown(int sig) {
if (listener_d) {
close(listener_d);
}
fprintf(stderr, "さようなら！\n");
exit(0);
}
void error(char *msg) {
fprintf(stderr, "%s: %s\n", msg, strerror(errno));
exit(1);
}
int open_listener_socket() {
int s = socket(PF_INET, SOCK_STREAM, 0);
if (s == -1) {
error("ソケットを開ません");
}
return s;
}
void bind_to_port(int socket, int port) {
struct sockaddr_in name;
name.sin_family = PF_INET;
name.sin_port = (in_port_t)htons(port);
name.sin_addr.s_addr = htonl(INADDR_ANY);
int reuse = 1;
if (setsockopt(socket, SOL_SOCKET, SO_REUSEADDR, (char *)&reuse, sizeof(int)) == -1) {
error("ソケットに再利用オプションを設定できません。");
}
int c = bind(socket, (struct sockaddr *) &name, sizeof(name));
if (c == -1) {
error("ソケットにバインドできません");
}
}
int say(int socket, char *s) {
int result = (int)send(socket, s, strlen(s), 0);
if (result == -1) {
fprintf(stderr, "%s: %s\n", "クライアントとの通信エラー", strerror(errno));
}
return result;
}
int main(int argc, const char * argv[]) {
catch_signal(SIGINT, handle_shutdown);
listener_d = open_listener_socket();
bind_to_port(listener_d, 30000);
if (listen(listener_d, 10) == -1) {
error("接続待ちできません");
}
puts("接続を待っています。");
while (true) {
struct sockaddr_storage cliend_addr;
unsigned int address_size = sizeof(cliend_addr);
int connect_d = accept(listener_d, (struct sockaddr*)&cliend_addr, &address_size);
if (connect_d == -1) {
error("第二のソケットを開ません");
}
if (say(connect_d, "Knock! Knock!\r\n") != -1) {
char msg[255];
read_in(connect_d, msg, sizeof(msg));
if (strncmp(msg, "Who's there?", 12)) {
say(connect_d, "入力が違います\r\n");
} else {
say(connect_d, "Oscar\r\n");
read_in(connect_d, msg, sizeof(msg));
if (strncmp(msg, "Oscar who?", 10)) {
say(connect_d, "入力が違います\r\n");
} else {
say(connect_d, "oscar silly question, you get a silly answer");
}
}
}
close(connect_d);
}
return 0;
}
入出力結果(Terminal)
$ ./server
接続を待っています。
C-c C-cさようなら！
$
0 コメント:
コメントを投稿
関連コンテンツ
|
__label__pos
| 0.997693 |
oci_new_cursor
(PHP 5, PHP 7, PECL OCI8 >= 1.1.0)
oci_new_cursorAllocates and returns a new cursor (statement handle)
Descrizione
resource oci_new_cursor ( resource $connection )
Allocates a new statement handle on the specified connection.
Elenco dei parametri
connection
An Oracle connection identifier, returned by oci_connect() or oci_pconnect().
Valori restituiti
Returns a new statement handle, or FALSE on error.
Esempi
Example #1 Binding a REF CURSOR in an Oracle stored procedure call
<?php
// Precreate:
// create or replace procedure myproc(myrc out sys_refcursor) as
// begin
// open myrc for select first_name from employees;
// end;
$conn oci_connect("hr""hrpwd""localhost/XE");
if (!
$conn) {
$m oci_error();
trigger_error(htmlentities($m['message']), E_USER_ERROR);
}
$curs oci_new_cursor($conn);
$stid oci_parse($conn"begin myproc(:cursbv); end;");
oci_bind_by_name($stid":cursbv"$curs, -1OCI_B_CURSOR);
oci_execute($stid);
oci_execute($curs); // Execute the REF CURSOR like a normal statement id
while (($row oci_fetch_array($cursOCI_ASSOC+OCI_RETURN_NULLS)) != false) {
echo
$row['FIRST_NAME'] . "<br />\n";
}
oci_free_statement($stid);
oci_free_statement($curs);
oci_close($conn);
?>
Note
Nota:
In PHP versions before 5.0.0 you must use ocinewcursor() instead. This name still can be used, it was left as alias of oci_new_cursor() for downwards compatability. This, however, is deprecated and not recommended.
add a note add a note
User Contributed Notes 3 notes
up
6
mgumiel at mgait dot com
3 years ago
Some packages in oracle are functions, and that functions returns a cursor.
For example:
CREATE FUNCTION F_Function( p1 char(2), p2 int)
RETURN SYS_REFCURSOR
AS
my_cursor SYS_REFCURSOR;
BEGIN
OPEN my_cursor FOR SELECT * FROM allitems
WHERE (cod=p1)
AND (Number=p2);
RETURN my_cursor;
END F_Function;
Here is the code that allows to obtain data from a function that returns a cursor.
<pre>
<?php
$conn
=oci_connect("server", "user", "pass");
if (!
$conn) {
$e = oci_error();
trigger_error(htmlentities($e['message']), E_USER_ERROR);
}
//You must asign before.
$p1 = '03';
$p2 = 2012016191;
$stid = oci_parse($conn, 'begin :cursor := server.PKG_package.F_Function(:p1,:p2); end;');
$p_cursor = oci_new_cursor($conn);
//Send parameters variable value lenght
oci_bind_by_name($stid, ':p1', $p1,2);
oci_bind_by_name($stid, ':p2', $p2,10);
//Bind Cursor put -1
oci_bind_by_name($stid, ':cursor', $p_cursor, -1, OCI_B_CURSOR);
// Execute Statement
oci_execute($stid);
oci_execute($p_cursor, OCI_DEFAULT);
oci_fetch_all($p_cursor, $cursor, null, null, OCI_FETCHSTATEMENT_BY_ROW);
echo
'<br>';
print_r($cursor);
?>
up
-2
sixd at php dot net
7 years ago
Because OCI8 uses "prefetching" to greatly improve returning query results, but Oracle doesn't support prefetching for REF CURSORs, application performance using REF CURSORs can be greatly improved by writing a PL/SQL function that pulls data from the REF CURSOR and PIPEs the output. The new function can be queried in a SELECT as if it were a table. See http://blogs.oracle.com/opal/2008/11/
converting_ref_cursor_to_pipe.html
up
-3
sixd at php dot net
6 years ago
Oracle 11.2 introduced support for REF CURSOR prefetching
To Top
|
__label__pos
| 0.613246 |
101 What is IPTV? Discover the Exciting IPTV Definition
WHAT IS IPTV
1. Introduction to IPTV
• Definition of IPTV
• Evolution of television services
2. How IPTV Works
• Transmission of IPTV signals
• Different components involved
3. Advantages of IPTV
• Variety of channels and content
• On-demand viewing
• Interactive features
4. Disadvantages of IPTV
• Dependence on internet connection
• Potential for buffering issues
5. Comparison with Traditional TV
• Cost-effectiveness
• Flexibility in content consumption
6. Legal Considerations
• Copyright issues
• Licensing agreements
7. Popular IPTV Services
• Netflix
• Hulu
• Sling TV
8. Future Trends in IPTV
• Integration with smart home technology
• Enhanced user experience
9. Impact of IPTV on Traditional Broadcasting
• Shift in consumer preferences
• Challenges faced by traditional broadcasters
10. IPTV Regulations
• Regulatory bodies governing IPTV services
• Compliance requirements for providers
11. Security Concerns
• Risks of unauthorized access
• Measures to protect user privacy
12. Global Adoption of IPTV
• Regional variations in popularity
• Growth projections
13. Consumer Behavior Towards IPTV
• Demographic preferences
• Usage patterns
14. Conclusion
15. FAQs
what is iptv
what is iptv
What Is IPTV? Understanding the Basics
Television has come a long way since its inception, evolving from traditional broadcast methods to digital platforms. One such innovation that has revolutionized the way we consume television content is IPTV.
How IPTV Works
IPTV, or Internet Protocol Television, delivers television services over the internet rather than through traditional satellite or cable formats. This is achieved through the transmission of IPTV signals, which are decoded by a set-top box or a compatible device connected to a television.
Advantages of IPTV
IPTV offers a multitude of benefits to consumers. Firstly, it provides access to a vast array of channels and content from around the world. Additionally, IPTV allows for on-demand viewing, enabling users to watch their favorite programs at their convenience. Moreover, IPTV often incorporates interactive features such as video-on-demand, interactive advertising, and personalized recommendations.
Disadvantages of IPTV
Despite its advantages, IPTV also has its drawbacks. One of the main concerns is its dependence on a stable internet connection. In areas with poor connectivity, users may experience buffering issues or interruptions in service. Furthermore, some IPTV services may be subject to legal scrutiny due to copyright infringement or licensing violations.
Comparison with Traditional TV
Compared to traditional television services, IPTV offers greater flexibility and cost-effectiveness. With IPTV, users can choose from a wider range of subscription options and customize their viewing experience according to their preferences. Additionally, IPTV eliminates the need for expensive hardware installations, making it a more accessible option for consumers.
Legal Considerations
Legal considerations play a significant role in the operation of IPTV services. Providers must adhere to copyright laws and licensing agreements to ensure that they are not infringing on intellectual property rights. Failure to comply with these regulations can result in legal repercussions and fines.
Popular IPTV Services
Some of the most popular IPTV services include Netflix, Hulu, and Sling TV. These platforms offer a diverse selection of content, ranging from movies and TV shows to live sports events and news broadcasts.
Future Trends in IPTV
Looking ahead, the future of IPTV looks promising, with advancements in technology expected to enhance the user experience further. Integration with smart home technology, such as voice-controlled devices and IoT sensors, will enable seamless integration between IPTV services and other connected devices.
Impact of IPTV on Traditional Broadcasting
The rise of IPTV has had a significant impact on traditional broadcasting companies, forcing them to adapt to changing consumer preferences and technological advancements. As more viewers opt for IPTV services, traditional broadcasters are facing challenges in retaining their audience and generating revenue.
IPTV Regulations
Regulatory bodies play a crucial role in overseeing IPTV services and ensuring compliance with industry standards. Providers must obtain licenses and adhere to regulations set forth by regulatory authorities to operate legally.
Security Concerns
Security is a major concern for IPTV users, as unauthorized access to content can pose a threat to user privacy and data security. To mitigate these risks, providers implement encryption and authentication measures to protect their users’ information.
Global Adoption of IPTV
While IPTV has gained widespread popularity in many parts of the world, its adoption varies regionally. Factors such as internet penetration rates, disposable income levels, and cultural preferences influence the uptake of IPTV services in different countries.
Consumer Behavior Towards IPTV
Understanding consumer behavior is essential for IPTV providers to tailor their services to meet the needs and preferences of their target audience. Demographic factors, such as age, gender, and income, influence how users interact with IPTV platforms and the types of content they consume.
Ready to leap into the future of television? Contact us via WhatsApp to get your free trial now and unlock the door to endless entertainment with IPTV. Your journey into a world of streaming without boundaries begins today!
Upgrade Your Entertainment Now!
Unlock a world of limitless entertainment with our IPTV server. Experience high-quality streaming, interactive features, and endless customization. Join us for a seamless, cost-effective viewing adventure.
Try it today and redefine your entertainment experience!
Dive into the vibrant world of Albanian TV with LIONIPTV! Discover your favorite shows, news, and cultural insights. Don’t wait, experience Albania like never before!
Conclusion
In conclusion, IPTV represents a significant advancement in the field of television technology, offering consumers a more flexible and personalized viewing experience. Despite some challenges and legal considerations, IPTV continues to gain popularity globally, shaping the future of television entertainment.
what is iptv
what is iptv
FAQs
1. Is IPTV legal?Yes, IPTV is legal as long as providers adhere to copyright laws and licensing agreements.
2. Can I watch live sports on IPTV?Many IPTV services offer live sports channels and events as part of their subscription packages.
3. Do I need a special device for IPTV?While some IPTV services require a set-top box or compatible device, many can be accessed using a smart TV, computer, or mobile device.
4. Is IPTV cheaper than cable or satellite TV?IPTV can be more cost-effective than traditional cable or satellite TV services, depending on the subscription package and provider.
5. Are there any security risks associated with IPTV?Like any online service, there are potential security risks associated with IPTV, such as unauthorized access and data breaches. However, providers implement security measures to protect their users’ information
Leave a Reply
Your email address will not be published. Required fields are marked *
|
__label__pos
| 0.659404 |
blob: aec0f19597a94b8d0df2d3781d93f9fb55bbec3b [file] [log] [blame]
// SPDX-License-Identifier: GPL-2.0+
/*
* FireWire Serial driver
*
* Copyright (C) 2012 Peter Hurley <[email protected]>
*/
#define pr_fmt(fmt) KBUILD_MODNAME ": " fmt
#include <linux/sched.h>
#include <linux/slab.h>
#include <linux/device.h>
#include <linux/mod_devicetable.h>
#include <linux/rculist.h>
#include <linux/workqueue.h>
#include <linux/ratelimit.h>
#include <linux/bug.h>
#include <linux/uaccess.h>
#include "fwserial.h"
inline u64 be32_to_u64(__be32 hi, __be32 lo)
{
return ((u64)be32_to_cpu(hi) << 32 | be32_to_cpu(lo));
}
#define LINUX_VENDOR_ID 0xd00d1eU /* same id used in card root directory */
#define FWSERIAL_VERSION 0x00e81cU /* must be unique within LINUX_VENDOR_ID */
/* configurable options */
static int num_ttys = 4; /* # of std ttys to create per fw_card */
/* - doubles as loopback port index */
static bool auto_connect = true; /* try to VIRT_CABLE to every peer */
static bool create_loop_dev = true; /* create a loopback device for each card */
module_param_named(ttys, num_ttys, int, 0644);
module_param_named(auto, auto_connect, bool, 0644);
module_param_named(loop, create_loop_dev, bool, 0644);
/*
* Threshold below which the tty is woken for writing
* - should be equal to WAKEUP_CHARS in drivers/tty/n_tty.c because
* even if the writer is woken, n_tty_poll() won't set EPOLLOUT until
* our fifo is below this level
*/
#define WAKEUP_CHARS 256
/**
* fwserial_list: list of every fw_serial created for each fw_card
* See discussion in fwserial_probe.
*/
static LIST_HEAD(fwserial_list);
static DEFINE_MUTEX(fwserial_list_mutex);
/**
* port_table: array of tty ports allocated to each fw_card
*
* tty ports are allocated during probe when an fw_serial is first
* created for a given fw_card. Ports are allocated in a contiguous block,
* each block consisting of 'num_ports' ports.
*/
static struct fwtty_port *port_table[MAX_TOTAL_PORTS];
static DEFINE_MUTEX(port_table_lock);
static bool port_table_corrupt;
#define FWTTY_INVALID_INDEX MAX_TOTAL_PORTS
#define loop_idx(port) (((port)->index) / num_ports)
#define table_idx(loop) ((loop) * num_ports + num_ttys)
/* total # of tty ports created per fw_card */
static int num_ports;
/* slab used as pool for struct fwtty_transactions */
static struct kmem_cache *fwtty_txn_cache;
struct tty_driver *fwtty_driver;
static struct tty_driver *fwloop_driver;
static struct dentry *fwserial_debugfs;
struct fwtty_transaction;
typedef void (*fwtty_transaction_cb)(struct fw_card *card, int rcode,
void *data, size_t length,
struct fwtty_transaction *txn);
struct fwtty_transaction {
struct fw_transaction fw_txn;
fwtty_transaction_cb callback;
struct fwtty_port *port;
union {
struct dma_pending dma_pended;
};
};
#define to_device(a, b) (a->b)
#define fwtty_err(p, fmt, ...) \
dev_err(to_device(p, device), fmt, ##__VA_ARGS__)
#define fwtty_info(p, fmt, ...) \
dev_info(to_device(p, device), fmt, ##__VA_ARGS__)
#define fwtty_notice(p, fmt, ...) \
dev_notice(to_device(p, device), fmt, ##__VA_ARGS__)
#define fwtty_dbg(p, fmt, ...) \
dev_dbg(to_device(p, device), "%s: " fmt, __func__, ##__VA_ARGS__)
#define fwtty_err_ratelimited(p, fmt, ...) \
dev_err_ratelimited(to_device(p, device), fmt, ##__VA_ARGS__)
#ifdef DEBUG
static inline void debug_short_write(struct fwtty_port *port, int c, int n)
{
int avail;
if (n < c) {
spin_lock_bh(&port->lock);
avail = dma_fifo_avail(&port->tx_fifo);
spin_unlock_bh(&port->lock);
fwtty_dbg(port, "short write: avail:%d req:%d wrote:%d\n",
avail, c, n);
}
}
#else
#define debug_short_write(port, c, n)
#endif
static struct fwtty_peer *__fwserial_peer_by_node_id(struct fw_card *card,
int generation, int id);
#ifdef FWTTY_PROFILING
static void fwtty_profile_fifo(struct fwtty_port *port, unsigned int *stat)
{
spin_lock_bh(&port->lock);
fwtty_profile_data(stat, dma_fifo_avail(&port->tx_fifo));
spin_unlock_bh(&port->lock);
}
static void fwtty_dump_profile(struct seq_file *m, struct stats *stats)
{
/* for each stat, print sum of 0 to 2^k, then individually */
int k = 4;
unsigned int sum;
int j;
char t[10];
snprintf(t, 10, "< %d", 1 << k);
seq_printf(m, "\n%14s %6s", " ", t);
for (j = k + 1; j < DISTRIBUTION_MAX_INDEX; ++j)
seq_printf(m, "%6d", 1 << j);
++k;
for (j = 0, sum = 0; j <= k; ++j)
sum += stats->reads[j];
seq_printf(m, "\n%14s: %6d", "reads", sum);
for (j = k + 1; j <= DISTRIBUTION_MAX_INDEX; ++j)
seq_printf(m, "%6d", stats->reads[j]);
for (j = 0, sum = 0; j <= k; ++j)
sum += stats->writes[j];
seq_printf(m, "\n%14s: %6d", "writes", sum);
for (j = k + 1; j <= DISTRIBUTION_MAX_INDEX; ++j)
seq_printf(m, "%6d", stats->writes[j]);
for (j = 0, sum = 0; j <= k; ++j)
sum += stats->txns[j];
seq_printf(m, "\n%14s: %6d", "txns", sum);
for (j = k + 1; j <= DISTRIBUTION_MAX_INDEX; ++j)
seq_printf(m, "%6d", stats->txns[j]);
for (j = 0, sum = 0; j <= k; ++j)
sum += stats->unthrottle[j];
seq_printf(m, "\n%14s: %6d", "avail @ unthr", sum);
for (j = k + 1; j <= DISTRIBUTION_MAX_INDEX; ++j)
seq_printf(m, "%6d", stats->unthrottle[j]);
}
#else
#define fwtty_profile_fifo(port, stat)
#define fwtty_dump_profile(m, stats)
#endif
/*
* Returns the max receive packet size for the given node
* Devices which are OHCI v1.0/ v1.1/ v1.2-draft or RFC 2734 compliant
* are required by specification to support max_rec of 8 (512 bytes) or more.
