[sf-perl] not understanding program behavior with alarm()

[Francisco Obispo]

Perl's system() call performs a fork() on the specified process,

you might want to try exec() instead..




On Feb 3, 2012, at 2:43 PM, David Alban wrote:

> greetings,
> 
> my code is not doing what i expect.  i have:
> 
> $ cat junk.test.alarm
> #!/usr/bin/perl
> 
> use strict;
> use warnings;
> 
> eval {
>  local $SIG{ ALRM } = sub { die "alarm\n"; };
>  alarm 5;
>  system 'bash', "$ENV{ HOME }/junk.sh";
>  alarm 0;
> };
> 
> if ( $@ ) {
>  if ( $@ eq "alarm\n" ) {
>    die "timed out!\n";
>  }
>  else {
>    die "non-timeout error: $!\n";
>  }
> }
> 
> print "right before exit stmt\n";
> 
> exit 0;
> 
> and:
> 
> $ cat ~/junk.sh
> #!/bin/bash
> 
> export PATH=/sbin:/bin:/usr/sbin:/usr/bin
> 
> i=0
> while true ; do
>  i=$(( $i + 1 ))
>  echo $i
>  sleep 1
> done
> 
> exit 0
> 
> when i run, i get:
> 
> $ perl junk.test.alarm
> $ perl junk.test.alarm
> 1
> 2
> 3
> 4
> 5
> timed out!
> 
> dalban at srwd00reg008             Fri Feb 03 22:41:37
> ~
> $ 6
> 7
> 8
> 9
> 10
> 11
> 12
> 13
> 
> with junk.sh continuing until i kill it from the command line.
> 
> what am i missing?
> 
> should the die() stop junk.sh?
> 
> thanks,
> david
> -- 
> Live in a world of your own, but always welcome visitors.
> ***
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