5

以下代码运行 2 个孩子，他们将等待 10 秒并终止。父母坐在一个循环中，等待孩子终止：

#!/usr/bin/perl

use strict;
use warnings;
use POSIX ":sys_wait_h";

sub func
# {{{
{
      my $i = shift;
      print "$i started\n";
      $| = 1;
      sleep(10);
      print "$i finished\n";
}
# }}}

my $n = 2;
my @children_pids;

for (my $i = 0; $i < $n; $i++) {
      if ((my $pid = fork()) == 0) {
            func($i);
            exit(0);
      } else {
            $children_pids[$i] = $pid;
      }
}

my $stillWaiting;
do {
      $stillWaiting = 0;
      for (my $i = 0; $i < $n; ++$i) {
            if ($children_pids[$i] > 0)
            {
                  if (waitpid($children_pids[$i], WNOHANG) != 0) {
                        # Child is done
                        print "child done\n";
                        $children_pids[$i] = 0;
                  } else {
                        # Still waiting on this child
                        #print "waiting\n";
                        $stillWaiting = 1;
                  }
            }
            #Give up timeslice and prevent hard loop: this may not work on all flavors of Unix
            sleep(0);
      }
} while ($stillWaiting);

print "parent finished\n";

代码基于这个答案：Multiple fork() Concurrency

它工作正常，但父循环正在消耗处理器时间。top命令给出了这个：

在此处输入图像描述

这里的答案是：

作为额外的奖励，循环将在waitpid孩子运行时阻塞，因此您在等待时不需要繁忙的循环。

但对我来说，它不会阻塞。怎么了？

4

2 回答 2

7

您正在传递WNOHANG标志，这使得调用非阻塞。删除此标志，waitpid并将在 0% CPU 处等待，直到孩子退出。

如果您采用这种方法，您可以简化代码。在孩子完成之前无需循环，因为阻塞waitpid调用将为您执行此操作：

for (my $i = 0; $i < $n; ++$i) {
    if ($children_pids[$i] > 0) {
          waitpid($children_pids[$i], 0);
          print "child done\n";
          $children_pids[$i] = 0;
    }
}

或者，将sleep呼叫更改为等待一秒钟。然后你的程序将每秒检查一次完成的子进程，而不会增加 CPU 使用率。

于 2013-06-07T09:21:01.993 回答
0

由于您的父线程在等待它的子线程时实际上并没有做任何事情，我会将其简化为

#!/usr/bin/perl    
use strict;
use warnings;
$| = 1; # autoflush

sub func{
    my($i) = @_;
    print "$i started\n";
    sleep(10);
    print "$i finished\n";
}

my $n = 2;
my @children_pids;

for my $i ( 0..($n-1) ) { # faster, and clearer than the C-style for loop
    my $pid = fork;
    die "Unable to fork" unless defined $pid; # check for errors
    if ( $pid == 0) { # child
        func($i);
        exit(0); # may need to be POSIX::_exit()
    } else { # parent
        push @children_pids, $pid; # don't allow undef or 0 on the list
    }
}

# while( @children_pids ){
#    waitpid shift @children_pids, 0;
# }

waitpid $_, 0 for @children_pids;

print "parent finished\n";

如果你的 perl 是用 IThreads 编译的，你可以使用线程模块。（Windows 上的fork仿真
需要 IThreads ）

使用线程还可以更轻松地执行您最初尝试的操作，在线程完成时加入线程。

use strict;
use warnings;
use threads (); # not using async
$| = 1; # autoflush

sub func{
    my($i) = @_;
    print "$i started\n";
    sleep(rand(10)); # randomize the order of completion for this example
    print "$i finished\n";
    return $i; # <===
}

my $n = 10;

for my $i ( 0..($n-1) ){
   my $thread = threads->create( \&func, $i ); # ask for scalar value
   die "unable to create thread $i" unless defined $thread;
}

while( threads->list ){
    # join the threads that are done
    for my $thread ( threads->list(threads::joinable) ){
        print 'thread-id: ',  $thread->tid, ' returned: ', $thread->join, "\n";
    }

    # be nice to other threads and processes
    threads->yield;

    # allows the threads to "bunch up" for this example
    # not necessary for real code.
    sleep 2;
}

0 started
1 started
2 started
3 started
4 started
5 started
6 started
7 started
7 finished
8 started
9 started
2 finished
thread-id: 3 returned: 2
thread-id: 8 returned: 7
8 finished
5 finished
thread-id: 6 returned: 5
thread-id: 9 returned: 8
1 finished
thread-id: 2 returned: 1
3 finished
6 finished
9 finished
4 finished
thread-id: 4 returned: 3
thread-id: 5 returned: 4
thread-id: 7 returned: 6
thread-id: 10 returned: 9
0 finished
thread-id: 1 returned: 0
于 2013-06-09T03:02:35.627 回答