*/
static inline int device_max_receive(struct fw_device *fw_device)
{
/* see IEEE 1394-2008 table 8-8 */
return min(2 << fw_device->max_rec, 4096);
}
static void fwtty_log_tx_error(struct fwtty_port *port, int rcode)
{
switch (rcode) {
case RCODE_SEND_ERROR:
fwtty_err_ratelimited(port, "card busy\n");
break;
case RCODE_ADDRESS_ERROR:
fwtty_err_ratelimited(port, "bad unit addr or write length\n");
break;
case RCODE_DATA_ERROR:
fwtty_err_ratelimited(port, "failed rx\n");
break;
case RCODE_NO_ACK:
fwtty_err_ratelimited(port, "missing ack\n");
break;
case RCODE_BUSY:
fwtty_err_ratelimited(port, "remote busy\n");
break;
default:
fwtty_err_ratelimited(port, "failed tx: %d\n", rcode);
}
}
static void fwtty_common_callback(struct fw_card *card, int rcode,
void *payload, size_t len, void *cb_data)
{
struct fwtty_transaction *txn = cb_data;
struct fwtty_port *port = txn->port;
if (port && rcode != RCODE_COMPLETE)
fwtty_log_tx_error(port, rcode);
if (txn->callback)
txn->callback(card, rcode, payload, len, txn);
kmem_cache_free(fwtty_txn_cache, txn);
}
static int fwtty_send_data_async(struct fwtty_peer *peer, int tcode,
unsigned long long addr, void *payload,
size_t len, fwtty_transaction_cb callback,
struct fwtty_port *port)
{
struct fwtty_transaction *txn;
int generation;
txn = kmem_cache_alloc(fwtty_txn_cache, GFP_ATOMIC);
if (!txn)
return -ENOMEM;
txn->callback = callback;
txn->port = port;
generation = peer->generation;
smp_rmb();
fw_send_request(peer->serial->card, &txn->fw_txn, tcode,
peer->node_id, generation, peer->speed, addr, payload,
len, fwtty_common_callback, txn);
return 0;
}
static void fwtty_send_txn_async(struct fwtty_peer *peer,
struct fwtty_transaction *txn, int tcode,
unsigned long long addr, void *payload,
size_t len, fwtty_transaction_cb callback,
struct fwtty_port *port)
{
int generation;
txn->callback = callback;
txn->port = port;
generation = peer->generation;
smp_rmb();
fw_send_request(peer->serial->card, &txn->fw_txn, tcode,
peer->node_id, generation, peer->speed, addr, payload,
len, fwtty_common_callback, txn);
}
static void __fwtty_restart_tx(struct fwtty_port *port)
{
int len, avail;
len = dma_fifo_out_level(&port->tx_fifo);
if (len)
schedule_delayed_work(&port->drain, 0);
avail = dma_fifo_avail(&port->tx_fifo);
fwtty_dbg(port, "fifo len: %d avail: %d\n", len, avail);
}
static void fwtty_restart_tx(struct fwtty_port *port)
{
spin_lock_bh(&port->lock);
__fwtty_restart_tx(port);
spin_unlock_bh(&port->lock);
}
/**
* fwtty_update_port_status - decodes & dispatches line status changes
*
* Note: in loopback, the port->lock is being held. Only use functions that
* don't attempt to reclaim the port->lock.
*/
static void fwtty_update_port_status(struct fwtty_port *port,
unsigned int status)
{
unsigned int delta;
struct tty_struct *tty;
/* simulated LSR/MSR status from remote */
status &= ~MCTRL_MASK;
delta = (port->mstatus ^ status) & ~MCTRL_MASK;
delta &= ~(status & TIOCM_RNG);
port->mstatus = status;
if (delta & TIOCM_RNG)
++port->icount.rng;
if (delta & TIOCM_DSR)
++port->icount.dsr;
if (delta & TIOCM_CAR)
++port->icount.dcd;
if (delta & TIOCM_CTS)
++port->icount.cts;
fwtty_dbg(port, "status: %x delta: %x\n", status, delta);
if (delta & TIOCM_CAR) {
tty = tty_port_tty_get(&port->port);
if (tty && !C_CLOCAL(tty)) {
if (status & TIOCM_CAR)
wake_up_interruptible(&port->port.open_wait);
else
schedule_work(&port->hangup);
}
tty_kref_put(tty);
}
if (delta & TIOCM_CTS) {
tty = tty_port_tty_get(&port->port);
if (tty && C_CRTSCTS(tty)) {
if (tty->hw_stopped) {
if (status & TIOCM_CTS) {
tty->hw_stopped = 0;
if (port->loopback)
__fwtty_restart_tx(port);
else
fwtty_restart_tx(port);
}
} else {
if (~status & TIOCM_CTS)
tty->hw_stopped = 1;
}
}
tty_kref_put(tty);
} else if (delta & OOB_TX_THROTTLE) {
tty = tty_port_tty_get(&port->port);
if (tty) {
if (tty->hw_stopped) {
if (~status & OOB_TX_THROTTLE) {
tty->hw_stopped = 0;
if (port->loopback)
__fwtty_restart_tx(port);
else
fwtty_restart_tx(port);
}
} else {
if (status & OOB_TX_THROTTLE)
tty->hw_stopped = 1;
}
}
tty_kref_put(tty);
}
if (delta & (UART_LSR_BI << 24)) {
if (status & (UART_LSR_BI << 24)) {
port->break_last = jiffies;
schedule_delayed_work(&port->emit_breaks, 0);
} else {
/* run emit_breaks one last time (if pending) */
mod_delayed_work(system_wq, &port->emit_breaks, 0);
}
}
if (delta & (TIOCM_DSR | TIOCM_CAR | TIOCM_CTS | TIOCM_RNG))
wake_up_interruptible(&port->port.delta_msr_wait);
}
/**
* __fwtty_port_line_status - generate 'line status' for indicated port
*
* This function returns a remote 'MSR' state based on the local 'MCR' state,
* as if a null modem cable was attached. The actual status is a mangling
* of TIOCM_* bits suitable for sending to a peer's status_addr.
*
* Note: caller must be holding port lock
*/
static unsigned int __fwtty_port_line_status(struct fwtty_port *port)
{
unsigned int status = 0;
/* TODO: add module param to tie RNG to DTR as well */
if (port->mctrl & TIOCM_DTR)
status |= TIOCM_DSR | TIOCM_CAR;
if (port->mctrl & TIOCM_RTS)
status |= TIOCM_CTS;
if (port->mctrl & OOB_RX_THROTTLE)
status |= OOB_TX_THROTTLE;
/* emulate BRK as add'l line status */
if (port->break_ctl)
status |= UART_LSR_BI << 24;
return status;
}
/**
* __fwtty_write_port_status - send the port line status to peer
*
* Note: caller must be holding the port lock.
*/
static int __fwtty_write_port_status(struct fwtty_port *port)
{
struct fwtty_peer *peer;
int err = -ENOENT;
unsigned int status = __fwtty_port_line_status(port);
rcu_read_lock();
peer = rcu_dereference(port->peer);
if (peer) {
err = fwtty_send_data_async(peer, TCODE_WRITE_QUADLET_REQUEST,
peer->status_addr, &status,
sizeof(status), NULL, port);
}
rcu_read_unlock();
return err;
}
/**
* fwtty_write_port_status - same as above but locked by port lock
*/
static int fwtty_write_port_status(struct fwtty_port *port)
{
int err;
spin_lock_bh(&port->lock);
err = __fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
return err;
}
static void fwtty_throttle_port(struct fwtty_port *port)
{
struct tty_struct *tty;
unsigned int old;
tty = tty_port_tty_get(&port->port);
if (!tty)
return;
spin_lock_bh(&port->lock);
old = port->mctrl;
port->mctrl |= OOB_RX_THROTTLE;
if (C_CRTSCTS(tty))
port->mctrl &= ~TIOCM_RTS;
if (~old & OOB_RX_THROTTLE)
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
tty_kref_put(tty);
}
/**
* fwtty_do_hangup - wait for ldisc to deliver all pending rx; only then hangup
*
* When the remote has finished tx, and all in-flight rx has been received and
* and pushed to the flip buffer, the remote may close its device. This will
* drop DTR on the remote which will drop carrier here. Typically, the tty is
* hung up when carrier is dropped or lost.
*
* However, there is a race between the hang up and the line discipline
* delivering its data to the reader. A hangup will cause the ldisc to flush
* (ie., clear) the read buffer and flip buffer. Because of firewire's
* relatively high throughput, the ldisc frequently lags well behind the driver,
* resulting in lost data (which has already been received and written to
* the flip buffer) when the remote closes its end.
*
* Unfortunately, since the flip buffer offers no direct method for determining
* if it holds data, ensuring the ldisc has delivered all data is problematic.
*/
/* FIXME: drop this workaround when __tty_hangup waits for ldisc completion */
static void fwtty_do_hangup(struct work_struct *work)
{
struct fwtty_port *port = to_port(work, hangup);
struct tty_struct *tty;
schedule_timeout_uninterruptible(msecs_to_jiffies(50));
tty = tty_port_tty_get(&port->port);
if (tty)
tty_vhangup(tty);
tty_kref_put(tty);
}
static void fwtty_emit_breaks(struct work_struct *work)
{
struct fwtty_port *port = to_port(to_delayed_work(work), emit_breaks);
static const char buf[16];
unsigned long now = jiffies;
unsigned long elapsed = now - port->break_last;
int n, t, c, brk = 0;
/* generate breaks at the line rate (but at least 1) */
n = (elapsed * port->cps) / HZ + 1;
port->break_last = now;
fwtty_dbg(port, "sending %d brks\n", n);
while (n) {
t = min(n, 16);
c = tty_insert_flip_string_fixed_flag(&port->port, buf,
TTY_BREAK, t);
n -= c;
brk += c;
if (c < t)
break;
}
tty_flip_buffer_push(&port->port);
if (port->mstatus & (UART_LSR_BI << 24))
schedule_delayed_work(&port->emit_breaks, FREQ_BREAKS);
port->icount.brk += brk;
}
static int fwtty_rx(struct fwtty_port *port, unsigned char *data, size_t len)
{
int c, n = len;
unsigned int lsr;
int err = 0;
fwtty_dbg(port, "%d\n", n);
fwtty_profile_data(port->stats.reads, n);
if (port->write_only) {
n = 0;
goto out;
}
/* disregard break status; breaks are generated by emit_breaks work */
lsr = (port->mstatus >> 24) & ~UART_LSR_BI;
if (port->overrun)
lsr |= UART_LSR_OE;
if (lsr & UART_LSR_OE)
++port->icount.overrun;
lsr &= port->status_mask;
if (lsr & ~port->ignore_mask & UART_LSR_OE) {
if (!tty_insert_flip_char(&port->port, 0, TTY_OVERRUN)) {
err = -EIO;
goto out;
}
}
port->overrun = false;
if (lsr & port->ignore_mask & ~UART_LSR_OE) {
/* TODO: don't drop SAK and Magic SysRq here */
n = 0;
goto out;
}
c = tty_insert_flip_string_fixed_flag(&port->port, data, TTY_NORMAL, n);
if (c > 0)
tty_flip_buffer_push(&port->port);
n -= c;
if (n) {
port->overrun = true;
err = -EIO;
fwtty_err_ratelimited(port, "flip buffer overrun\n");
} else {
/* throttle the sender if remaining flip buffer space has
* reached high watermark to avoid losing data which may be
* in-flight. Since the AR request context is 32k, that much
* data may have _already_ been acked.
*/
if (tty_buffer_space_avail(&port->port) < HIGH_WATERMARK)
fwtty_throttle_port(port);
}
out:
port->icount.rx += len;
port->stats.lost += n;
return err;
}
/**
* fwtty_port_handler - bus address handler for port reads/writes
* @parameters: fw_address_callback_t as specified by firewire core interface
*
* This handler is responsible for handling inbound read/write dma from remotes.
*/
static void fwtty_port_handler(struct fw_card *card,
struct fw_request *request,
int tcode, int destination, int source,
int generation,
unsigned long long addr,
void *data, size_t len,
void *callback_data)
{
struct fwtty_port *port = callback_data;
struct fwtty_peer *peer;
int err;
int rcode;
/* Only accept rx from the peer virtual-cabled to this port */
rcu_read_lock();
peer = __fwserial_peer_by_node_id(card, generation, source);
rcu_read_unlock();
if (!peer || peer != rcu_access_pointer(port->peer)) {
rcode = RCODE_ADDRESS_ERROR;
fwtty_err_ratelimited(port, "ignoring unauthenticated data\n");
goto respond;
}
switch (tcode) {
case TCODE_WRITE_QUADLET_REQUEST:
if (addr != port->rx_handler.offset || len != 4) {
rcode = RCODE_ADDRESS_ERROR;
} else {
fwtty_update_port_status(port, *(unsigned int *)data);
rcode = RCODE_COMPLETE;
}
break;
case TCODE_WRITE_BLOCK_REQUEST:
if (addr != port->rx_handler.offset + 4 ||
len > port->rx_handler.length - 4) {
rcode = RCODE_ADDRESS_ERROR;
} else {
err = fwtty_rx(port, data, len);
switch (err) {
case 0:
rcode = RCODE_COMPLETE;
break;
case -EIO:
rcode = RCODE_DATA_ERROR;
break;
default:
rcode = RCODE_CONFLICT_ERROR;
break;
}
}
break;
default:
rcode = RCODE_TYPE_ERROR;
}
respond:
fw_send_response(card, request, rcode);
}
/**
* fwtty_tx_complete - callback for tx dma
* @data: ignored, has no meaning for write txns
* @length: ignored, has no meaning for write txns
*
* The writer must be woken here if the fifo has been emptied because it
* may have slept if chars_in_buffer was != 0
*/
static void fwtty_tx_complete(struct fw_card *card, int rcode,
void *data, size_t length,
struct fwtty_transaction *txn)
{
struct fwtty_port *port = txn->port;
int len;
fwtty_dbg(port, "rcode: %d\n", rcode);
switch (rcode) {
case RCODE_COMPLETE:
spin_lock_bh(&port->lock);
dma_fifo_out_complete(&port->tx_fifo, &txn->dma_pended);
len = dma_fifo_level(&port->tx_fifo);
spin_unlock_bh(&port->lock);
port->icount.tx += txn->dma_pended.len;
break;
default:
/* TODO: implement retries */
spin_lock_bh(&port->lock);
dma_fifo_out_complete(&port->tx_fifo, &txn->dma_pended);
len = dma_fifo_level(&port->tx_fifo);
spin_unlock_bh(&port->lock);
port->stats.dropped += txn->dma_pended.len;
}
if (len < WAKEUP_CHARS)
tty_port_tty_wakeup(&port->port);
}
static int fwtty_tx(struct fwtty_port *port, bool drain)
{
struct fwtty_peer *peer;
struct fwtty_transaction *txn;
struct tty_struct *tty;
int n, len;
tty = tty_port_tty_get(&port->port);
if (!tty)
return -ENOENT;
rcu_read_lock();
peer = rcu_dereference(port->peer);
if (!peer) {
n = -EIO;
goto out;
}
if (test_and_set_bit(IN_TX, &port->flags)) {
n = -EALREADY;
goto out;
}
/* try to write as many dma transactions out as possible */
n = -EAGAIN;
while (!tty->stopped && !tty->hw_stopped &&
!test_bit(STOP_TX, &port->flags)) {
txn = kmem_cache_alloc(fwtty_txn_cache, GFP_ATOMIC);
if (!txn) {
n = -ENOMEM;
break;
}
spin_lock_bh(&port->lock);
n = dma_fifo_out_pend(&port->tx_fifo, &txn->dma_pended);
spin_unlock_bh(&port->lock);
fwtty_dbg(port, "out: %u rem: %d\n", txn->dma_pended.len, n);
if (n < 0) {
kmem_cache_free(fwtty_txn_cache, txn);
if (n == -EAGAIN) {
++port->stats.tx_stall;
} else if (n == -ENODATA) {
fwtty_profile_data(port->stats.txns, 0);
} else {
++port->stats.fifo_errs;
fwtty_err_ratelimited(port, "fifo err: %d\n",
n);
}
break;
}
fwtty_profile_data(port->stats.txns, txn->dma_pended.len);
fwtty_send_txn_async(peer, txn, TCODE_WRITE_BLOCK_REQUEST,
peer->fifo_addr, txn->dma_pended.data,
txn->dma_pended.len, fwtty_tx_complete,
port);
++port->stats.sent;
/*
* Stop tx if the 'last view' of the fifo is empty or if
* this is the writer and there's not enough data to bother
*/
if (n == 0 || (!drain && n < WRITER_MINIMUM))
break;
}
if (n >= 0 || n == -EAGAIN || n == -ENOMEM || n == -ENODATA) {
spin_lock_bh(&port->lock);
len = dma_fifo_out_level(&port->tx_fifo);
if (len) {
unsigned long delay = (n == -ENOMEM) ? HZ : 1;
schedule_delayed_work(&port->drain, delay);
}
len = dma_fifo_level(&port->tx_fifo);
spin_unlock_bh(&port->lock);
/* wakeup the writer */
if (drain && len < WAKEUP_CHARS)
tty_wakeup(tty);
}
clear_bit(IN_TX, &port->flags);
wake_up_interruptible(&port->wait_tx);
out:
rcu_read_unlock();
tty_kref_put(tty);
return n;
}
static void fwtty_drain_tx(struct work_struct *work)
{
struct fwtty_port *port = to_port(to_delayed_work(work), drain);
fwtty_tx(port, true);
}
static void fwtty_write_xchar(struct fwtty_port *port, char ch)
{
struct fwtty_peer *peer;
++port->stats.xchars;
fwtty_dbg(port, "%02x\n", ch);
rcu_read_lock();
peer = rcu_dereference(port->peer);
if (peer) {
fwtty_send_data_async(peer, TCODE_WRITE_BLOCK_REQUEST,
peer->fifo_addr, &ch, sizeof(ch),
NULL, port);
}
rcu_read_unlock();
}
static struct fwtty_port *fwtty_port_get(unsigned int index)
{
struct fwtty_port *port;
if (index >= MAX_TOTAL_PORTS)
return NULL;
mutex_lock(&port_table_lock);
port = port_table[index];
if (port)
kref_get(&port->serial->kref);
mutex_unlock(&port_table_lock);
return port;
}
static int fwtty_ports_add(struct fw_serial *serial)
{
int err = -EBUSY;
int i, j;
if (port_table_corrupt)
return err;
mutex_lock(&port_table_lock);
for (i = 0; i + num_ports <= MAX_TOTAL_PORTS; i += num_ports) {
if (!port_table[i]) {
for (j = 0; j < num_ports; ++i, ++j) {
serial->ports[j]->index = i;
port_table[i] = serial->ports[j];
}
err = 0;
break;
}
}
mutex_unlock(&port_table_lock);
return err;
}
static void fwserial_destroy(struct kref *kref)
{
struct fw_serial *serial = to_serial(kref, kref);
struct fwtty_port **ports = serial->ports;
int j, i = ports[0]->index;
synchronize_rcu();
mutex_lock(&port_table_lock);
for (j = 0; j < num_ports; ++i, ++j) {
port_table_corrupt |= port_table[i] != ports[j];
WARN_ONCE(port_table_corrupt, "port_table[%d]: %p != ports[%d]: %p",
i, port_table[i], j, ports[j]);
port_table[i] = NULL;
}
mutex_unlock(&port_table_lock);
for (j = 0; j < num_ports; ++j) {
fw_core_remove_address_handler(&ports[j]->rx_handler);
tty_port_destroy(&ports[j]->port);
kfree(ports[j]);
}
kfree(serial);
}
static void fwtty_port_put(struct fwtty_port *port)
{
kref_put(&port->serial->kref, fwserial_destroy);
}
static void fwtty_port_dtr_rts(struct tty_port *tty_port, int on)
{
struct fwtty_port *port = to_port(tty_port, port);
fwtty_dbg(port, "on/off: %d\n", on);
spin_lock_bh(&port->lock);
/* Don't change carrier state if this is a console */
if (!port->port.console) {
if (on)
port->mctrl |= TIOCM_DTR | TIOCM_RTS;
else
port->mctrl &= ~(TIOCM_DTR | TIOCM_RTS);
}
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
}
/**
* fwtty_port_carrier_raised: required tty_port operation
*
* This port operation is polled after a tty has been opened and is waiting for
* carrier detect -- see drivers/tty/tty_port:tty_port_block_til_ready().
