perl - 等待子进程超时

Question

我想等待两个进程(或更多)完成，或者等待计时器倒计时。计时器倒计时将在终端窗口中输出。例如:

use warnings;
use strict;
use feature qw(say);

my $pid1 = fork();
if ( $pid1 == 0 ) {
    runTask1();
    exit 0;
}

my $pid2 = fork();
if ( $pid2 == 0 ) {
    runTask2();
    exit 0;
}

my $timer = startTimer();

say "Waiting for child processes..";
my $counter = 20;
my $i       = 0;
while (1) {
    my $pid = wait;
    last if ( $pid == -1 );
    if ( $pid == $timer ) {
        $counter--;
        say $counter;
        $timer = startTimer();
    } else {
        say "Child with PID=$pid finished..";
        $i++;
        last if $i == 2;
    }
}

say "Done.";

sub startTimer {

    my $pidTimer = fork();
    if ( $pidTimer == 0 ) {
        sleep 1;
        exit 0;
    }
    return $pidTimer;
}

sub runTask1 {
    sleep 10;
    exit 0;
}

sub runTask2 {
    sleep 5;
    exit 0;
}

我对这种方法的担忧是我为计时器创建了一个 fork 的子进程，这看起来有点矫枉过正。这是必要的还是有更简单的方法？

Answer 1

看看alarm() - 在指定的超时后触发终止信号ALRM。这要么会终止进程，要么您可以使用:

    $SIG{'ALRM'} = \&some_sub_to_handle_alarms;

alarm() 不会传播到 fork 进程，因此您可以将其设置在“父进程”上，这样它就会在等待子进程时超时。

你可能不需要 fork 你的 pidTimer - 你可以让你的主进程处于循环中。

以下演示:

use strict;
use warnings;
use feature qw(say);

my $pid1 = fork();
if ( $pid1 == 0 ) {    # Simulated Task 1
    sleep 10;
    exit 0;
}

my $pid2 = fork();
if ( $pid2 == 0 ) {    # Simulated Task 2
    sleep 5;
    exit 0;
}

say "Waiting for child processes..";
my $counter = 20;
local $SIG{ALRM} = sub {
    say --$counter;
    alarm 1;
};
alarm 1;

while ((my $pid = wait) != -1) {
    say "Child with PID=$pid finished..";
}

alarm 0;
say "Done.";

输出:

Waiting for child processes..
19
18
17
16
Child with PID=55240 finished..
15
14
13
12
11
Child with PID=55239 finished..
Done.