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too many zombies with fork

Omicron
Deleted

Nov 9, 2000, 6:46 AM

Post #1 of 4 (1987 views)

too many zombies with fork	Can't Post

Hello,

I am having a problem with a PERL script.

Any help would be great fully received,
I have Perl script that uses a “for” to launch many processes children using “fork”.
The problem is the child processes in my script are spawning zombies. (I closed every Child process with exit)
The result of these zombies is that they fill up the process table allowing no new processes to be created.

I have tried the following code to fix to problem but to no avail.

$SIG(CHILD) = sub {wait}

DrZed
User

Nov 27, 2000, 1:38 PM

Post #2 of 4 (1988 views)

Re: too many zombies with fork [In reply to]	Can't Post

Well.... either your not actually exiting each child, or the child processes are forking more processes before they exit.

If you can show your code, it would help.

perlplexer
Deleted

Nov 28, 2000, 5:49 AM

Post #3 of 4 (1987 views)

Re: too many zombies with fork [In reply to]	Can't Post

Well, I guess you can refer to Perl Cookbook since someone put it online (hopefully they've got a licence for it ;) )
http://hrg.dhtp.kiae.ru/corvin/perlbs/cookbook/ch16_20.htm

Danni
Deleted

Mar 27, 2001, 12:54 PM

Post #4 of 4 (1960 views)

Re: too many zombies with fork [In reply to]	Can't Post

Theres a few methods you can use to prevent zombie processes.

Basic Method
This method uses a blocking wait call to wait on the child processes. Kind of defeats the point, however.

if ($pid = fork()) {
# The Parent
waitpid ($pid,0);

} else {
# The Child
}

SIGCHLD Method
This method allows multiple processes at one time, much faster but ive found it to cause out of memory errors. Insert before the fork() call.

use POSIX ":sys_wait_h";
$SIG{CHLD} = /&waitchild;
sub waitchild {
while(waitpid(-1,&WNOHANG) > 0) {}
$SIG{CHLD} = /&waitchild;
}

Hope this helps.

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