*/
static int fwtty_port_carrier_raised(struct tty_port *tty_port)
{
struct fwtty_port *port = to_port(tty_port, port);
int rc;
rc = (port->mstatus & TIOCM_CAR);
fwtty_dbg(port, "%d\n", rc);
return rc;
}
static unsigned int set_termios(struct fwtty_port *port, struct tty_struct *tty)
{
unsigned int baud, frame;
baud = tty_termios_baud_rate(&tty->termios);
tty_termios_encode_baud_rate(&tty->termios, baud, baud);
/* compute bit count of 2 frames */
frame = 12 + ((C_CSTOPB(tty)) ? 4 : 2) + ((C_PARENB(tty)) ? 2 : 0);
switch (C_CSIZE(tty)) {
case CS5:
frame -= (C_CSTOPB(tty)) ? 1 : 0;
break;
case CS6:
frame += 2;
break;
case CS7:
frame += 4;
break;
case CS8:
frame += 6;
break;
}
port->cps = (baud << 1) / frame;
port->status_mask = UART_LSR_OE;
if (_I_FLAG(tty, BRKINT | PARMRK))
port->status_mask |= UART_LSR_BI;
port->ignore_mask = 0;
if (I_IGNBRK(tty)) {
port->ignore_mask |= UART_LSR_BI;
if (I_IGNPAR(tty))
port->ignore_mask |= UART_LSR_OE;
}
port->write_only = !C_CREAD(tty);
/* turn off echo and newline xlat if loopback */
if (port->loopback) {
tty->termios.c_lflag &= ~(ECHO | ECHOE | ECHOK | ECHOKE |
ECHONL | ECHOPRT | ECHOCTL);
tty->termios.c_oflag &= ~ONLCR;
}
return baud;
}
static int fwtty_port_activate(struct tty_port *tty_port,
struct tty_struct *tty)
{
struct fwtty_port *port = to_port(tty_port, port);
unsigned int baud;
int err;
set_bit(TTY_IO_ERROR, &tty->flags);
err = dma_fifo_alloc(&port->tx_fifo, FWTTY_PORT_TXFIFO_LEN,
cache_line_size(),
port->max_payload,
FWTTY_PORT_MAX_PEND_DMA,
GFP_KERNEL);
if (err)
return err;
spin_lock_bh(&port->lock);
baud = set_termios(port, tty);
/* if console, don't change carrier state */
if (!port->port.console) {
port->mctrl = 0;
if (baud != 0)
port->mctrl = TIOCM_DTR | TIOCM_RTS;
}
if (C_CRTSCTS(tty) && ~port->mstatus & TIOCM_CTS)
tty->hw_stopped = 1;
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
clear_bit(TTY_IO_ERROR, &tty->flags);
return 0;
}
/**
* fwtty_port_shutdown
*
* Note: the tty port core ensures this is not the console and
* manages TTY_IO_ERROR properly
*/
static void fwtty_port_shutdown(struct tty_port *tty_port)
{
struct fwtty_port *port = to_port(tty_port, port);
/* TODO: cancel outstanding transactions */
cancel_delayed_work_sync(&port->emit_breaks);
cancel_delayed_work_sync(&port->drain);
spin_lock_bh(&port->lock);
port->flags = 0;
port->break_ctl = 0;
port->overrun = 0;
__fwtty_write_port_status(port);
dma_fifo_free(&port->tx_fifo);
spin_unlock_bh(&port->lock);
}
static int fwtty_open(struct tty_struct *tty, struct file *fp)
{
struct fwtty_port *port = tty->driver_data;
return tty_port_open(&port->port, tty, fp);
}
static void fwtty_close(struct tty_struct *tty, struct file *fp)
{
struct fwtty_port *port = tty->driver_data;
tty_port_close(&port->port, tty, fp);
}
static void fwtty_hangup(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
tty_port_hangup(&port->port);
}
static void fwtty_cleanup(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
tty->driver_data = NULL;
fwtty_port_put(port);
}
static int fwtty_install(struct tty_driver *driver, struct tty_struct *tty)
{
struct fwtty_port *port = fwtty_port_get(tty->index);
int err;
err = tty_standard_install(driver, tty);
if (!err)
tty->driver_data = port;
else
fwtty_port_put(port);
return err;
}
static int fwloop_install(struct tty_driver *driver, struct tty_struct *tty)
{
struct fwtty_port *port = fwtty_port_get(table_idx(tty->index));
int err;
err = tty_standard_install(driver, tty);
if (!err)
tty->driver_data = port;
else
fwtty_port_put(port);
return err;
}
static int fwtty_write(struct tty_struct *tty, const unsigned char *buf, int c)
{
struct fwtty_port *port = tty->driver_data;
int n, len;
fwtty_dbg(port, "%d\n", c);
fwtty_profile_data(port->stats.writes, c);
spin_lock_bh(&port->lock);
n = dma_fifo_in(&port->tx_fifo, buf, c);
len = dma_fifo_out_level(&port->tx_fifo);
if (len < DRAIN_THRESHOLD)
schedule_delayed_work(&port->drain, 1);
spin_unlock_bh(&port->lock);
if (len >= DRAIN_THRESHOLD)
fwtty_tx(port, false);
debug_short_write(port, c, n);
return (n < 0) ? 0 : n;
}
static int fwtty_write_room(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
int n;
spin_lock_bh(&port->lock);
n = dma_fifo_avail(&port->tx_fifo);
spin_unlock_bh(&port->lock);
fwtty_dbg(port, "%d\n", n);
return n;
}
static int fwtty_chars_in_buffer(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
int n;
spin_lock_bh(&port->lock);
n = dma_fifo_level(&port->tx_fifo);
spin_unlock_bh(&port->lock);
fwtty_dbg(port, "%d\n", n);
return n;
}
static void fwtty_send_xchar(struct tty_struct *tty, char ch)
{
struct fwtty_port *port = tty->driver_data;
fwtty_dbg(port, "%02x\n", ch);
fwtty_write_xchar(port, ch);
}
static void fwtty_throttle(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
/*
* Ignore throttling (but not unthrottling).
* It only makes sense to throttle when data will no longer be
* accepted by the tty flip buffer. For example, it is
* possible for received data to overflow the tty buffer long
* before the line discipline ever has a chance to throttle the driver.
* Additionally, the driver may have already completed the I/O
* but the tty buffer is still emptying, so the line discipline is
* throttling and unthrottling nothing.
*/
++port->stats.throttled;
}
static void fwtty_unthrottle(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
fwtty_dbg(port, "CRTSCTS: %d\n", C_CRTSCTS(tty) != 0);
fwtty_profile_fifo(port, port->stats.unthrottle);
spin_lock_bh(&port->lock);
port->mctrl &= ~OOB_RX_THROTTLE;
if (C_CRTSCTS(tty))
port->mctrl |= TIOCM_RTS;
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
}
static int check_msr_delta(struct fwtty_port *port, unsigned long mask,
struct async_icount *prev)
{
struct async_icount now;
int delta;
now = port->icount;
delta = ((mask & TIOCM_RNG && prev->rng != now.rng) ||
(mask & TIOCM_DSR && prev->dsr != now.dsr) ||
(mask & TIOCM_CAR && prev->dcd != now.dcd) ||
(mask & TIOCM_CTS && prev->cts != now.cts));
*prev = now;
return delta;
}
static int wait_msr_change(struct fwtty_port *port, unsigned long mask)
{
struct async_icount prev;
prev = port->icount;
return wait_event_interruptible(port->port.delta_msr_wait,
check_msr_delta(port, mask, &prev));
}
static int get_serial_info(struct tty_struct *tty,
struct serial_struct *ss)
{
struct fwtty_port *port = tty->driver_data;
mutex_lock(&port->port.mutex);
ss->type = PORT_UNKNOWN;
ss->line = port->port.tty->index;
ss->flags = port->port.flags;
ss->xmit_fifo_size = FWTTY_PORT_TXFIFO_LEN;
ss->baud_base = 400000000;
ss->close_delay = port->port.close_delay;
mutex_unlock(&port->port.mutex);
return 0;
}
static int set_serial_info(struct tty_struct *tty,
struct serial_struct *ss)
{
struct fwtty_port *port = tty->driver_data;
if (ss->irq != 0 || ss->port != 0 || ss->custom_divisor != 0 ||
ss->baud_base != 400000000)
return -EPERM;
mutex_lock(&port->port.mutex);
if (!capable(CAP_SYS_ADMIN)) {
if (((ss->flags & ~ASYNC_USR_MASK) !=
(port->port.flags & ~ASYNC_USR_MASK))) {
mutex_unlock(&port->port.mutex);
return -EPERM;
}
}
port->port.close_delay = ss->close_delay * HZ / 100;
mutex_unlock(&port->port.mutex);
return 0;
}
static int fwtty_ioctl(struct tty_struct *tty, unsigned int cmd,
unsigned long arg)
{
struct fwtty_port *port = tty->driver_data;
int err;
switch (cmd) {
case TIOCMIWAIT:
err = wait_msr_change(port, arg);
break;
default:
err = -ENOIOCTLCMD;
}
return err;
}
static void fwtty_set_termios(struct tty_struct *tty, struct ktermios *old)
{
struct fwtty_port *port = tty->driver_data;
unsigned int baud;
spin_lock_bh(&port->lock);
baud = set_termios(port, tty);
if ((baud == 0) && (old->c_cflag & CBAUD)) {
port->mctrl &= ~(TIOCM_DTR | TIOCM_RTS);
} else if ((baud != 0) && !(old->c_cflag & CBAUD)) {
if (C_CRTSCTS(tty) || !tty_throttled(tty))
port->mctrl |= TIOCM_DTR | TIOCM_RTS;
else
port->mctrl |= TIOCM_DTR;
}
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
if (old->c_cflag & CRTSCTS) {
if (!C_CRTSCTS(tty)) {
tty->hw_stopped = 0;
fwtty_restart_tx(port);
}
} else if (C_CRTSCTS(tty) && ~port->mstatus & TIOCM_CTS) {
tty->hw_stopped = 1;
}
}
/**
* fwtty_break_ctl - start/stop sending breaks
*
* Signals the remote to start or stop generating simulated breaks.
* First, stop dequeueing from the fifo and wait for writer/drain to leave tx
* before signalling the break line status. This guarantees any pending rx will
* be queued to the line discipline before break is simulated on the remote.
* Conversely, turning off break_ctl requires signalling the line status change,
* then enabling tx.
*/
static int fwtty_break_ctl(struct tty_struct *tty, int state)
{
struct fwtty_port *port = tty->driver_data;
long ret;
fwtty_dbg(port, "%d\n", state);
if (state == -1) {
set_bit(STOP_TX, &port->flags);
ret = wait_event_interruptible_timeout(port->wait_tx,
!test_bit(IN_TX, &port->flags),
10);
if (ret == 0 || ret == -ERESTARTSYS) {
clear_bit(STOP_TX, &port->flags);
fwtty_restart_tx(port);
return -EINTR;
}
}
spin_lock_bh(&port->lock);
port->break_ctl = (state == -1);
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
if (state == 0) {
spin_lock_bh(&port->lock);
dma_fifo_reset(&port->tx_fifo);
clear_bit(STOP_TX, &port->flags);
spin_unlock_bh(&port->lock);
}
return 0;
}
static int fwtty_tiocmget(struct tty_struct *tty)
{
struct fwtty_port *port = tty->driver_data;
unsigned int tiocm;
spin_lock_bh(&port->lock);
tiocm = (port->mctrl & MCTRL_MASK) | (port->mstatus & ~MCTRL_MASK);
spin_unlock_bh(&port->lock);
fwtty_dbg(port, "%x\n", tiocm);
return tiocm;
}
static int fwtty_tiocmset(struct tty_struct *tty,
unsigned int set, unsigned int clear)
{
struct fwtty_port *port = tty->driver_data;
fwtty_dbg(port, "set: %x clear: %x\n", set, clear);
/* TODO: simulate loopback if TIOCM_LOOP set */
spin_lock_bh(&port->lock);
port->mctrl &= ~(clear & MCTRL_MASK & 0xffff);
port->mctrl |= set & MCTRL_MASK & 0xffff;
__fwtty_write_port_status(port);
spin_unlock_bh(&port->lock);
return 0;
}
static int fwtty_get_icount(struct tty_struct *tty,
struct serial_icounter_struct *icount)
{
struct fwtty_port *port = tty->driver_data;
struct stats stats;
memcpy(&stats, &port->stats, sizeof(stats));
if (port->port.console)
(*port->fwcon_ops->stats)(&stats, port->con_data);
icount->cts = port->icount.cts;
icount->dsr = port->icount.dsr;
icount->rng = port->icount.rng;
icount->dcd = port->icount.dcd;
icount->rx = port->icount.rx;
icount->tx = port->icount.tx + stats.xchars;
icount->frame = port->icount.frame;
icount->overrun = port->icount.overrun;
icount->parity = port->icount.parity;
icount->brk = port->icount.brk;
icount->buf_overrun = port->icount.overrun;
return 0;
}
static void fwtty_proc_show_port(struct seq_file *m, struct fwtty_port *port)
{
struct stats stats;
memcpy(&stats, &port->stats, sizeof(stats));
if (port->port.console)
(*port->fwcon_ops->stats)(&stats, port->con_data);
seq_printf(m, " addr:%012llx tx:%d rx:%d", port->rx_handler.offset,
port->icount.tx + stats.xchars, port->icount.rx);
seq_printf(m, " cts:%d dsr:%d rng:%d dcd:%d", port->icount.cts,
port->icount.dsr, port->icount.rng, port->icount.dcd);
seq_printf(m, " fe:%d oe:%d pe:%d brk:%d", port->icount.frame,
port->icount.overrun, port->icount.parity, port->icount.brk);
}
static void fwtty_debugfs_show_port(struct seq_file *m, struct fwtty_port *port)
{
struct stats stats;
memcpy(&stats, &port->stats, sizeof(stats));
if (port->port.console)
(*port->fwcon_ops->stats)(&stats, port->con_data);
seq_printf(m, " dr:%d st:%d err:%d lost:%d", stats.dropped,
stats.tx_stall, stats.fifo_errs, stats.lost);
seq_printf(m, " pkts:%d thr:%d", stats.sent, stats.throttled);
if (port->port.console) {
seq_puts(m, "\n ");
(*port->fwcon_ops->proc_show)(m, port->con_data);
}
fwtty_dump_profile(m, &port->stats);
}
static void fwtty_debugfs_show_peer(struct seq_file *m, struct fwtty_peer *peer)
{
int generation = peer->generation;
smp_rmb();
seq_printf(m, " %s:", dev_name(&peer->unit->device));
seq_printf(m, " node:%04x gen:%d", peer->node_id, generation);
seq_printf(m, " sp:%d max:%d guid:%016llx", peer->speed,
peer->max_payload, (unsigned long long)peer->guid);
seq_printf(m, " mgmt:%012llx", (unsigned long long)peer->mgmt_addr);
seq_printf(m, " addr:%012llx", (unsigned long long)peer->status_addr);
seq_putc(m, '\n');
}
static int fwtty_proc_show(struct seq_file *m, void *v)
{
struct fwtty_port *port;
int i;
seq_puts(m, "fwserinfo: 1.0 driver: 1.0\n");
for (i = 0; i < MAX_TOTAL_PORTS && (port = fwtty_port_get(i)); ++i) {
seq_printf(m, "%2d:", i);
if (capable(CAP_SYS_ADMIN))
fwtty_proc_show_port(m, port);
fwtty_port_put(port);
seq_puts(m, "\n");
}
return 0;
}
static int fwtty_stats_show(struct seq_file *m, void *v)
{
struct fw_serial *serial = m->private;
struct fwtty_port *port;
int i;
for (i = 0; i < num_ports; ++i) {
port = fwtty_port_get(serial->ports[i]->index);
if (port) {
seq_printf(m, "%2d:", port->index);
fwtty_proc_show_port(m, port);
fwtty_debugfs_show_port(m, port);
fwtty_port_put(port);
seq_puts(m, "\n");
}
}
return 0;
}
DEFINE_SHOW_ATTRIBUTE(fwtty_stats);
static int fwtty_peers_show(struct seq_file *m, void *v)
{
struct fw_serial *serial = m->private;
struct fwtty_peer *peer;
rcu_read_lock();
seq_printf(m, "card: %s guid: %016llx\n",
dev_name(serial->card->device),
(unsigned long long)serial->card->guid);
list_for_each_entry_rcu(peer, &serial->peer_list, list)
fwtty_debugfs_show_peer(m, peer);
rcu_read_unlock();
return 0;
}
DEFINE_SHOW_ATTRIBUTE(fwtty_peers);
static const struct tty_port_operations fwtty_port_ops = {
.dtr_rts = fwtty_port_dtr_rts,
.carrier_raised = fwtty_port_carrier_raised,
.shutdown = fwtty_port_shutdown,
.activate = fwtty_port_activate,
};
static const struct tty_operations fwtty_ops = {
.open = fwtty_open,
.close = fwtty_close,
.hangup = fwtty_hangup,
.cleanup = fwtty_cleanup,
.install = fwtty_install,
.write = fwtty_write,
.write_room = fwtty_write_room,
.chars_in_buffer = fwtty_chars_in_buffer,
.send_xchar = fwtty_send_xchar,
.throttle = fwtty_throttle,
.unthrottle = fwtty_unthrottle,
.ioctl = fwtty_ioctl,
.set_termios = fwtty_set_termios,
.break_ctl = fwtty_break_ctl,
.tiocmget = fwtty_tiocmget,
.tiocmset = fwtty_tiocmset,
.get_icount = fwtty_get_icount,
.set_serial = set_serial_info,
.get_serial = get_serial_info,
.proc_show = fwtty_proc_show,
};
static const struct tty_operations fwloop_ops = {
.open = fwtty_open,
.close = fwtty_close,
.hangup = fwtty_hangup,
.cleanup = fwtty_cleanup,
.install = fwloop_install,
.write = fwtty_write,
.write_room = fwtty_write_room,
.chars_in_buffer = fwtty_chars_in_buffer,
.send_xchar = fwtty_send_xchar,
.throttle = fwtty_throttle,
.unthrottle = fwtty_unthrottle,
.ioctl = fwtty_ioctl,
.set_termios = fwtty_set_termios,
.break_ctl = fwtty_break_ctl,
.tiocmget = fwtty_tiocmget,
.tiocmset = fwtty_tiocmset,
.get_icount = fwtty_get_icount,
.set_serial = set_serial_info,
.get_serial = get_serial_info,
};
static inline int mgmt_pkt_expected_len(__be16 code)
{
static const struct fwserial_mgmt_pkt pkt;
switch (be16_to_cpu(code)) {
case FWSC_VIRT_CABLE_PLUG:
return sizeof(pkt.hdr) + sizeof(pkt.plug_req);
case FWSC_VIRT_CABLE_PLUG_RSP: /* | FWSC_RSP_OK */
return sizeof(pkt.hdr) + sizeof(pkt.plug_rsp);
case FWSC_VIRT_CABLE_UNPLUG:
case FWSC_VIRT_CABLE_UNPLUG_RSP:
case FWSC_VIRT_CABLE_PLUG_RSP | FWSC_RSP_NACK:
case FWSC_VIRT_CABLE_UNPLUG_RSP | FWSC_RSP_NACK:
return sizeof(pkt.hdr);
default:
return -1;
}
}
static inline void fill_plug_params(struct virt_plug_params *params,
struct fwtty_port *port)
{
u64 status_addr = port->rx_handler.offset;
u64 fifo_addr = port->rx_handler.offset + 4;
size_t fifo_len = port->rx_handler.length - 4;
params->status_hi = cpu_to_be32(status_addr >> 32);
params->status_lo = cpu_to_be32(status_addr);
params->fifo_hi = cpu_to_be32(fifo_addr >> 32);
params->fifo_lo = cpu_to_be32(fifo_addr);
params->fifo_len = cpu_to_be32(fifo_len);
}
static inline void fill_plug_req(struct fwserial_mgmt_pkt *pkt,
struct fwtty_port *port)
{
pkt->hdr.code = cpu_to_be16(FWSC_VIRT_CABLE_PLUG);
pkt->hdr.len = cpu_to_be16(mgmt_pkt_expected_len(pkt->hdr.code));
fill_plug_params(&pkt->plug_req, port);
}
static inline void fill_plug_rsp_ok(struct fwserial_mgmt_pkt *pkt,
struct fwtty_port *port)
{
pkt->hdr.code = cpu_to_be16(FWSC_VIRT_CABLE_PLUG_RSP);
pkt->hdr.len = cpu_to_be16(mgmt_pkt_expected_len(pkt->hdr.code));
fill_plug_params(&pkt->plug_rsp, port);
}
static inline void fill_plug_rsp_nack(struct fwserial_mgmt_pkt *pkt)
{
pkt->hdr.code = cpu_to_be16(FWSC_VIRT_CABLE_PLUG_RSP | FWSC_RSP_NACK);
pkt->hdr.len = cpu_to_be16(mgmt_pkt_expected_len(pkt->hdr.code));
}
static inline void fill_unplug_rsp_nack(struct fwserial_mgmt_pkt *pkt)
{
pkt->hdr.code = cpu_to_be16(FWSC_VIRT_CABLE_UNPLUG_RSP | FWSC_RSP_NACK);
pkt->hdr.len = cpu_to_be16(mgmt_pkt_expected_len(pkt->hdr.code));
}
static inline void fill_unplug_rsp_ok(struct fwserial_mgmt_pkt *pkt)
{
pkt->hdr.code = cpu_to_be16(FWSC_VIRT_CABLE_UNPLUG_RSP);
pkt->hdr.len = cpu_to_be16(mgmt_pkt_expected_len(pkt->hdr.code));
}
static void fwserial_virt_plug_complete(struct fwtty_peer *peer,
