fork

I am trying to run a background process in perl. I create a child process, which is used to call another perl script. I want to run few lines of code parallely with this child process. And after the child process is done.I want to print a line of code. Main script #!/usr/bin/perl $|=1; print "before the child process\n"; my $pid = fork(); if (defined $pid) { system("perl testing.pl"); } print "before wait command\n"; wait(); print "after 20 secs of waiting\n"; testing.pl #!/usr/bin/perl print "inside testing\n"; sleep(20); Expected output before the child process before wait command (should

I have a doubt in the following piece of code and its behaviour: #include <stdio.h> #include <unistd.h> #include <stdlib.h> #define N 5 #define nt 1 int pm[N][2],pid[N]; int i,j,size; char s[100]; int count=0; int main() { for(i=1;i<N;i++) { printf("\n i=%d",i); if(pid[i]=fork() == -1) { printf("\n fork not wrking"); exit(1); } else if(pid[i]>0) { printf(" \n pid:%3d\n",pid[i]); break; } } return 0; } I initially thought that the code will spawn a process and skip out of the loop. Thereby, 1 spawns 2 and skips out. 2 spawns 3 and skips out 3 spawns 4 and skips out 4 spawns 5 and skips out. I

Can someone explain to me why Process.fork makes stuff so much slower in Ruby? I'm using Ruby 2.3.1 on OS X El Capitan. require 'time' require 'benchmark' def do_stuff 50000.times { Time.parse(Time.utc(2016).iso8601) } end puts Benchmark.measure { do_stuff } # => 1.660000 0.010000 1.670000 ( 1.675466) Process.fork do puts Benchmark.measure { do_stuff } # => 3.170000 6.250000 9.420000 ( 9.508235) end EDIT: Just noticed that running that code on Linux (tested Debian or Ubuntu) does not result in a negative performance impact. user513951 "Why does Process.fork make stuff slower in Ruby on OS X?"

为什么要做持久化存储? 持久化存储是将 Redis 存储在内存中的数据存储在硬盘中，实现数据的永久保存。我们都知道 Redis 是一个基于内存的 nosql 数据库，内存存储很容易造成数据的丢失，因为当服务器关机等一些异常情况都会导致存储在内存中的数据丢失。 持久化存储分类 在 Redis 中，持久化存储分为两种。一种是 aof 日志追加的方式，另外一种是 rdb 数据快照的方式。 RDB持久化存储 什么是RDB持久化存储 RDB持久化存储即是将redis存在内存中的数据以快照的形式保存在本地磁盘中。 .RDB持久化存储分为自动备份和手动备份 1.手动备份通过 save 命令和 bgsave 命令。save是同步阻塞，而 bgsave 是非阻塞(阻塞实际发生在 fork 的子进程中)。因此，在我们实际过程中大多是使用bgsave命令实现备份. redis> SAVE OK redis> BGSAVE Background saving started 2.自动备份 a.修改配置项 save m n即表示在 m 秒内执行了 n 次命令则进行备份. b.当Redis 从服务器项主服务器发送复制请求时，主服务器则会使用 bgsave命令生成 rbd 文件，然后传输给从服务器. c.当执行 debug reload 命令时也会使用 save 命令生成rdb文件. d.当使用

I have a program which runs two different operations and i'd like to share variables between them. At the moment, i'm using threads instead of fork processes but i'm having problems in sharing variables even if I declared them as volatile. I tried with boost doing: boost::thread collisions_thread2(boost::bind(function_thread2); by declaring the shared variables as volatile, but it seems that function_thread2() function is not able to see changes on the shared variables. What i'd like to do is something like: thread1: while(true){ //..do somet stuff check variable1 } thread2: while(true){ do

Here is a code: int i=0; pid_t pid; puts("Hello, World!"); puts(""); pid = fork(); if(pid) i=42; printf("%p\n", &i); printf("%d\n", i); puts(""); And output Hello, World! 0x7fffc2490278 42 0x7fffc2490278 0 Program print Hello, World! one time, so child process wasn't start from the beginning and it wasn't re-define variables. Adresses of variables are same. So they are same. But I change i's value in parent process which is executed first, it didn't change for child process. Why? Adresses of variables are same. So they are same. But I change i's value in parent process which is executed first,

I have a perl script, script.pl which, when run, does a fork, the parent process outputs its pid to a file then exits while the child process outputs something to STOUT and then goes into a while loop. $pid = fork(); if ( ! defined $pid ) { die "Failed to fork."; } #Parent process elsif($pid) { if(!open (PID, ">>running_PIDs")) { warn "Error opening file to append PID"; } print PID "$pid \n"; close PID; } #child process else { print "Output started"; while($loopControl) { #Do some stuff } } This works fine when I call it locally ie: perl script.pl. The script prints out some things then

My project A (hosted on GitHub) somehow (not sure how) forgot that it was originally forked from another open-source project B. By 'forgot', I mean, when creating a pull request, I cannot choose B as a target for sending the pull request. Is there some way I tell GitHub that A is a fork of B? (I can create a PR by forking B into A', then merging A into A' and sending the PR from A' to B which naturally works but, naturally, I don't like it) svlasov No, there is no way to turn an existing repo into a fork. You can permanently switch to A' and abandon A , it's a one-time procedure. And there is

I want to run programs in linux by a button click an therefore I wrote a function execute : void execute(const char* program_call, const char* param ) { pid_t child = vfork(); if(child == 0) // child process { int child_pid = getpid(); char *args[2]; // arguments for exec args[0] = (char*)program_call; // first argument is program_call args[1] = (char*)param; // close all opened file descriptors: const char* prefix = "/proc/"; const char* suffix = "/fd/"; char child_proc_dir[16]; sprintf(child_proc_dir,"%s%d%s",prefix,child_pid, suffix); DIR *dir; struct dirent *ent; if ((dir = opendir (child