Perl fork和kill - kill(0,$ pid)总是返回1,并且不能杀死孩子

Question

我在Linux主机上运行perl脚本.我正在尝试编写一个分叉的脚本,其中孩子启动一个永远需要的程序,父母在5秒后超时,杀死了孩子.这是我有的:

my $start = time();
my $timeOut = 5;

my $childPid = fork();
if ($childPid) {
    # I am the parent, and $childPid is my child
    while (kill(0, $childPid)) {
        if (time() > $start + $timeOut) {
            $numKilled = kill(SIGTERM, $childPid);
            printf("numKilled: %s\n", $numKilled);
        }
        sleep 1;
    }
}
else {
    # I am the child - do something that blocks forever
    `adb -s 410bf6c1 logcat`;
    exit;
}

输出:

aschirma@graphics9-lnx:~$ perl perl-fork-test.pl 
numKilled: 1
numKilled: 1
numKilled: 1
numKilled: 1
numKilled: 1
...

我期望的行为是我看到"numKilled:1"恰好一次,并且子进程(及其任何子进程)在大约5秒后被杀死.但我从实验中看到,孩子和孩子都没有被杀.该kill(SIGTERM, $childPid)显得无能为力.

我怎么能真的杀了孩子？

Answer 1

应该是这样的.这不遵循最佳做法,但它应该可以帮助您解决问题...

#!/usr/bin/perl

use warnings;
use strict;
use POSIX qw(:sys_wait_h);

my $timeOut = 5;
$SIG{ALRM} = \&timeout;
$SIG{CHLD} = 'IGNORE',
alarm($timeOut);

my $childPid = fork();
if ($childPid) {
    while(1) {
        print "[$$]: parent...\n"; 
        sleep(2); 
    }
}else {
    # I am the child - do something that blocks forever
    while(1){
        print "[$$]: child...\n";
        sleep(2);
    }
    exit;
}

sub timeout {
    print "killing $childPid\n";
    print "###\n" . `ps -ef | grep -v grep | grep perl` . "###\n";
    if ( ! (waitpid($childPid, WNOHANG)) ) {
        print "killing $childPid...\n";
        kill 9, $childPid;
        die "[$$]: exiting\n";
    }
}

OUTPUT:

$ forktest.pl
[24118]: parent...
[24119]: child...
[24118]: parent...
[24119]: child...
[24118]: parent...
[24119]: child...
killing 24119
###
cblack   24118 12548  0 14:12 pts/8    00:00:00 /usr/bin/perl ./forktest.pl
cblack   24119 24118  0 14:12 pts/8    00:00:00 /usr/bin/perl ./forktest.pl
###
killing 24119...
[24118]: exiting