struct virt_plug_params *params)
{
struct fwtty_port *port = peer->port;
peer->status_addr = be32_to_u64(params->status_hi, params->status_lo);
peer->fifo_addr = be32_to_u64(params->fifo_hi, params->fifo_lo);
peer->fifo_len = be32_to_cpu(params->fifo_len);
peer_set_state(peer, FWPS_ATTACHED);
/* reconfigure tx_fifo optimally for this peer */
spin_lock_bh(&port->lock);
port->max_payload = min(peer->max_payload, peer->fifo_len);
dma_fifo_change_tx_limit(&port->tx_fifo, port->max_payload);
spin_unlock_bh(&peer->port->lock);
if (port->port.console && port->fwcon_ops->notify)
(*port->fwcon_ops->notify)(FWCON_NOTIFY_ATTACH, port->con_data);
fwtty_info(&peer->unit, "peer (guid:%016llx) connected on %s\n",
(unsigned long long)peer->guid, dev_name(port->device));
}
static inline int fwserial_send_mgmt_sync(struct fwtty_peer *peer,
struct fwserial_mgmt_pkt *pkt)
{
int generation;
int rcode, tries = 5;
do {
generation = peer->generation;
smp_rmb();
rcode = fw_run_transaction(peer->serial->card,
TCODE_WRITE_BLOCK_REQUEST,
peer->node_id,
generation, peer->speed,
peer->mgmt_addr,
pkt, be16_to_cpu(pkt->hdr.len));
if (rcode == RCODE_BUSY || rcode == RCODE_SEND_ERROR ||
rcode == RCODE_GENERATION) {
fwtty_dbg(&peer->unit, "mgmt write error: %d\n", rcode);
continue;
} else {
break;
}
} while (--tries > 0);
return rcode;
}
/**
* fwserial_claim_port - attempt to claim port @ index for peer
*
* Returns ptr to claimed port or error code (as ERR_PTR())
* Can sleep - must be called from process context
*/
static struct fwtty_port *fwserial_claim_port(struct fwtty_peer *peer,
int index)
{
struct fwtty_port *port;
if (index < 0 || index >= num_ports)
return ERR_PTR(-EINVAL);
/* must guarantee that previous port releases have completed */
synchronize_rcu();
port = peer->serial->ports[index];
spin_lock_bh(&port->lock);
if (!rcu_access_pointer(port->peer))
rcu_assign_pointer(port->peer, peer);
else
port = ERR_PTR(-EBUSY);
spin_unlock_bh(&port->lock);
return port;
}
/**
* fwserial_find_port - find avail port and claim for peer
*
* Returns ptr to claimed port or NULL if none avail
* Can sleep - must be called from process context
*/
static struct fwtty_port *fwserial_find_port(struct fwtty_peer *peer)
{
struct fwtty_port **ports = peer->serial->ports;
int i;
/* must guarantee that previous port releases have completed */
synchronize_rcu();
/* TODO: implement optional GUID-to-specific port # matching */
/* find an unattached port (but not the loopback port, if present) */
for (i = 0; i < num_ttys; ++i) {
spin_lock_bh(&ports[i]->lock);
if (!ports[i]->peer) {
/* claim port */
rcu_assign_pointer(ports[i]->peer, peer);
spin_unlock_bh(&ports[i]->lock);
return ports[i];
}
spin_unlock_bh(&ports[i]->lock);
}
return NULL;
}
static void fwserial_release_port(struct fwtty_port *port, bool reset)
{
/* drop carrier (and all other line status) */
if (reset)
fwtty_update_port_status(port, 0);
spin_lock_bh(&port->lock);
/* reset dma fifo max transmission size back to S100 */
port->max_payload = link_speed_to_max_payload(SCODE_100);
dma_fifo_change_tx_limit(&port->tx_fifo, port->max_payload);
RCU_INIT_POINTER(port->peer, NULL);
spin_unlock_bh(&port->lock);
if (port->port.console && port->fwcon_ops->notify)
(*port->fwcon_ops->notify)(FWCON_NOTIFY_DETACH, port->con_data);
}
static void fwserial_plug_timeout(struct timer_list *t)
{
struct fwtty_peer *peer = from_timer(peer, t, timer);
struct fwtty_port *port;
spin_lock_bh(&peer->lock);
if (peer->state != FWPS_PLUG_PENDING) {
spin_unlock_bh(&peer->lock);
return;
}
port = peer_revert_state(peer);
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, false);
}
/**
* fwserial_connect_peer - initiate virtual cable with peer
*
* Returns 0 if VIRT_CABLE_PLUG request was successfully sent,
* otherwise error code. Must be called from process context.
*/
static int fwserial_connect_peer(struct fwtty_peer *peer)
{
struct fwtty_port *port;
struct fwserial_mgmt_pkt *pkt;
int err, rcode;
pkt = kmalloc(sizeof(*pkt), GFP_KERNEL);
if (!pkt)
return -ENOMEM;
port = fwserial_find_port(peer);
if (!port) {
fwtty_err(&peer->unit, "avail ports in use\n");
err = -EBUSY;
goto free_pkt;
}
spin_lock_bh(&peer->lock);
/* only initiate VIRT_CABLE_PLUG if peer is currently not attached */
if (peer->state != FWPS_NOT_ATTACHED) {
err = -EBUSY;
goto release_port;
}
peer->port = port;
peer_set_state(peer, FWPS_PLUG_PENDING);
fill_plug_req(pkt, peer->port);
mod_timer(&peer->timer, jiffies + VIRT_CABLE_PLUG_TIMEOUT);
spin_unlock_bh(&peer->lock);
rcode = fwserial_send_mgmt_sync(peer, pkt);
spin_lock_bh(&peer->lock);
if (peer->state == FWPS_PLUG_PENDING && rcode != RCODE_COMPLETE) {
if (rcode == RCODE_CONFLICT_ERROR)
err = -EAGAIN;
else
err = -EIO;
goto cancel_timer;
}
spin_unlock_bh(&peer->lock);
kfree(pkt);
return 0;
cancel_timer:
del_timer(&peer->timer);
peer_revert_state(peer);
release_port:
spin_unlock_bh(&peer->lock);
fwserial_release_port(port, false);
free_pkt:
kfree(pkt);
return err;
}
/**
* fwserial_close_port -
* HUP the tty (if the tty exists) and unregister the tty device.
* Only used by the unit driver upon unit removal to disconnect and
* cleanup all attached ports
*
* The port reference is put by fwtty_cleanup (if a reference was
* ever taken).
*/
static void fwserial_close_port(struct tty_driver *driver,
struct fwtty_port *port)
{
struct tty_struct *tty;
mutex_lock(&port->port.mutex);
tty = tty_port_tty_get(&port->port);
if (tty) {
tty_vhangup(tty);
tty_kref_put(tty);
}
mutex_unlock(&port->port.mutex);
if (driver == fwloop_driver)
tty_unregister_device(driver, loop_idx(port));
else
tty_unregister_device(driver, port->index);
}
/**
* fwserial_lookup - finds first fw_serial associated with card
* @card: fw_card to match
*
* NB: caller must be holding fwserial_list_mutex
*/
static struct fw_serial *fwserial_lookup(struct fw_card *card)
{
struct fw_serial *serial;
list_for_each_entry(serial, &fwserial_list, list) {
if (card == serial->card)
return serial;
}
return NULL;
}
/**
* __fwserial_lookup_rcu - finds first fw_serial associated with card
* @card: fw_card to match
*
* NB: caller must be inside rcu_read_lock() section
*/
static struct fw_serial *__fwserial_lookup_rcu(struct fw_card *card)
{
struct fw_serial *serial;
list_for_each_entry_rcu(serial, &fwserial_list, list) {
if (card == serial->card)
return serial;
}
return NULL;
}
/**
* __fwserial_peer_by_node_id - finds a peer matching the given generation + id
*
* If a matching peer could not be found for the specified generation/node id,
* this could be because:
* a) the generation has changed and one of the nodes hasn't updated yet
* b) the remote node has created its remote unit device before this
* local node has created its corresponding remote unit device
* In either case, the remote node should retry
*
* Note: caller must be in rcu_read_lock() section
*/
static struct fwtty_peer *__fwserial_peer_by_node_id(struct fw_card *card,
int generation, int id)
{
struct fw_serial *serial;
struct fwtty_peer *peer;
serial = __fwserial_lookup_rcu(card);
if (!serial) {
/*
* Something is very wrong - there should be a matching
* fw_serial structure for every fw_card. Maybe the remote node
* has created its remote unit device before this driver has
* been probed for any unit devices...
*/
fwtty_err(card, "unknown card (guid %016llx)\n",
(unsigned long long)card->guid);
return NULL;
}
list_for_each_entry_rcu(peer, &serial->peer_list, list) {
int g = peer->generation;
smp_rmb();
if (generation == g && id == peer->node_id)
return peer;
}
return NULL;
}
#ifdef DEBUG
static void __dump_peer_list(struct fw_card *card)
{
struct fw_serial *serial;
struct fwtty_peer *peer;
serial = __fwserial_lookup_rcu(card);
if (!serial)
return;
list_for_each_entry_rcu(peer, &serial->peer_list, list) {
int g = peer->generation;
smp_rmb();
fwtty_dbg(card, "peer(%d:%x) guid: %016llx\n",
g, peer->node_id, (unsigned long long)peer->guid);
}
}
#else
#define __dump_peer_list(s)
#endif
static void fwserial_auto_connect(struct work_struct *work)
{
struct fwtty_peer *peer = to_peer(to_delayed_work(work), connect);
int err;
err = fwserial_connect_peer(peer);
if (err == -EAGAIN && ++peer->connect_retries < MAX_CONNECT_RETRIES)
schedule_delayed_work(&peer->connect, CONNECT_RETRY_DELAY);
}
static void fwserial_peer_workfn(struct work_struct *work)
{
struct fwtty_peer *peer = to_peer(work, work);
peer->workfn(work);
}
/**
* fwserial_add_peer - add a newly probed 'serial' unit device as a 'peer'
* @serial: aggregate representing the specific fw_card to add the peer to
* @unit: 'peer' to create and add to peer_list of serial
*
* Adds a 'peer' (ie, a local or remote 'serial' unit device) to the list of
* peers for a specific fw_card. Optionally, auto-attach this peer to an
* available tty port. This function is called either directly or indirectly
* as a result of a 'serial' unit device being created & probed.
*
* Note: this function is serialized with fwserial_remove_peer() by the
* fwserial_list_mutex held in fwserial_probe().
*
* A 1:1 correspondence between an fw_unit and an fwtty_peer is maintained
* via the dev_set_drvdata() for the device of the fw_unit.
*/
static int fwserial_add_peer(struct fw_serial *serial, struct fw_unit *unit)
{
struct device *dev = &unit->device;
struct fw_device *parent = fw_parent_device(unit);
struct fwtty_peer *peer;
struct fw_csr_iterator ci;
int key, val;
int generation;
peer = kzalloc(sizeof(*peer), GFP_KERNEL);
if (!peer)
return -ENOMEM;
peer_set_state(peer, FWPS_NOT_ATTACHED);
dev_set_drvdata(dev, peer);
peer->unit = unit;
peer->guid = (u64)parent->config_rom[3] << 32 | parent->config_rom[4];
peer->speed = parent->max_speed;
peer->max_payload = min(device_max_receive(parent),
link_speed_to_max_payload(peer->speed));
generation = parent->generation;
smp_rmb();
peer->node_id = parent->node_id;
smp_wmb();
peer->generation = generation;
/* retrieve the mgmt bus addr from the unit directory */
fw_csr_iterator_init(&ci, unit->directory);
while (fw_csr_iterator_next(&ci, &key, &val)) {
if (key == (CSR_OFFSET | CSR_DEPENDENT_INFO)) {
peer->mgmt_addr = CSR_REGISTER_BASE + 4 * val;
break;
}
}
if (peer->mgmt_addr == 0ULL) {
/*
* No mgmt address effectively disables VIRT_CABLE_PLUG -
* this peer will not be able to attach to a remote
*/
peer_set_state(peer, FWPS_NO_MGMT_ADDR);
}
spin_lock_init(&peer->lock);
peer->port = NULL;
timer_setup(&peer->timer, fwserial_plug_timeout, 0);
INIT_WORK(&peer->work, fwserial_peer_workfn);
INIT_DELAYED_WORK(&peer->connect, fwserial_auto_connect);
/* associate peer with specific fw_card */
peer->serial = serial;
list_add_rcu(&peer->list, &serial->peer_list);
fwtty_info(&peer->unit, "peer added (guid:%016llx)\n",
(unsigned long long)peer->guid);
/* identify the local unit & virt cable to loopback port */
if (parent->is_local) {
serial->self = peer;
if (create_loop_dev) {
struct fwtty_port *port;
port = fwserial_claim_port(peer, num_ttys);
if (!IS_ERR(port)) {
struct virt_plug_params params;
spin_lock_bh(&peer->lock);
peer->port = port;
fill_plug_params(¶ms, port);
fwserial_virt_plug_complete(peer, ¶ms);
spin_unlock_bh(&peer->lock);
fwtty_write_port_status(port);
}
}
} else if (auto_connect) {
/* auto-attach to remote units only (if policy allows) */
schedule_delayed_work(&peer->connect, 1);
}
return 0;
}
/**
* fwserial_remove_peer - remove a 'serial' unit device as a 'peer'
*
* Remove a 'peer' from its list of peers. This function is only
* called by fwserial_remove() on bus removal of the unit device.
*
* Note: this function is serialized with fwserial_add_peer() by the
* fwserial_list_mutex held in fwserial_remove().
*/
static void fwserial_remove_peer(struct fwtty_peer *peer)
{
struct fwtty_port *port;
spin_lock_bh(&peer->lock);
peer_set_state(peer, FWPS_GONE);
spin_unlock_bh(&peer->lock);
cancel_delayed_work_sync(&peer->connect);
cancel_work_sync(&peer->work);
spin_lock_bh(&peer->lock);
/* if this unit is the local unit, clear link */
if (peer == peer->serial->self)
peer->serial->self = NULL;
/* cancel the request timeout timer (if running) */
del_timer(&peer->timer);
port = peer->port;
peer->port = NULL;
list_del_rcu(&peer->list);
fwtty_info(&peer->unit, "peer removed (guid:%016llx)\n",
(unsigned long long)peer->guid);
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, true);
synchronize_rcu();
kfree(peer);
}
/**
* fwserial_create - init everything to create TTYs for a specific fw_card
* @unit: fw_unit for first 'serial' unit device probed for this fw_card
*
* This function inits the aggregate structure (an fw_serial instance)
* used to manage the TTY ports registered by a specific fw_card. Also, the
* unit device is added as the first 'peer'.
*
* This unit device may represent a local unit device (as specified by the
* config ROM unit directory) or it may represent a remote unit device
* (as specified by the reading of the remote node's config ROM).
*
* Returns 0 to indicate "ownership" of the unit device, or a negative errno
* value to indicate which error.
*/
static int fwserial_create(struct fw_unit *unit)
{
struct fw_device *parent = fw_parent_device(unit);
struct fw_card *card = parent->card;
struct fw_serial *serial;
struct fwtty_port *port;
struct device *tty_dev;
int i, j;
int err;
serial = kzalloc(sizeof(*serial), GFP_KERNEL);
if (!serial)
return -ENOMEM;
kref_init(&serial->kref);
serial->card = card;
INIT_LIST_HEAD(&serial->peer_list);
for (i = 0; i < num_ports; ++i) {
port = kzalloc(sizeof(*port), GFP_KERNEL);
if (!port) {
err = -ENOMEM;
goto free_ports;
}
tty_port_init(&port->port);
port->index = FWTTY_INVALID_INDEX;
port->port.ops = &fwtty_port_ops;
port->serial = serial;
tty_buffer_set_limit(&port->port, 128 * 1024);
spin_lock_init(&port->lock);
INIT_DELAYED_WORK(&port->drain, fwtty_drain_tx);
INIT_DELAYED_WORK(&port->emit_breaks, fwtty_emit_breaks);
INIT_WORK(&port->hangup, fwtty_do_hangup);
init_waitqueue_head(&port->wait_tx);
port->max_payload = link_speed_to_max_payload(SCODE_100);
dma_fifo_init(&port->tx_fifo);
RCU_INIT_POINTER(port->peer, NULL);
serial->ports[i] = port;
/* get unique bus addr region for port's status & recv fifo */
port->rx_handler.length = FWTTY_PORT_RXFIFO_LEN + 4;
port->rx_handler.address_callback = fwtty_port_handler;
port->rx_handler.callback_data = port;
/*
* XXX: use custom memory region above cpu physical memory addrs
* this will ease porting to 64-bit firewire adapters
*/
err = fw_core_add_address_handler(&port->rx_handler,
&fw_high_memory_region);
if (err) {
kfree(port);
goto free_ports;
}
}
/* preserve i for error cleanup */
err = fwtty_ports_add(serial);
if (err) {
fwtty_err(&unit, "no space in port table\n");
goto free_ports;
}
for (j = 0; j < num_ttys; ++j) {
tty_dev = tty_port_register_device(&serial->ports[j]->port,
fwtty_driver,
serial->ports[j]->index,
card->device);
if (IS_ERR(tty_dev)) {
err = PTR_ERR(tty_dev);
fwtty_err(&unit, "register tty device error (%d)\n",
err);
goto unregister_ttys;
}
serial->ports[j]->device = tty_dev;
}
/* preserve j for error cleanup */
if (create_loop_dev) {
struct device *loop_dev;
loop_dev = tty_port_register_device(&serial->ports[j]->port,
fwloop_driver,
loop_idx(serial->ports[j]),
card->device);
if (IS_ERR(loop_dev)) {
err = PTR_ERR(loop_dev);
fwtty_err(&unit, "create loop device failed (%d)\n",
err);
goto unregister_ttys;
}
serial->ports[j]->device = loop_dev;
serial->ports[j]->loopback = true;
}
if (!IS_ERR_OR_NULL(fwserial_debugfs)) {
serial->debugfs = debugfs_create_dir(dev_name(&unit->device),
fwserial_debugfs);
if (!IS_ERR_OR_NULL(serial->debugfs)) {
debugfs_create_file("peers", 0444, serial->debugfs,
serial, &fwtty_peers_fops);
debugfs_create_file("stats", 0444, serial->debugfs,
serial, &fwtty_stats_fops);
}
}
list_add_rcu(&serial->list, &fwserial_list);
fwtty_notice(&unit, "TTY over FireWire on device %s (guid %016llx)\n",
dev_name(card->device), (unsigned long long)card->guid);
err = fwserial_add_peer(serial, unit);
if (!err)
return 0;
fwtty_err(&unit, "unable to add peer unit device (%d)\n", err);
/* fall-through to error processing */
debugfs_remove_recursive(serial->debugfs);
list_del_rcu(&serial->list);
if (create_loop_dev)
tty_unregister_device(fwloop_driver,
loop_idx(serial->ports[j]));
unregister_ttys:
for (--j; j >= 0; --j)
tty_unregister_device(fwtty_driver, serial->ports[j]->index);
kref_put(&serial->kref, fwserial_destroy);
return err;
free_ports:
for (--i; i >= 0; --i) {
tty_port_destroy(&serial->ports[i]->port);
kfree(serial->ports[i]);
}
kfree(serial);
return err;
}
/**
* fwserial_probe: bus probe function for firewire 'serial' unit devices
*
* A 'serial' unit device is created and probed as a result of:
* - declaring a ieee1394 bus id table for 'devices' matching a fabricated
* 'serial' unit specifier id
* - adding a unit directory to the config ROM(s) for a 'serial' unit
*
* The firewire core registers unit devices by enumerating unit directories
* of a node's config ROM after reading the config ROM when a new node is
* added to the bus topology after a bus reset.
*
* The practical implications of this are:
* - this probe is called for both local and remote nodes that have a 'serial'
* unit directory in their config ROM (that matches the specifiers in
* fwserial_id_table).
* - no specific order is enforced for local vs. remote unit devices
*
* This unit driver copes with the lack of specific order in the same way the
* firewire net driver does -- each probe, for either a local or remote unit
* device, is treated as a 'peer' (has a struct fwtty_peer instance) and the
* first peer created for a given fw_card (tracked by the global fwserial_list)
* creates the underlying TTYs (aggregated in a fw_serial instance).
*
* NB: an early attempt to differentiate local & remote unit devices by creating
* peers only for remote units and fw_serial instances (with their
* associated TTY devices) only for local units was discarded. Managing
* the peer lifetimes on device removal proved too complicated.
*
* fwserial_probe/fwserial_remove are effectively serialized by the
* fwserial_list_mutex. This is necessary because the addition of the first peer
* for a given fw_card will trigger the creation of the fw_serial for that
* fw_card, which must not simultaneously contend with the removal of the
* last peer for a given fw_card triggering the destruction of the same
* fw_serial for the same fw_card.
*/
static int fwserial_probe(struct fw_unit *unit,
const struct ieee1394_device_id *id)
{
struct fw_serial *serial;
int err;
mutex_lock(&fwserial_list_mutex);
serial = fwserial_lookup(fw_parent_device(unit)->card);
if (!serial)
err = fwserial_create(unit);
else
err = fwserial_add_peer(serial, unit);
mutex_unlock(&fwserial_list_mutex);
return err;
}
/**
* fwserial_remove: bus removal function for firewire 'serial' unit devices
*
* The corresponding 'peer' for this unit device is removed from the list of
* peers for the associated fw_serial (which has a 1:1 correspondence with a
* specific fw_card). If this is the last peer being removed, then trigger
* the destruction of the underlying TTYs.
*/
static void fwserial_remove(struct fw_unit *unit)
{
struct fwtty_peer *peer = dev_get_drvdata(&unit->device);
struct fw_serial *serial = peer->serial;
int i;
mutex_lock(&fwserial_list_mutex);
fwserial_remove_peer(peer);
if (list_empty(&serial->peer_list)) {
/* unlink from the fwserial_list here */
list_del_rcu(&serial->list);
debugfs_remove_recursive(serial->debugfs);
for (i = 0; i < num_ttys; ++i)
fwserial_close_port(fwtty_driver, serial->ports[i]);
if (create_loop_dev)
fwserial_close_port(fwloop_driver, serial->ports[i]);
kref_put(&serial->kref, fwserial_destroy);
}
mutex_unlock(&fwserial_list_mutex);
}
/**
* fwserial_update: bus update function for 'firewire' serial unit devices
*
* Updates the new node_id and bus generation for this peer. Note that locking
* is unnecessary; but careful memory barrier usage is important to enforce the
* load and store order of generation & node_id.
*
* The fw-core orders the write of node_id before generation in the parent
* fw_device to ensure that a stale node_id cannot be used with a current
* bus generation. So the generation value must be read before the node_id.
*
* In turn, this orders the write of node_id before generation in the peer to
* also ensure a stale node_id cannot be used with a current bus generation.
*/
static void fwserial_update(struct fw_unit *unit)
{
struct fw_device *parent = fw_parent_device(unit);
struct fwtty_peer *peer = dev_get_drvdata(&unit->device);
int generation;
generation = parent->generation;
smp_rmb();
peer->node_id = parent->node_id;
smp_wmb();
peer->generation = generation;
}
static const struct ieee1394_device_id fwserial_id_table[] = {
{
.match_flags = IEEE1394_MATCH_SPECIFIER_ID |
IEEE1394_MATCH_VERSION,
.specifier_id = LINUX_VENDOR_ID,
.version = FWSERIAL_VERSION,
},
{ }
};
static struct fw_driver fwserial_driver = {
.driver = {
.owner = THIS_MODULE,
.name = KBUILD_MODNAME,
.bus = &fw_bus_type,
},
.probe = fwserial_probe,
.update = fwserial_update,
.remove = fwserial_remove,
.id_table = fwserial_id_table,
};
#define FW_UNIT_SPECIFIER(id) ((CSR_SPECIFIER_ID << 24) | (id))
#define FW_UNIT_VERSION(ver) ((CSR_VERSION << 24) | (ver))
#define FW_UNIT_ADDRESS(ofs) (((CSR_OFFSET | CSR_DEPENDENT_INFO) << 24) \
| (((ofs) - CSR_REGISTER_BASE) >> 2))
/* XXX: config ROM definitons could be improved with semi-automated offset
* and length calculation
*/
#define FW_ROM_LEN(quads) ((quads) << 16)
#define FW_ROM_DESCRIPTOR(ofs) (((CSR_LEAF | CSR_DESCRIPTOR) << 24) | (ofs))
struct fwserial_unit_directory_data {
u32 len_crc;
u32 unit_specifier;
u32 unit_sw_version;
u32 unit_addr_offset;
u32 desc1_ofs;
u32 desc1_len_crc;
u32 desc1_data[5];
} __packed;
static struct fwserial_unit_directory_data fwserial_unit_directory_data = {
.len_crc = FW_ROM_LEN(4),
.unit_specifier = FW_UNIT_SPECIFIER(LINUX_VENDOR_ID),
.unit_sw_version = FW_UNIT_VERSION(FWSERIAL_VERSION),
.desc1_ofs = FW_ROM_DESCRIPTOR(1),
.desc1_len_crc = FW_ROM_LEN(5),
.desc1_data = {
0x00000000, /* type = text */
0x00000000, /* enc = ASCII, lang EN */
0x4c696e75, /* 'Linux TTY' */
0x78205454,
0x59000000,
},
};
static struct fw_descriptor fwserial_unit_directory = {
.length = sizeof(fwserial_unit_directory_data) / sizeof(u32),
.key = (CSR_DIRECTORY | CSR_UNIT) << 24,
.data = (u32 *)&fwserial_unit_directory_data,
};
/*
* The management address is in the unit space region but above other known
* address users (to keep wild writes from causing havoc)
*/
static const struct fw_address_region fwserial_mgmt_addr_region = {
.start = CSR_REGISTER_BASE + 0x1e0000ULL,
.end = 0x1000000000000ULL,
};
static struct fw_address_handler fwserial_mgmt_addr_handler;
/**
* fwserial_handle_plug_req - handle VIRT_CABLE_PLUG request work
* @work: ptr to peer->work
*
* Attempts to complete the VIRT_CABLE_PLUG handshake sequence for this peer.
*
* This checks for a collided request-- ie, that a VIRT_CABLE_PLUG request was
* already sent to this peer. If so, the collision is resolved by comparing
* guid values; the loser sends the plug response.
*
* Note: if an error prevents a response, don't do anything -- the
* remote will timeout its request.
*/
static void fwserial_handle_plug_req(struct work_struct *work)
{
struct fwtty_peer *peer = to_peer(work, work);
struct virt_plug_params *plug_req = &peer->work_params.plug_req;
struct fwtty_port *port;
struct fwserial_mgmt_pkt *pkt;
int rcode;
pkt = kmalloc(sizeof(*pkt), GFP_KERNEL);
if (!pkt)
return;
port = fwserial_find_port(peer);
spin_lock_bh(&peer->lock);
switch (peer->state) {
case FWPS_NOT_ATTACHED:
if (!port) {
fwtty_err(&peer->unit, "no more ports avail\n");
fill_plug_rsp_nack(pkt);
} else {
peer->port = port;
fill_plug_rsp_ok(pkt, peer->port);
peer_set_state(peer, FWPS_PLUG_RESPONDING);
/* don't release claimed port */
port = NULL;
}
break;
case FWPS_PLUG_PENDING:
if (peer->serial->card->guid > peer->guid)
goto cleanup;
/* We lost - hijack the already-claimed port and send ok */
del_timer(&peer->timer);
fill_plug_rsp_ok(pkt, peer->port);
peer_set_state(peer, FWPS_PLUG_RESPONDING);
break;
default:
fill_plug_rsp_nack(pkt);
}
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, false);
rcode = fwserial_send_mgmt_sync(peer, pkt);
spin_lock_bh(&peer->lock);
if (peer->state == FWPS_PLUG_RESPONDING) {
if (rcode == RCODE_COMPLETE) {
struct fwtty_port *tmp = peer->port;
fwserial_virt_plug_complete(peer, plug_req);
spin_unlock_bh(&peer->lock);
fwtty_write_port_status(tmp);
spin_lock_bh(&peer->lock);
} else {
fwtty_err(&peer->unit, "PLUG_RSP error (%d)\n", rcode);
port = peer_revert_state(peer);
}
}
cleanup:
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, false);
kfree(pkt);
}
static void fwserial_handle_unplug_req(struct work_struct *work)
{
struct fwtty_peer *peer = to_peer(work, work);
struct fwtty_port *port = NULL;
struct fwserial_mgmt_pkt *pkt;
int rcode;
pkt = kmalloc(sizeof(*pkt), GFP_KERNEL);
if (!pkt)
return;
spin_lock_bh(&peer->lock);
switch (peer->state) {
case FWPS_ATTACHED:
fill_unplug_rsp_ok(pkt);
peer_set_state(peer, FWPS_UNPLUG_RESPONDING);
break;
case FWPS_UNPLUG_PENDING:
if (peer->serial->card->guid > peer->guid)
goto cleanup;
/* We lost - send unplug rsp */
del_timer(&peer->timer);
fill_unplug_rsp_ok(pkt);
peer_set_state(peer, FWPS_UNPLUG_RESPONDING);
break;
default:
fill_unplug_rsp_nack(pkt);
}
spin_unlock_bh(&peer->lock);
rcode = fwserial_send_mgmt_sync(peer, pkt);
spin_lock_bh(&peer->lock);
if (peer->state == FWPS_UNPLUG_RESPONDING) {
if (rcode != RCODE_COMPLETE)
fwtty_err(&peer->unit, "UNPLUG_RSP error (%d)\n",
rcode);
port = peer_revert_state(peer);
}
cleanup:
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, true);
kfree(pkt);
}
static int fwserial_parse_mgmt_write(struct fwtty_peer *peer,
struct fwserial_mgmt_pkt *pkt,
unsigned long long addr,
size_t len)
{
struct fwtty_port *port = NULL;
bool reset = false;
int rcode;
if (addr != fwserial_mgmt_addr_handler.offset || len < sizeof(pkt->hdr))
return RCODE_ADDRESS_ERROR;
if (len != be16_to_cpu(pkt->hdr.len) ||
len != mgmt_pkt_expected_len(pkt->hdr.code))
return RCODE_DATA_ERROR;
spin_lock_bh(&peer->lock);
if (peer->state == FWPS_GONE) {
/*
* This should never happen - it would mean that the
* remote unit that just wrote this transaction was
* already removed from the bus -- and the removal was
* processed before we rec'd this transaction
*/
fwtty_err(&peer->unit, "peer already removed\n");
spin_unlock_bh(&peer->lock);
return RCODE_ADDRESS_ERROR;
}
rcode = RCODE_COMPLETE;
fwtty_dbg(&peer->unit, "mgmt: hdr.code: %04hx\n", pkt->hdr.code);
switch (be16_to_cpu(pkt->hdr.code) & FWSC_CODE_MASK) {
case FWSC_VIRT_CABLE_PLUG:
if (work_pending(&peer->work)) {
fwtty_err(&peer->unit, "plug req: busy\n");
rcode = RCODE_CONFLICT_ERROR;
} else {
peer->work_params.plug_req = pkt->plug_req;
peer->workfn = fwserial_handle_plug_req;
queue_work(system_unbound_wq, &peer->work);
}
break;
case FWSC_VIRT_CABLE_PLUG_RSP:
if (peer->state != FWPS_PLUG_PENDING) {
rcode = RCODE_CONFLICT_ERROR;
} else if (be16_to_cpu(pkt->hdr.code) & FWSC_RSP_NACK) {
fwtty_notice(&peer->unit, "NACK plug rsp\n");
port = peer_revert_state(peer);
} else {
struct fwtty_port *tmp = peer->port;
fwserial_virt_plug_complete(peer, &pkt->plug_rsp);
spin_unlock_bh(&peer->lock);
fwtty_write_port_status(tmp);
spin_lock_bh(&peer->lock);
}
break;
case FWSC_VIRT_CABLE_UNPLUG:
if (work_pending(&peer->work)) {
fwtty_err(&peer->unit, "unplug req: busy\n");
rcode = RCODE_CONFLICT_ERROR;
} else {
peer->workfn = fwserial_handle_unplug_req;
queue_work(system_unbound_wq, &peer->work);
}
break;
case FWSC_VIRT_CABLE_UNPLUG_RSP:
if (peer->state != FWPS_UNPLUG_PENDING) {
rcode = RCODE_CONFLICT_ERROR;
} else {
if (be16_to_cpu(pkt->hdr.code) & FWSC_RSP_NACK)
fwtty_notice(&peer->unit, "NACK unplug?\n");
port = peer_revert_state(peer);
reset = true;
}
break;
default:
fwtty_err(&peer->unit, "unknown mgmt code %d\n",
be16_to_cpu(pkt->hdr.code));
rcode = RCODE_DATA_ERROR;
}
spin_unlock_bh(&peer->lock);
if (port)
fwserial_release_port(port, reset);
return rcode;
}
/**
* fwserial_mgmt_handler: bus address handler for mgmt requests
* @parameters: fw_address_callback_t as specified by firewire core interface
*
* This handler is responsible for handling virtual cable requests from remotes
* for all cards.
*/
static void fwserial_mgmt_handler(struct fw_card *card,
struct fw_request *request,
int tcode, int destination, int source,
int generation,
unsigned long long addr,
void *data, size_t len,
void *callback_data)
{
struct fwserial_mgmt_pkt *pkt = data;
struct fwtty_peer *peer;
int rcode;
rcu_read_lock();
peer = __fwserial_peer_by_node_id(card, generation, source);
if (!peer) {
fwtty_dbg(card, "peer(%d:%x) not found\n", generation, source);
__dump_peer_list(card);
rcode = RCODE_CONFLICT_ERROR;
} else {
switch (tcode) {
case TCODE_WRITE_BLOCK_REQUEST:
rcode = fwserial_parse_mgmt_write(peer, pkt, addr, len);
break;
default:
rcode = RCODE_TYPE_ERROR;
}
}
rcu_read_unlock();
fw_send_response(card, request, rcode);
}
static int __init fwserial_init(void)
{
int err, num_loops = !!(create_loop_dev);
/* XXX: placeholder for a "firewire" debugfs node */
fwserial_debugfs = debugfs_create_dir(KBUILD_MODNAME, NULL);
/* num_ttys/num_ports must not be set above the static alloc avail */
if (num_ttys + num_loops > MAX_CARD_PORTS)
num_ttys = MAX_CARD_PORTS - num_loops;
num_ports = num_ttys + num_loops;
fwtty_driver = tty_alloc_driver(MAX_TOTAL_PORTS
|
__label__pos
| 0.997648 |
15 Aug
Water ripple effect done in HTML 5
I am sure you have seen this effect before, a picture distorted more or less realistically while you move your mouse over it causing the “water” to “ripple”:
water_ripples
In the early 2000s this was done with Java Applets causing a hell of a lot of troubles – from missing and wrong Java versions, over high CPU usage to crashing browsers.
Then Flash came around the corner solving a lot of these problems, but introducing new ones like gaping security holes.
But since all major browsers support HTML 5 Canvas now (Yes, even Internet Explorer starting from IE9) this can be done with Javascript only, which makes things a lot easier and running out of the box (See also: Snow effect done in HTML 5 and Starfield animation done in HTML 5).
Although this effect is nothing more than a gimmick and the practical applications are bare to none, building it was a great way to learn a ton about the Canvas element and its limitations.
The underlying algorithm could not be simpler, i basically copied it from this great explanation: http://freespace.virgin.net/hugo.elias/graphics/x_water.htm
I would never have thought that doing all this calculations on 170.000 pixels 40 times per second would be that efficient (My CPU usage is at only 10-14%), but it shows how much modern Javascript engines optimize your code on runtime.
So it seems that even the Canvas API is missing some features i loved and used in Flash (Like full matrices, direct filtering of image data) you can actually manipulate Canvas image data directly with Javascript and program your own filters and effects to be fast enough on moderately big images.
Enough yapping, here is the code, just copy the files onto a webserver (see notes below as to why): Javascript – Water Ripples
If you are using this on your website and you are a nice person you can show it by giving me a credit (backlink) somewhere on your page, thank you!
Some things to note:
• Runing the script from your local machine instead from a website will not work in Chrome due to the restrictive same-domain policy of the browser.
• Make sure the image is loaded from the same (sub)domain as the website containing the script, otherwise you will get security problems.
• The speed of the waves is mostly defined by the FPS rate and cannot be changed, it is hardwired with the algorithm.
• It can be that the mouse movement is not working in some browsers, i used a W3C working draft method for getting the mouse position. If you want to use this code in production you have to care for that issue yourself (Maybe with jQuery).
• Some calculations could be sped up by using bitwise operators instead of multiplication or division, but this needs the dampers to be hardcoded. It seems that with the newest browers there is no longer a difference in speed with bitwise operators over Math methods, so forget this. They have bad caveats you have to care for yourself anyway.
14 thoughts on “Water ripple effect done in HTML 5
1. hey there – i have been looking for a way to integrate this into a logo image for a while now. I am a TOTAL novice w/ HTML, and our website is a wordpress site. i have tried pasting your code into a “page” in a variety of ways, but none seem to work. wondering if you have any insite – i assumed that the only place to “customize” the code was in the source for the “Original image” at the end:
i.e. Original image:
Water effect:
WAS CHANGED TO
Original image:
Water effect:
…which seemed to give me the image i wanted, but i could not get the ripple effect to work.
Other than the change in image source, i just pasted your code into the area where coding usually lives –
thoughts>?
much appreciated
Jess
• Hi Jess,
it seems that stuff from your post has gone missing while posting, so i can only guess what your problem is.
I’ll just assume that you have a same-origin-policy issue here, because this seems to be the most likely:
The Script (Page) and the Image must be delivered from the same url (i.e. http://www.yourdomain.com) otherwise your browser will not allow the script to manipulate the canvas because of security reasons.
Please check that both resources are delivered via the same domain AND subdomain, and if you still have problems afterwards please don’t hesitate to write me again.
• Persistence pays off 🙂
Add “togglea = false” to the declarations at the top like this:
var image = document.getElementById(waterImageId),
canvas = document.getElementById(waterCanvasId),
context = canvas.getContext("2d"),
width = canvas.width,
height = canvas.height,
waterCache1 = [],
waterCache2 = [],
imageDataSource,
imageDataTarget,
togglea = false;
Then add this just before the for loop that does the animation (just before ” for (var x = 2; x < width + 2; x++) " etc):
if (togglea) {
var blaX = Math.floor(Math.random() * width) + 1;
var blaY = Math.floor(Math.random() * height) + 1;
waterCache1[blaX][blaY] = 127;
waterCache1[blaX + 1][blaY] = 127;
waterCache1[blaX - 1][blaY] = 127;
waterCache1[blaX][blaY + 1] = 127;
waterCache1[blaX][blaY - 1] = 127;
togglea = false;
} else if (Math.floor(Math.random() * 5) == 1) {
togglea = true;
}
Works like a charm 🙂
And I will definitely be linking here, Timewaster. 🙂 My architect at work pointed this page out to me and the water effect is perfect for a pet project I am working on. Much more direct and cleaner than the others I've found
• ah, i see you found the script in the source code of this page.
i only created that so one can see that the image above already has the effect applied to it, hence the really bad code and the “bla” variables and so on, i did not expect people would want to use the droplets…
i will add it to the official script, but with a bit better code.
2. Works in IE 11, but very slow effects; Chrome fails completely. Running Windows 10 Pro i7 Core w/8GB and 2GB Vid. I downloaded .zip and unzip to Desktop sub folder /Water/Javascript-Water_Ripples/… I run index.html and Chrome Inspect flags an ERROR: [index.html:58 Uncaught SecurityError: Failed to execute ‘getImageData’ on ‘CanvasRenderingContext2D’: The canvas has been tainted by cross-origin data.window.onload @ index.html:58 ]. I have checked permissions and moved to a C:\Water folder and still have issues? Your demo page works great with a quick response, so I know it should work. I might be something simple like me not enabling it in a WordPress content folder or failing to call a missing CSS. The images show up top and bottom, just does not respond to mouse ripples.
• hi jason,
that it is slow in IE11 is no surprise, but nothing i can do there. (there is no performance optimization i can do, it is a browser problem)
the problem in chrome is exactly what your browser told you: “The canvas has been tainted by cross-origin data”, this means that a resource (the image) that the script tried to copy onto the canvas element did not came from the same domain as the script (website).
in this case running it locally from your machine is the problem, since both resources (files) did not came from a domain the browser interprets them as foreign to each other.
this means the script can only run when you put it on a webserver and the image and the script are delivered by the same (sub)domain.
if you want to run it locally you need to use a browser which is not as strict on the same origin policy.
3. Hello,
Thank you for hosting this resource. I am working on an HTML5 game that uses a water rippling effect like this in the background. I am going through your code and trying to understand the logic, but unfortunately, the page that you list as your source for the algorithm for this is no longer around. I imagine it must have been good, because you are not the only person who has referred to it as a source. Would you be willing, publicly or privately, to talk me through what’s going on, mostly in your manipulatePixel function? I haven’t worked on anything down to a pixel-by-pixel level before, and I’m not sure exactly how the numeric values are translating into a meaningful ripple effect.
Thanks!
4. How does one go about positioning the image in the document. I can apply a simple center tag but applying anything like a div with a position around both the image and canvas have no effect?
• what the script uses is just an image tag and an canvas tag, both are blocking elements and thus can be positioned like any other div. if it is not working for you i presume you have an error in your css somewhere.
Leave a Reply
Your email address will not be published. Required fields are marked *
This site uses Akismet to reduce spam. Learn how your comment data is processed.
|
__label__pos
| 0.589838 |
Job Queues & CQRS - The pattern that you need to scale to a million request / minute
Santiago Quinteros - CEO & CTO - Software on the road
By:
Santiago Quinteros
Job Queues
A job queue is a mechanism for managing and processing tasks or "jobs" asynchronously. It allows applications to offload tasks not immediately necessary for the primary workflow to a separate service or process, which handles these tasks in the background.
This approach is beneficial for tasks that are resource-intensive, time-consuming, or not critical to the system's immediate response, such as sending emails, generating reports, or performing batch data processing.
How It Works
1. Job Creation: The application creates a job to perform a specific task. This job is then added to the queue, essentially a list of tasks waiting to be processed.
2. Queue Management: The job queue holds the jobs until they can be processed. Jobs can be prioritized, delayed, or scheduled for future execution based on the application's needs.
3. Worker Processes: Separate processes or "workers" continuously monitor the job queue for new jobs to process. Once a worker picks a job from the queue, it executes the task associated with that job.
4. Job Processing: The worker processes the job according to its specified task. This processing happens asynchronously, meaning the application can continue to operate and respond to other requests while the job is being handled in the background.
5. Completion and Callbacks: Once a job is completed, the worker can notify the application of its completion, update its status, and optionally trigger callbacks or follow-up actions.
Benefits of Using a Job Queue
1. Improved Application Performance: By offloading heavy or time-consuming tasks to a job queue, the main application thread remains free to handle incoming requests, improving overall performance and responsiveness.
2. Scalability: Job queues can distribute tasks across multiple workers or servers, making it easier to scale the application horizontally as the workload increases.
3. Reliability: Job queues can implement features like retry mechanisms for failed jobs, ensuring that all tasks are eventually processed even during temporary failures.
4. Flexibility: Developers can schedule jobs to run at specific times, manage the priority of tasks, and adjust the number of workers based on the load, offering greater control over how and when jobs are processed.
Worker
A worker is a dedicated process or thread whose primary purpose is to monitor a job queue and execute the tasks or jobs it finds.
This concept is central to asynchronous processing architectures, where specific tasks are offloaded from the main application flow to be processed in the background, improving efficiency and user experience. Workers are the backbone of this system, ensuring that tasks are executed promptly and orderly.
Key Characteristics of a Worker
Autonomous Operation: Workers run independently of the main application process, often on separate threads or different servers. This separation allows the main application to remain responsive to user requests while workers handle resource-intensive or time-consuming tasks.
Continuous Polling: Workers continuously poll or listen to the job queue for new tasks. Once a task is identified, a worker picks it up, marks it as being processed, and starts the execution of the associated task.
Task Execution: Workers are responsible for executing tasks they pick from the queue. These tasks can range from sending emails, processing files, and generating reports, to any other asynchronous operation that can be performed independently of the main application workflow.
Error Handling and Retries: Workers often have built-in mechanisms for handling errors or failures. If a task fails, a worker can retry the task according to specified rules or policies, ensuring that temporary issues do not lead to task failure.
Scalability: Multiple workers can operate concurrently, allowing the system to scale by adding more workers based on the volume of tasks in the queue. This scalability ensures that the system can handle increasing workloads efficiently.
How Workers Fit into System Architecture
In a typical job queue architecture, workers play a crucial role in balancing the load and ensuring the system's smooth operation.
Here’s how they fit into the overall architecture:
1. Task Offloading: The main application offloads tasks to a job queue instead of executing them synchronously, ensuring the application remains responsive.
2. Monitoring: Workers monitor this queue constantly, waiting for new tasks to be added.
3. Processing: When a worker finds a new task in the queue, it retrieves the task, processes it, and upon completion, removes it from the queue or marks it as completed.
4. Feedback Loop: After processing a task, workers can update the system or application about the task's status, facilitating a feedback loop where the outcome of background processes can influence the application's state or user experience.
Workers are designed to be robust and resilient, capable of gracefully handling system failures, network issues, and other unforeseen errors.
This resilience and the ability to scale by adding more workers make the worker and job queue pattern a powerful tool for building scalable, efficient, and responsive applications.
Implementation in node.js using Mongodb and Agenda.js
Step 1: Set Up MongoDB
Before using Agenda.js, you need a MongoDB instance since Agenda uses MongoDB to store job data. If you still need to get MongoDB installed and running, you must set it up. You can install MongoDB locally or use a cloud-based solution like MongoDB Atlas.
1. Local MongoDB Installation: Follow the MongoDB installation guide for your operating system on the official MongoDB documentation.
2. Cloud-based MongoDB: Sign up for MongoDB Atlas and create a cluster. Atlas offers a free tier for development purposes.
Once your MongoDB instance is ready, note your connection string. You'll need it to connect Agenda.js to your database.
Step 2: Initialize Your Node.js Project
If you haven't already, create a new Node.js project and initialize it:
mkdir my-agenda-app
cd my-agenda-app
npm init -y
Step 3: Install Agenda.js
Install Agenda.js and the MongoDB driver by running:
npm install agenda mongodb
Step 4: Set Up Agenda.js
Now, you'll set up Agenda.js in your Node.js application. Create a file named agendaSetup.js and initialize Agenda with your MongoDB connection:
const Agenda = require('agenda');
const connectionOpts = {
db: { address: 'mongodb://localhost:27017/agendaDb', collection: 'jobs' },
processEvery: '30 seconds'
};
const agenda = new Agenda(connectionOpts);
module.exports = agenda;
If you're using a different database or host, replace 'mongodb://localhost:27017/agendaDb' with your MongoDB connection string.
Step 5: Define Jobs
With Agenda, you define jobs by specifying a name and a function that gets called when the job is run. In the same or a different file, define a job like so:
const agenda = require('./agendaSetup');
agenda.define('say hello', async job => {
console.log('Hello, World!');
});
Step 6: Schedule Jobs
To schedule jobs, you need to start the agenda and then schedule your defined jobs according to your needs. You can do this in an app.js file or at the end of your agendaSetup.js file:
(async function() { // IIFE to use async/await
await agenda.start();
await agenda.every('1 hour', 'say hello');
console.log('Job scheduled to say hello every hour.');
})();
Step 7: Running Your Application
Run your application using Node.js:
node app.js
Advance use case - Sending a welcome email after user sign up
In your sign up endpoint just call agenda and schedule a job
app.post('/signup', async (req, res) => {
const { email } = req.body;
// Here you would add logic to save the user to your database
// Schedule the 'send welcome email' job
await agenda.schedule('in 2 minutes', 'send welcome email', { email });
res.status(200).send('User signed up successfully, welcome email scheduled.');
});
And have the job defined
// Define the 'send welcome email' job
agenda.define('send welcome email', async job => {
const { email } = job.attrs.data;
console.log(`Sending welcome email to ${email}`);
// Here you would integrate with your email service
});
The CQRS Pattern and Integration with Job Queues
Command Query Responsibility Segregation (CQRS) is a software architectural pattern that separates the operations of reading data (queries) from the operations of updating data (commands), allowing them to scale independently and optimize performance, complexity, and security for each operation type.
Integrating job queues with the CQRS pattern can enhance its effectiveness, particularly on the command side of the architecture. This integration brings several benefits, improving the system's scalability, reliability, and responsiveness.
Understanding CQRS
CQRS is based on the principle that the models used to update information do not have to be the same as those used to read information. This separation allows for system design flexibility and can improve performance and scalability. The pattern fits well with event-driven architectures and domain-driven design (DDD), where it can provide clear boundaries and responsibilities within the system.
Benefits of Integrating Job Queues with CQRS
1. Improved Scalability: By using job queues to handle commands, you can offload the execution of these commands to background workers. This allows the system to handle a high volume of write requests more efficiently by spreading the load across multiple workers and resources, enhancing the scalability of the command model.
2. Enhanced Performance: Separating commands and queries allows each to be optimized for specific roles. Job queues can further optimize command execution by ensuring that write operations do not block read operations, thus improving the application's overall performance.
3. Increased Reliability and Fault Tolerance: Job queues can automatically retry failed commands, improving the system's reliability. This is particularly important for operations that must not fail, such as financial transactions or critical data updates. Using job queues ensures that commands can be retried or postponed until they can be completed.
4. Asynchronous Processing: Integrating job queues allows commands to be processed asynchronously, significantly improving the user experience by making the UI more responsive. Users can receive immediate feedback for their actions, even if the underlying command is processed in the background.
5. Event Sourcing Compatibility: CQRS often complements Event Sourcing, where changes to the application state are stored as a sequence of events. Job queues can efficiently handle generating and processing these events, facilitating a robust event-driven architecture.
Implementation Considerations
• Command Handling: In a CQRS-based system integrated with job queues, commands are dispatched to the job queue instead of being executed directly. This decouples the command's issuance from its execution, allowing for more flexible and scalable processing.
• Consistency: While job queues and CQRS can improve performance and scalability, they also introduce eventual consistency into the system. This means the system might only partially reflect the results of a command. Designing your system to handle or mitigate the effects of eventual consistency is crucial.
• Error Handling: Robust error handling and retry mechanisms should be in place to manage failed commands during execution. This ensures that the system can recover gracefully from errors without losing data or corrupting the application state.
CQRS Example
To demonstrate a minimal reproducible example of a CQRS architecture using Express and Agenda.js, let's create a simple application. This app will have a command to "create a user" and a query to "get user details". The "create a user" command will be processed asynchronously using Agenda.js.
Setup Your Project
Initialize a new Node.js project (if you haven't already):
mkdir cqrs-agenda-example
cd cqrs-agenda-example
npm init -y
Install necessary packages:
npm install express agenda mongodb body-parser
Implementing the Command Side with Agenda.js
Set up Express and Agenda.js (app.js):
const express = require('express');
const bodyParser = require('body-parser');
const { MongoClient } = require('mongodb');
const Agenda = require('agenda');
const app = express();
const port = 3000;
app.use(bodyParser.json());
const mongoConnectionString = 'mongodb://127.0.0.1/agenda';
// Initialize MongoDB connection and Agenda
const client = new MongoClient(mongoConnectionString);
const agenda = new Agenda({ db: { address: mongoConnectionString } });
// Placeholder for users' data storage
const users = {};
// Define a job for creating a user in Agenda
agenda.define('create user', async (job) => {
const { userId, userName } = job.attrs.data;
// Simulate user creation delay
await new Promise(resolve => setTimeout(resolve, 1000));
users[userId] = { userId, userName };
console.log(`User created: ${userName}`);
});
(async function() { // Self-invoking async function to ensure proper startup
await client.connect();
await agenda.start();
console.log('Agenda started');
})();
// Command API to create a user
app.post('/users', async (req, res) => {
const { userId, userName } = req.body;
await agenda.schedule('in 2 seconds', 'create user', { userId, userName });
res.send({ message: `User creation scheduled for ${userName}` });
});
// Query API to get a user
app.get('/users/:userId', (req, res) => {
const { userId } = req.params;
const user = users[userId];
if (user) {
res.send(user);
} else {
res.status(404).send({ message: 'User not found' });
}
});
app.listen(port, () => {
console.log(`Example app listening at http://localhost:${port}`);
});
Explanation
• MongoDB and Agenda Setup: This example connects to MongoDB, initializes Agenda with the connection, and defines a job for creating a user. The users object acts as a simple in-memory store.
• Command Endpoint: The POST /users endpoint receives a userId and userName, schedules a "create user" job with Agenda, and responds immediately, acknowledging the scheduling.
• Query Endpoint: The GET /users/:userId endpoint looks up and returns the user's details from the in-memory store. If the user doesn't exist, it returns a 404 error.
• Asynchronous Job Processing: The "create user" job simulates a delay, mimicking a time-consuming task like sending a welcome email or processing additional data. Once the job runs, it adds the user to the in-memory store.
Running the Example
Make sure MongoDB is running locally.
Start your application with node app.js.
Use a tool like Postman or curl to test the command and query endpoints:
To create a user: POST http://localhost:3000/users with JSON body {"userId": "1", "userName": "John Doe"}.
To get a user: GET http://localhost:3000/users/1.
This example illustrates a basic CQRS pattern with asynchronous command processing using Express and Agenda.js.
It demonstrates how commands can be handled separately from queries, allowing for more scalable and responsive applications.
Advance example - CQRS for web scrapping
For this example, we'll design a simple CQRS-based application that schedules web scraping tasks using Playwright, tracks the status of these jobs, and retrieves their results.
This will involve creating a command to schedule a scraping job, and queries to check job status and get results. We'll use Express.js for the web server, Agenda.js for job queueing, and Playwright for web scraping.
Setup
Initialize a new Node.js project:
mkdir cqrs-scraping
cd cqrs-scraping
npm init -y
Install necessary packages:
npm install express agenda mongodb body-parser playwright
Implementation
Set up Express and Agenda.js (server.js):
const express = require('express');
const bodyParser = require('body-parser');
const { MongoClient } = require('mongodb');
const Agenda = require('agenda');
const { chromium } = require('playwright');
const app = express();
app.use(bodyParser.json());
const mongoConnectionString = 'mongodb://127.0.0.1/agenda';
const agenda = new Agenda({ db: { address: mongoConnectionString } });
const jobsResult = {}; // Store job results keyed by job ID
// Define a job for web scraping
agenda.define('web scraping', async (job) => {
const { url } = job.attrs.data;
const browser = await chromium.launch();
const page = await browser.newPage();
await page.goto(url);
const content = await page.content(); // Simplified scraping logic
await browser.close();
// Store result with job ID for retrieval
jobsResult[job.attrs._id] = content;
console.log(`Scraping completed for job ${job.attrs._id}`);
});
(async function() {
await agenda.start();
console.log('Agenda started');
})();
// Endpoint to schedule web scraping
app.post('/scrape', async (req, res) => {
const { url } = req.body;
const job = await agenda.now('web scraping', { url });
res.send({ message: 'Scraping job scheduled', jobId: job.attrs._id });
});
// Endpoint to check job status
app.get('/status/:jobId', (req, res) => {
const { jobId } = req.params;
if (jobsResult[jobId]) {
res.send({ status: 'Completed' });
} else {
res.send({ status: 'In Progress' });
}
});
// Endpoint to get job result
app.get('/result/:jobId', (req, res) => {
const { jobId } = req.params;
const result = jobsResult[jobId];
if (result) {
res.send({ result });
} else {
res.status(404).send({ message: 'Result not found' });
}
});
const port = 3000;
app.listen(port, () => console.log(`Server running on port ${port}`));
How It Works
• Web Scraping Job: The agenda.define function defines a job for web scraping. It uses Playwright to navigate to the specified URL and store the page content in jobsResult, keyed by the job's unique ID.
• Scheduling Endpoint (/scrape): This endpoint schedules a new web scraping job for the given URL and returns the job ID. Clients can use this ID to check the job's status and retrieve results.
• Status Checking Endpoint (/status/:jobId): Clients can use this endpoint to check if the scraping job has been completed.
• Result Retrieval Endpoint (/result/:jobId): Once a job is completed, clients can retrieve the scraped content through this endpoint using the job ID.
Running the Example
Start MongoDB locally if it's not running already.
Run the server script:
node server.js
Usage
Schedule a web scraping job by sending a POST request to /scrape with a JSON body containing the URL to scrape.
Check the job status by sending a GET request to /status/:jobId using the job ID returned from the previous step.
Retrieve the job result by sending a GET request to /result/:jobId once the job is completed.
Conclusion
Job queues and the Command Query Responsibility Segregation (CQRS) pattern represent powerful architectural choices that can significantly enhance the scalability, performance, and maintainability of software systems, especially in complex, distributed environments like microservices.
When implemented thoughtfully, these patterns facilitate a high degree of decoupling between components, allowing for more granular scaling, improved fault tolerance, and greater flexibility in responding to changing requirements or workloads.
Get the latest articles in your inbox.
Join the other 2000+ savvy node.js developers who get article updates. You will receive only high-quality articles about Node.js, Cloud Computing and Javascript front-end frameworks.
santypk4
CEO at Softwareontheroad.com
|
__label__pos
| 0.96527 |
Beefy Boxes and Bandwidth Generously Provided by pair Networks
Problems? Is your data what you think it is?
PerlMonks
Re^5: I know what I mean. Why don't you?
by bigmacbear (Monk)
on May 20, 2006 at 00:31 UTC ( #550624=note: print w/replies, xml ) Need Help??
in reply to Re^4: I know what I mean. Why don't you?
in thread I know what I mean. Why don't you?
Global filehandles and two-argument open are additional red flags.
True, unless one is writing scripts that must run on older systems which came with perl 5.005_03 and can't be upgraded in a timely manner. There's still a lot of Solaris 8 out in the wild to which this applies.
• Comment on Re^5: I know what I mean. Why don't you?
Replies are listed 'Best First'.
Re^6: I know what I mean. Why don't you?
by Aristotle (Chancellor) on May 20, 2006 at 06:34 UTC
You can use Symbol::gensym on such old perls to get lexical filehandles (more awkward, but gets you the same benefits), and you could write a conditionally installed wrapper to emulate three-arg open (only awkward once, but only worthwhile when you can foresee value in making the code easily forward-upgradable). Whether the effort is justified will depend on your particular circumstances, of course (but see also my signature).
Makeshifts last the longest.
Log In?
Username:
Password:
What's my password?
Create A New User
Node Status?
node history
Node Type: note [id://550624]
help
Chatterbox?
and the web crawler heard nothing...
How do I use this? | Other CB clients
Other Users?
Others studying the Monastery: (12)
As of 2016-08-31 17:23 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?
The best thing I ever won in a lottery was:
Results (433 votes). Check out past polls.
|
__label__pos
| 0.504881 |
CSS and jQuery Tutorial: Fancy Apple-Style Icon Slide Out Navigation
Today I want to show you, how to create an Apple-style navigation menu with a twist. Although “fancy” and “Apple-style” don’t really go together, I thought that it’s time for something different. This menu looks very similar to the Apple-style navigation but it reveals some icons when hovering over it. The icons slide out from […]
From our sponsor: Looking for an intuitive whiteboard style project management tool? Give Shortcut a try for free.
Today I want to show you, how to create an Apple-style navigation menu with a twist. Although “fancy” and “Apple-style” don’t really go together, I thought that it’s time for something different.
This menu looks very similar to the Apple-style navigation but it reveals some icons when hovering over it. The icons slide out from the top and when the mouse leaves the link, the icon slides back under the link element. This creates a neat “card-shuffle” effect.
The icons used in this tutorial can be found on DryIcons.com. I am not allowed to redistribute the icons under the free license, so I did not include them in the downloadable ZIP file. But you can easily find them here. I did not rename them for the tutorial so that you can easily recreate the navigation without changing the names or the stylesheet.
Ok, let’s get started!
1. The HTML
The markup just consists of a div with an unordered list inside. The list elements contain a span for the icon and the link element:
<div class="navigation">
<ul class="menu" id="menu">
<li><span class="ipod"></span><a href="" class="first">Players</a></li>
<li><span class="video_camera"></span><a href="">Cameras</a></li>
<li><span class="television"></span><a href="">TVs</a></li>
<li><span class="monitor"></span><a href="">Screens</a></li>
<li><span class="toolbox"></span><a href="">Tools</a></li>
<li><span class="telephone"></span><a href="">Phones</a></li>
<li><span class="print"></span><a href="" class="last">Printers</a></li>
</ul>
</div>
2. The CSS
The style of the navigation container and the unordered list will be the following:
.navigation{
position:relative;
margin:0 auto;
width:915px;
}
ul.menu{
list-style:none;
font-family:"Verdana",sans-serif;
border-top:1px solid #bebebe;
margin:0px;
padding:0px;
float:left;
}
ul.menu li{
float:left;
}
Since we are making the list float, we will need some relative wrapper for the icons. You see, the icons which will be the background-images of the spans in our list, will have absolute positioning. That comes handy when you need to define a z-index, i.e. saying that some element is on top or under another one. Since we want the icons to appear on top and then dissapear under the link element, we will have to deal with z-indeces. And absolute positioning of elements in a relative container will makes things easier.
Now, let’s define the style for the link elements:
ul.menu li a{
text-decoration:none;
background:#7E7E7E url(../images/bgMenu.png) repeat-x top left;
padding:15px 0px;
width:128px;
color:#333333;
float:left;
text-align:center;
border-right:1px solid #a1a1a1;
border-left:1px solid #e8e8e8;
font-weight:bold;
font-size:13px;
-moz-box-shadow: 0 1px 3px #555;
-webkit-box-shadow: 0 1px 3px #555;
text-shadow: 0 1px 1px #fff;
}
We want to give the link elements a fixed width and some background gradient. The borders will create a nice inset effect.
The next hover class will be applied using jQuery, since the :hover pseudo class creates an unwanted effect: When the icon slides out it covers the link for a few milliseconds, making the hover style dissapear in that time. That is perceived as a flicker and we will avoid that by defining a class that we will simply add with jQuery when we do the icon slide out effect:
ul.menu li a.hover{
background-image:none;
color:#fff;
text-shadow: 0 -1px 1px #000;
}
The first and last link element should have a rounded border on the respective side, so we define two classes for those:
ul.menu li a.first{
-moz-border-radius:0px 0px 0px 10px;
-webkit-border-bottom-left-radius: 10px;
border-left:none;
}
ul.menu li a.last{
-moz-border-radius:0px 0px 10px 0px;
-webkit-border-bottom-right-radius: 10px;
}
The common style for all the icon spans will be the following:
ul.menu li span{
width:64px;
height:64px;
background-repeat:no-repeat;
background-color:transparent;
position:absolute;
z-index:-1;
top:80px;
cursor:pointer;
}
The single styles for the specific icons will contain the background-image and the x-position:
ul.menu li span.ipod{
background-image:url(../icons/ipod.png);
left:33px; /*128/2 - 32(half of icon) +1 of border*/
}
ul.menu li span.video_camera{
background-image:url(../icons/video_camera.png);
left:163px; /* plus 128 + 2px of border*/
}
ul.menu li span.television{
background-image:url(../icons/television.png);
left:293px;
}
ul.menu li span.monitor{
background-image:url(../icons/monitor.png);
left:423px;
}
ul.menu li span.toolbox{
background-image:url(../icons/toolbox.png);
left:553px;
}
ul.menu li span.telephone{
background-image:url(../icons/telephone.png);
left:683px;
}
ul.menu li span.print{
background-image:url(../icons/print.png);
left:813px;
}
As you can see, we are positioning the icons in such a way, that they are centered inside of the list element. The top position is 80px initially since we want to show them to the user when the page get’s loaded. Then we will hide them in a stair-like fashion to create an awesome effect!
3. The JavaScript
First, we want to create the effect of the icons dissapearing in a stair-like fashion and then we will define the hover function for the list elements:
$(function() {
var d=1000;
$('#menu span').each(function(){
$(this).stop().animate({
'top':'-17px'
},d+=250);
});
$('#menu > li').hover(
function () {
var $this = $(this);
$('a',$this).addClass('hover');
$('span',$this).stop().animate({
'top':'40px'
},300).css({
'zIndex':'10'
});
},
function () {
var $this = $(this);
$('a',$this).removeClass('hover');
$('span',$this).stop().animate({
'top':'-17px'
},800).css({
'zIndex':'-1'
});
}
);
});
When hovering, we add the class “hover” to the link element, make the icon appear from the top, and increase the z-Index, so that the icon stays on top of the link. When the mouse goes out, we do exactly the opposite, which creates the effect that the icon dissapears behind the link element. For that we will allow more time (800 milliseconds) because it’s such a fun effect that the user might want to enjoy a little bit!
And that’s all!
I hope you like and enjoy it!
Message from TestkingJoin cisco training to learn jQuery and other useful web applications. Learn how to create fancy apple style slid out navigation in jquery using ccvp self study guide and ccda video tutorials.
Tagged with:
Mary Lou
ML is a freelance web designer and developer with a passion for interaction design. She studied Cognitive Science and Computational Logic and has a weakness for the smell of freshly ground peppercorns.
http://www.codrops.com
Stay up to date with the latest web design and development news and relevant updates from Codrops.
Feedback 38
Comments are closed.
1. Why doesn’t appear icons when I put all files out of /Fancy/ in directly in /localhost/? It seems like it can’s see icons or smth else… I’ve changed all links in style.css, etc. Maybe someone knows?
2. @Konstantin The icons are not included in the ZIP file since I am not allowed to redistribute them under the free license. You can download them from dryicons.com. There is as well a direct link to the icon set at the beginning of this post. Hope it helps, cheers, ML
3. Ok! Thks! May be my english isn’t great. so I wrote smth wrong!) Now I put all files,except “icons”, out of /Fancy/ to the root, and it works! I’ve even spread out all icons from “icon” straight to /Fancy/ and it works! But I don’t wunderstand why it doesn’t work if I, for example, rename “Fancy” folder to “icons”??It’s mystic!)) But your message I’ve understood!)Thank you!
4. Dear Mary Lou, that tutorial is absolutely fantastic.
I was wondering whether it is possible to integrate a sub-menu in slide-out for each of the tabs. If yes, can you tell me how…
Take Care
5. This is pretty neat. I love how it animates form the front and goes away form the back of the buttons. That’s slick!
6. Lovely menu bar, working perfectly but asked for a change by client. What changes should i do if i dont want the icons to glide uowards when the page loads?
7. thanks a ton Mary for this wonderful thought of Iconic process, i’m amazed at the things we could do with CSS, ur tutorials are like cake walk, especially this one , i’m a teacher myself but never knew that learning would be such an ease if we made it, thanks again 🙂 looking fwd for more such adorable tutorials
8. Great tutorial, thanks for sharing it. Question: Is there a way to modify the code so that the icons appear/disappear on hovering, but NOT all at once when each page is loaded?
9. Good afternoon,
follow the step by step your tutorial, but I can not animate the menu. When I move the mouse over the menu, nothing happens. No icons appear. What could it be? I’m working with localhost before sending it to the server.
Camilla M.
|
__label__pos
| 0.779045 |
Artuino Assembly Code UCLA
FratBro23
Category:
Other
Price: $5 USD
Question description
ARDUINO CODE
Problem 1.
Write a program largeHalf.c that takes in a list of 8 integers, splits the list into two halves (first 4
elements, last 4 elements), sums the elements in each half, selects the half with the larger sum, and then
repeats this process with the selected list until there is only one element selected. The program should
print out the half selected at each step. If the two halves have equal sum, the program should tell the
user this and should always pick the first half (the one with a smaller starting index, see the example
output below).
Your program must include the following functions:
int sumArray(int array[], int startIdx, int len) - Returns the sum of the
elements in array with index startIdx to startIdx+len.
void printArray(int array[], int startIdx, int len) - Print the elements
in array with index startIdx to startIdx+len.
We suggest implementing and testing each of these functions before completing the full program.
EXAMPLE:
(˜)$ a.out
Enter 8 numbers: 1 2 3 4 5 6 7 8
Larger half is 5 6 7 8
Larger half is 7 8
Larger half is 8
(˜)$ a.out
Enter 8 numbers: 1 2 3 4 5 4 1 0
The two halves are equal, picking 1 2 3 4
Larger half is 3 4
Larger half is 4
(˜)$ a.out
Enter 8 numbers: 1 4 5 0 -10 10 1 2
Larger half is 1 4 5 0
The two halves are equal, picking 1 4
Larger half is 4
(˜)$
Problem 2.
Write a program sortLen.c that takes in a set of user supplied words and sorts them by their length.
Before taking in the words, the program will prompt the user for the total number of words they want
to enter. The program will then prompt the user for each word, one by one. After taking in the words,
the program will convert them all to lowercase. Then the program will sort the words by their length
(from shortest to longest word). It will do so by using a simple sorting algorithm.
This algorithm works by repeatedly scanning through the list and checking if adjacent words are in the
correct order (if the shorter word precedes the longer word). If the adjacent words are not in the correct
order, it will swap them. After scanning through the list once, it is possible that not all words are sorted,
for example, if we started with words of length 1 4 3 2, when we check the first and second word, they
have the appropriate order (1 < 4), so no swap is made. Then when we continue scanning, we see that
the second and third words have lengths 4 and 3, respectively. Thus, we’ll swap these two words, and
the lengths will now be 1 3 4 2. Next we need to check the third and fourth words, with lengths 4 and
2, and we’ll need to swap these words. After this scan of the list, the words have been reordered, so the
lengths, in order, are now 1 3 2 4. The overall list of words is not yet sorted by length. However, if we
repeatedly scan through the list in this manner until the algorithm does not need to swap any adjacent
entries, then the whole list will be sorted. Thus, the basic algorithm is as follows:
a. Scan through the list, starting from the first word, comparing adjacent words. If the order of the
two words is incorrect (if the longer word precedes the shorter word), swap them. Note: if the
words are the same length, no swap is necessary.
b. Repeat step a until we can scan through the list without swapping any consecutive words. (At
this point the list is sorted!)
Your program must include the following functions:
void lowercaseString(char string[]) - Converts all of the letter characters in the
array string into lowercase.
int stringLen(char string[]) - Returns the length of the string array string.
void swapStrings(char string1[], char string2[]) - Swaps the string arrays
string1 and string2.
We suggest implementing and testing each of these functions before completing the full program.
EXAMPLE:
(˜)$ ./a.out
How many words are you planning to enter? 3
Enter word 1: This
Enter word 2: is
Enter word 3: FUN
Sorted words:
is
fun
this
(˜)$ ./a.out
How many words are you planning to enter? 5
Enter word 1: Sorting
Enter word 2: is
Enter word 3: NOT
Enter word 4: too
Enter word 5: difficult
Sorted words:
is
not
too
sorting
difficult
(˜)$ ./a.out
How many words are you planning to enter? 4
Enter word 1: 1
Enter word 2: LAST
Enter word 3: example
Enter word 4: rUn
Sorted words:
1
run
last
example
Problem 3.
Write a program integral.c that calculates rectangular and trapezoidal integral approximations for
a function. Your program will ask the user for the starting and ending time-points of the integration
and the number of time points (polygons) to use in the approximation. In addition to printing out the
estimated integral, the program will provide the error in the approximation. The figure below provides
an illustration of integration and the approximation methods you will be implementing.
...
...
f(t)
t
t=a t=b
Δt
...
...
f(t)
t
t=a t=b
f(a+Δt/2)
Δt
...
...
f(t)
t
t=a t=b
f(a+Δt) f(a+2*Δt)
∫f(t)dt
a
b ∫f(t)dt
a
b
Rectangle Approx of ∫f(t)dt
a
b
Trapezoid Approx of
0 2
1
In the figure on the left, when we integrate a function we are taking the area under the curve between
t = a and t = b (in blue), with the caveat that areas calculated for ranges of t for which f (t) < 0 are
subtracted from areas calculated for ranges of t for which f (t) > 0. Thus, in the figure, the integral of
f (t) over the range a to b is equal the blue area labeled 2 subtracted from the blue area labeled 1.
We can approximate this integral computationally (instead of using analytical integration for an exact
answer, i.e. what you learn in calculus) by using a number of methods. Two straightforward methods
of approximation are the rectangle and trapezoid methods. With these methods, the integration range
is split into n segments and the area of these segments is approximated with a simple polygon. The
estimated areas of all of the n segments are signed as either positive or negative (based upon whether
they are above or below the t-axis) and are summed.
The center figure (green) demonstrates the rectangle method for n = 4. The width of the rectangle is
4t = (b
Tutor Answer
(Top Tutor) Daniel C.
(997)
School: UC Berkeley
PREMIUM TUTOR
Studypool has helped 1,244,100 students
Ask your homework questions. Receive quality answers!
Type your question here (or upload an image)
1828 tutors are online
Related Other questions
11/17/2013
11/17/2013
11/17/2013
Brown University
1271 Tutors
California Institute of Technology
2131 Tutors
Carnegie Mellon University
982 Tutors
Columbia University
1256 Tutors
Dartmouth University
2113 Tutors
Emory University
2279 Tutors
Harvard University
599 Tutors
Massachusetts Institute of Technology
2319 Tutors
New York University
1645 Tutors
Notre Dam University
1911 Tutors
Oklahoma University
2122 Tutors
Pennsylvania State University
932 Tutors
Princeton University
1211 Tutors
Stanford University
983 Tutors
University of California
1282 Tutors
Oxford University
123 Tutors
Yale University
2325 Tutors
|
__label__pos
| 0.818411 |
1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.
Question about transfering Euros between Accounts
Discussion in 'Archive' started by Lisek, Dec 20, 2011.
Thread Status:
Not open for further replies.
1. Lisek
Lisek User
Joined:
06.12.11
Messages:
21
Likes Received:
4
hello,
i would like to ask if there is any way to transfer my euros from my acc to another? i mean that i would like to buy pa on other acc.
thx for answer.
edit : damn, i should add this to questions and info
Last edited by a moderator: Dec 21, 2011
2. Sefyu
Sefyu User
Joined:
30.11.11
Messages:
19
Likes Received:
0
the answer is proly "no", cause otherwise people could sell each other items with real money.
3. Lisek
Lisek User
Joined:
06.12.11
Messages:
21
Likes Received:
4
? now they can do this aswell, it's called bank transfer...
4. Adamaris
Adamaris User
Joined:
28.11.11
Messages:
48
Likes Received:
0
the answer should be obviously a no, because people would abuse the facebook event and get 10€ extra from multiple accounts.
in fact, there is no point in transfering points from an account to another unless you want to abuse the 10€ reward.
Last edited by a moderator: Dec 21, 2011
5. Lisek
Lisek User
Joined:
06.12.11
Messages:
21
Likes Received:
4
and if i want to stop playing, and i want to give that € to my friend?
or... ive got another acc with buffer, and now i want to play my buffer as main char?
6. Adamaris
Adamaris User
Joined:
28.11.11
Messages:
48
Likes Received:
0
if you stop playing and you want to give that € to a friend, spend the € on items and give them to your friend, easy fix ;)
and if you got another account and want to spend money on it, an easy fix is to not charge 50€ in an account and instead charge the amount you need, all your questions have common sense solutions.
i don't see the point of needing € transfer between accounts, it has more "illegal" purposes than legit ones.
Last edited by a moderator: Dec 21, 2011
7. Hemigidius
Hemigidius User
Joined:
05.12.11
Messages:
12
Likes Received:
0
whats is this 10€ you talk about?
8. OteReload
OteReload User
Joined:
15.12.11
Messages:
102
Likes Received:
0
ppl are creating fake facebook accounts for the 10€ bonus
then buy somethings in the shop and sell it for adena
simple exploit.
9. Etherial
Etherial User
Joined:
03.12.11
Messages:
1,398
Likes Received:
0
we'd like to inform you that this topic is being closed, since it is inactive, and is being archived, as planned.
kind regards.
Thread Status:
Not open for further replies.
|
__label__pos
| 0.906869 |
How much Internet data does Valorant use? [ANSWERED]
how_much_internet_data_does_valorant_use
Valorant is quite a popular game in the gaming market.
It has a large number of players which keeps on increasing!
When a player plays the game, numerous questions come to the player's mind.
One of the most common questions that comes to the players' minds is regarding the internet data usage of Valorant.
They can't help but wonder, how much internet data does Valorant use?
On average, Valorant uses somewhere around 200 to 400 MB of internet data per hour which is higher than the average data usage of an online game.
Although the data usage of the game might fluctuate depending on a variety of factors, it generally stays under 500 MB.
However, at times, Valorant uses more than 500 MB of internet data in one hour.
In this article, I've answered this question in more detail.
I've also answered another related question that is quite common among the Valorant players.
So, without any further ado, let's get into it!
Internet Data Usage of Valorant
To find the actual internet data usage of the game, I did some research.
I asked all of my friends who played the game, to keep an eye on its data usage.
Meanwhile, I also did the same.
Every one of us played Valorant for an hour at a stretch and recorded its data usage.
I found out that the data usage of the game on the different devices was in the range of 150 to 400 MB per hour.
But, in most of the devices, the internet data used by the game in one hour was around 280 MB.
So, we can conclude that the internet data usage of Valorant is around 280 MB per hour.
From here, we can also calculate the data used by it in one week, month, or even more!
If you play Valorant for one hour every day for one week, the internet data consumed by the game would be around 1.91 GB in one week.
Similarly, if the game is played for one hour every day for one month, the internet data usage of Valorant in one month would be around 8.2 GB.
Now, what is the data usage per match in Valorant?
Each match in Valorant consumes around 50 to 180 MB of internet data.
At times, the game consumes more than 200 MB per match.
And, if one uses the voice chat, the data used by the game per match might be more than 180 MB.
You might now want to know about the data used by Valorant according to the duration for which you have played the game.
To help you with that, I've prepared a table.
I've also done all the required math so that you don't need to do so!
Following is the required table:
Playtime per day Data usage per day Data usage per week Data usage per month
1 hour 280 MB 1.91 GB 8.2 GB
2 hours 560 MB 3.82 GB 16.4 GB
3 hours 840 MB 5.74 GB 24.6 GB
4 hours 1.09 GB 7.65 GB 32.81 GB
5 hours 1.36 GB 9.57 GB 41.01 GB
6 hours 1.64 GB 11.48 GB 49.21 GB
7 hours 1.91 GB 13.39 GB 57.42 GB
Note: In the above table, the internet data usage per hour of Valorant has been taken as 280 MB.
Does Valorant use a lot of Internet data?
Compared to most of the other online games available in the market, Valorant relatively uses a slightly higher amount of internet data (around 280 MB per hour).
Conclusion
In this article, I've answered a common question regarding the internet data usage of Valorant.
I've also answered a related question that is quite common among the Valorant players.
Now, I'd like to hear from you:
Do you already play Valorant? If so, when did you start playing it?
Or, you might have a question.
Either way, feel free to let me know by dropping down a comment below!
I've also written an article about the internet data usage of League of Legends. You can read it here
Cheers,
Raj Oberoi
Raj Oberoi
Raj Oberoi is a gaming enthusiast who plays a wide variety of games. When not playing games, he loves to share his views and opinions about different games.
Post a Comment (0)
Previous Post Next Post
–>
|
__label__pos
| 0.584594 |
Video Screencast Help
Symantec Protection Engine: What is a Container File?
Created: 15 Jan 2013 • Updated: 12 Feb 2013 | 2 comments
InsentraCameronM's picture
This issue has been solved. See solution.
Hello,
I work with Symantec Scan Engine and Symantec Protection Engine. Can someone please define exactly what a container file is? Which file types are container files?
Cheers,
Cameron
Comments 2 CommentsJump to latest comment
ᗺrian's picture
Anything that can be compressed. So a ZIP or RAR file or anything similar.
Please click the "Mark as solution" link at bottom left on the post that best answers your question. This will benefit admins looking for a solution to the same problem.
TSE-JDavis's picture
The definition is fairly broad; basically anything that can contain binary data embedded in it. This can be PDF files, XLSX files and even JPEG files.
SOLUTION
|
__label__pos
| 0.766957 |
15 divided by what equals 59?
15 divided by what equals 59? If you're looking to solve this word problem then you're in the right place. If you have the number 15 and you want to divide it by something to get the answer 59 then this quick equation lesson will show you exactly how to find that missing number "something".
First of all, we can write this problem out and use the letter X to be the missing number we want to try and find:
15 / X = 59
The first step is to multiply both sides of this equation by the missing number X. We don't know what X is yet, so we do this by adding X in brackets:
15(X) / X = 59(X)
If you're new to equations this might look a little confusing but all we are really saying is that 15 is the same as 59 times X.
To find X, we need to divide both sides by our final answer, 59:
15 / 59 = 59(X) / 59
So, our final answer here to 15 divided by what equals 59 is:
0.2542 = X
In these answers we round them to a maximum of 4 decimal places because some calculations might have long decimal answers. If you want to check whether the answer is close, you can divide 15 by 0.2542:
15 / 0.2542 = 59.0087
Hopefully now you know exactly how to work out math problems like these yourself in future. Could I have just told you to divide 15 by 59? Yes, but aren't you glad you learned the process?
Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers.
Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "15 divided by what equals 59?". VisualFractions.com. Accessed on January 28, 2023. http://visualfractions.com/calculator/divided-by-what/15-divided-by-what-equals-59/.
• "15 divided by what equals 59?". VisualFractions.com, http://visualfractions.com/calculator/divided-by-what/15-divided-by-what-equals-59/. Accessed 28 January, 2023.
• 15 divided by what equals 59?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/divided-by-what/15-divided-by-what-equals-59/.
Divided by What Equals Calculator
|
__label__pos
| 0.686147 |
更新时间：2022-02-25 GMT+08:00
分享
目的端为OBS
JSON样例
"to-config-values": {
"configs": [
{
"inputs": [
{
"name": "toJobConfig.bucketName",
"value": "cdm"
},
{
"name": "toJobConfig.outputDirectory",
"value": "/obsfrom/advance/"
},
{
"name": "toJobConfig.outputFormat",
"value": "CSV_FILE"
},
{
"name": "toJobConfig.fieldSeparator",
"value": ","
},
{
"name": "toJobConfig.writeToTempFile",
"value": "false"
},
{
"name": "toJobConfig.validateMD5",
"value": "false"
},
{
"name": "toJobConfig.recordMD5Result",
"value": "false"
},
{
"name": "toJobConfig.encodeType",
"value": "UTF-8"
},
{
"name": "toJobConfig.markerFile",
"value": "finish.txt"
},
{
"name": "toJobConfig.duplicateFileOpType",
"value": "REPLACE"
},
{
"name": "toJobConfig.shouldClearTable",
"value": "false"
},
{
"name": "toJobConfig.columnList",
"value": "1&2"
},
{
"name": "toJobConfig.quoteChar",
"value": "false"
},
{
"name": "toJobConfig.encryption",
"value": "NONE"
},
{
"name": "toJobConfig.copyContentType",
"value": "false"
}
],
"name": "toJobConfig"
}
]
}
参数说明
参数
是否必选
类型
说明
toJobConfig.bucketName
String
OBS的桶名，例如“cdm”
toJobConfig.outputDirectory
String
数据写入路径，例如“data_dir”
toJobConfig.outputFormat
枚举
写入数据时所用的文件格式（二进制除外），支持以下文件格式：
• CSV_FILE：按照CSV格式写入数据。
• BINARY_FILE：二进制格式，不解析文件内容直接传输，CDM会原样写入文件，不改变原始文件格式。
当选择“BINARY_FILE”时，源端也必须为文件系统。
toJobConfig.fieldSeparator
String
列分割符号，当“toJobConfig.outputFormat”（文件格式）为“CSV_FILE”时此参数有效，默认值为：“,”
toJobConfig.lineSeparator
String
行分割符号，当“toJobConfig.outputFormat”（文件格式）为“CSV_FILE”时此参数有效，默认值为：“\r\n
toJobConfig.writeFileSize
String
源端为数据库时该参数有效，支持按大小分成多个文件存储，避免导出的文件过大，单位为MB。
toJobConfig.duplicateFileOpType
枚举
重复文件处理方式，只有文件名和文件大小都相同才会判定为重复文件。重复文件支持以下处理方式：
• REPLACE：替换重复文件。
• SKIP：跳过重复文件。
• ABANDON：发现重复文件停止任务。
toJobConfig.encryption
枚举
选择是否对上传的数据进行加密，以及加密方式：
• NONE：不加密，直接写入数据。
• KMS：使用数据加密服务中的KMS进行加密。如果启用KMS加密则无法进行数据的MD5校验。
• AES-256-GCM：使用长度为256bit的AES对称加密算法，目前加密算法只支持AES-256-GCM（NoPadding）。
toJobConfig.dek
String
数据加密密钥，“toJobConfig.encryption”（加密方式）选择“AES-256-GCM”时有该参数，密钥由长度64的十六进制数组成。
请您牢记这里配置的密钥，解密时的密钥与这里配置的必须一致。如果不一致系统不会报异常，只是解密出来的数据会错误。
toJobConfig.iv
String
初始化向量，“toJobConfig.encryption”（加密方式）选择“AES-256-GCM”时有该参数，初始化向量由长度32的十六进制数组成。
请您牢记这里配置的初始化向量，解密时的初始化向量与这里配置的必须一致。如果不一致系统不会报异常，只是解密出来的数据会错误。
toJobConfig.kmsID
String
上传时加密使用的密钥。需先在密钥管理服务中创建密钥。
toJobConfig.projectID
String
KMS密钥所属的项目ID。
toJobConfig.validateMD5
Boolean
选择是否校验MD5值，不能与KMS加密同时使用。使用二进制格式传输文件时，才能校验MD5值。
计算源文件的MD5值，并与OBS返回的MD5值进行校验。如果源端已经存在MD5文件，则直接读取源端的MD5文件与OBS返回的MD5值进行校验。
toJobConfig.recordMD5Result
Boolean
当选择校验MD5值时，这里配置是否记录校验结果。
toJobConfig.recordMD5Link
String
可以指定任意一个OBS连接，将MD5校验结果写入该连接的桶。
toJobConfig.recordMD5Bucket
String
写入MD5校验结果的OBS桶。
toJobConfig.recordMD5Directory
String
写入MD5校验结果的目录。
toJobConfig.encodeType
String
编码类型，例如：“UTF_8”“GBK”
toJobConfig.markerFile
String
当作业执行成功时，会在写入目录下生成一个标识文件，文件名由用户指定，不指定时默认关闭该功能。
toJobConfig.copyContentType
Boolean
“toJobConfig.outputFormat”（文件格式）为“BINARY_FILE”，且源端、目的端都为对象存储时，才有该参数。
选择“是”后，迁移对象文件时会复制源文件的Content-Type属性，主要用于静态网站的迁移场景。
归档存储的桶不支持设置Content-Type属性，所以如果开启了该参数，目的端选择写入的桶时，必须选择非归档存储的桶。
toJobConfig.quoteChar
Boolean
“toJobConfig.outputFormat”（文件格式）为“CSV_FILE”，才有该参数，用于将数据库的表迁移到文件系统的场景。
选择“是”时，如果源端数据表中的某一个字段内容包含字段分隔符或换行符，写入目的端时CDM会使用双引号（"）作为包围符将该字段内容括起来，作为一个整体存储，避免其中的字段分隔符误将一个字段分隔成两个，或者换行符误将字段换行。例如：数据库中某字段为hello,world，使用包围符后，导出到CSV文件的时候数据为"hello,world"。
toJobConfig.firstRowAsHeader
Boolean
“toJobConfig.outputFormat”（文件格式）为“CSV_FILE”时才有该参数。在迁移表到CSV文件时，CDM默认是不迁移表的标题行，如果该参数选择“是”，CDM在才会将表的标题行数据写入文件。
分享：
相关文档
相关产品
关闭导读
|
__label__pos
| 0.647703 |
Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post
-/+ Root Solutions
Announcements Posted on
1. Offline
ReputationRep:
\sqrt{10}cos(x-a)=2
The mark scheme lists the only solutions when 2 is divided by positive root 10
Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?
2. Offline
ReputationRep:
(Original post by pleasedtobeatyou)
\sqrt{10}cos(x-a)=2
The mark scheme lists the only solutions when 2 is divided by positive root 10
Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?
\sqrt4 = 2 \not= \pm 2
The value obtained from the square root is always taken as positive.
3. Offline
ReputationRep:
maybe there is some limitation on x, like 0\leq x\leq \pi
4. Offline
ReputationRep:
This is from Core 4 but in some Core 3 questions for example,
tan^2x = 6
There's a need to find the solutions for -root6 and +root6
5. Offline
ReputationRep:
(Original post by njl94)
maybe there is some limitation on x, like 0\leq x\leq \pi
Nope, the solutions for the negative root would also be acceptable
6. Offline
ReputationRep:
(Original post by pleasedtobeatyou)
This is from Core 4 but in some Core 3 questions for example,
tan^2x = 6
There's a need to find the solutions for -root6 and +root6
\tan^2 x = 6 \implies \tan x = \pm \sqrt{6}
It is the same as x^2 = 4 \implies x = \pm \sqrt4 = \pm 2
But x = \sqrt4 = 2 \not= \pm \sqrt2
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
that username has been taken, please choose another Forgotten your password?
2. this can't be left blank
this email is already registered. Forgotten your password?
3. this can't be left blank
6 characters or longer with both numbers and letters is safer
4. this can't be left empty
your full birthday is required
1. By joining you agree to our Ts and Cs, privacy policy and site rules
2. Slide to join now Processing…
Updated: June 12, 2012
New on TSR
Your TSR exam season toolkit
Everything you need to know about study help on TSR
Study resources
Article updates
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.
|
__label__pos
| 0.984249 |
Closing a TCP Asynch socket C#
I have a TCP socket application based on examples of asynch socket programming available on Microsoft site and an example on Code Project.
My client is a GPRS based device which does not generally do a clean close on the TCP connection. I need to remove these idle connections on my TCP server. Am maintaining a hashtable in the server that keeps a record of the unique unitID sent by the client and the associated socket. When the client reconnects -I need to course through the hashtable and close the socket related with the previous client ID and remove the entry in the hashtable and add this new entry with the current socket.
My question is how to cleanly and surely remove a socket and how to manage the hashtable while doing so
My code is as follows
private bool removePrevSocketIfPresent(Hashtable hashTableUnits, long unitID)
{
bool allWEll = false;
SocketAsyncEventArgs tempSocketToremove;
DataHoldingUserToken tempTokenToRemove;
//locking the hashtable for the whole operation
lock (TRInitModule.hashTableUnits.SyncRoot)
{
bool markForRemoval = false;
foreach (DictionaryEntry entry in TRInitModule.hashTableUnits)
{
Int64 currentUnitID = Convert.ToInt64(entry.Key);
if (currentUnitID == unitID)
{
try
{
tempSocketToremove = (SocketAsyncEventArgs)(entry.Value);
tempTokenToRemove = (DataHoldingUserToken)tempSocketToremove.UserToken;
tempSocketToremove.SocketError = SocketError.OperationAborted;
tempSocketToremove.AcceptSocket.Shutdown(SocketShutdown.Both);
//is the above line enough or do I need to do the below lines too?
// tempSocketToremove.AcceptSocket.Close();
// tempTokenToRemove.Reset();
//if (tempTokenToRemove.theDataHolder.dataMessageReceived != null)
//{
// tempTokenToRemove.CreateNewDataHolder();
//}
markForRemoval = true;
Console.WriteLine();
Console.WriteLine("One socket removed, count is = " + TRInitModule.hashTableUnits.Keys.Count);
allWEll = true;
}
catch (Exception eRT)
{
allWEll = false;
Console.WriteLine("ERROR IN REMOVE ITEM " + eRT.Message);
}
}
else
{
markForRemoval = false;
}
}
if (markForRemoval)
{
TRInitModule.hashTableUnits.Remove(unitID);
}
}
return allWEll;
}
Open in new window
MILIND_JOGAsked:
Who is Participating?
ambienceCommented:
For one, this defeats the purpose of a dictionary
foreach (DictionaryEntry entry in TRInitModule.hashTableUnits)
{
Int64 currentUnitID = Convert.ToInt64(entry.Key);
if (currentUnitID == unitID)
Open in new window
This is how it should be
if (TRInitModule.hashTableUnits.ContainsKey(unitID))
Also, "AcceptSocket.Shutdown(SocketShutdown.Both);" will try to flush buffers and most likely wont do anything if the socket has been idle for some time. You should call Socket.Close() because that would free the resources associated with the socket. In fact, you may choose not to call Shutdown() if a client will only establish one connection at a time and is found reconnecting.
0
MILIND_JOGAuthor Commented:
Please explain why Socket.Shutdown() is not needed.
Also, once I know that the hashtable contains the unitID - I need to first close the socket with that key - so how do I get to that socket without running a foreach loop.
Thanks for your help though.
0
ambienceCommented:
Shutdown() is needed to make sure any pending data is received and sent, but since your device is "re-connecting" it wouldnt work since its establishing a new connection. You call shutdown in following scenario for example
socket.send(.......)
socket.shutdown() -> make sure previous send is completed
socket.close()
-
This is how you should get the socket for dictionary.
var socketToRemove = TRInitModule.hashTableUnits[unitId];
0
Cloud Class® Course: CompTIA Cloud+
The CompTIA Cloud+ Basic training course will teach you about cloud concepts and models, data storage, networking, and network infrastructure.
MILIND_JOGAuthor Commented:
I also need to check for idle sockets ( more than 5 minutes) and remove them. Which is the best property to use to check for idle sockets?
0
ambienceCommented:
Well, a simple way is to keep a timestamp with the device specific data, like say
DateTime mLastActivityTime;
Update this every time you receive anything from the device.
You can then have a separate thread run at regular intervals to run over all connected devices and close those been idle for over 5 minutes.
0
MILIND_JOGAuthor Commented:
Wow that means I am on the right path. I have done just that. Please tell me if the closing is alright is code below. I have a list of all sockets plus a hashtable with unitID as key and the current live socket as I need to be able to send specific commands to the unit from time to time.
int listcount4 = Program.listOfSockets.Count;
SocketAsyncEventArgs[] array4Purge = new SocketAsyncEventArgs[listcount4];
Program.listOfSockets.CopyTo(array4Purge);
foreach (SocketAsyncEventArgs socket4 in array4Purge)
{
theUserTokenHere = (DataHoldingUserToken)socket4.UserToken;
//unitInKey = theUserTokenHere.unitID;
double timeSincePrevData = calcTimeInterval(theUserTokenHere);
if (timeSincePrevData > 0.8)
{
socket4.SocketError = SocketError.NotConnected;
socket4.AcceptSocket.Shutdown(SocketShutdown.Both);
Program.listOfSockets.Remove(socket4);
if (TRInitModule.hashTableUnits.ContainsValue(socket4))
{
lock (TRInitModule.hashTableUnits.SyncRoot)
{
TRInitModule.hashTableUnits.Remove(socket4);
}
}
}
}
Thanks !!
Milind
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment.
All Courses
From novice to tech pro — start learning today.
|
__label__pos
| 0.851312 |
Take the tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
This is one of my Mathematica Assignment problems. Assuming that the matrix $H$ is Hermitian and the matrix $U$ is unitary, prove that the matrix $A = U^{-1}HU$ is Hermitian. Can someone help? I'm unable to do it!
share|improve this question
I believe you meant to post this on Mathematics. This site is for the software program Mathematica, not mathematics. – rcollyer Aug 5 at 11:55
2
The definitions of Unitary and Hermitian matrices are $U^{-1} = U^*$ and $H=H^*$. You just need to check that $A^* = A$. The identity $(MN)^*=N^*M^*$ will be helpful. – Anthony Carapetis Aug 5 at 13:24
add comment
migrated from mathematica.stackexchange.com Aug 5 at 13:20
This question came from our site for users of Mathematica.
1 Answer
The matrix $H$ satisfies ${}^t \overline{H} = H$ and $U$ satisfies ${}^t \overline{U} U = I_{n}$, which means that $\left( {}^t \overline{U} \right)^{-1} = U$. Consider ${}^t \overline{A}$, you have
$${}^t \overline{A} = {}^t \overline{U} {}^t \overline{H} {}^t \overline{U^{-1}} $$
Using the previous equalities, you get :
$${}^t \overline{A} = U^{-1} H U = A$$
So $A$ is hermitian.
share|improve this answer
add comment
Your Answer
discard
By posting your answer, you agree to the privacy policy and terms of service.
Not the answer you're looking for? Browse other questions tagged or ask your own question.
|
__label__pos
| 0.994568 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.