in reply to Re: timeout process problems
in thread timeout process problems

Just as a FYI, I was just poking around my code and realized they child processes were NOT dying with a certain command and becoming zombies. Sure enough there were 56 zombie processes. On the 57th timeout, no child process or zombie would spawn so I said to myself, "maybe there's a limit on zombies somewhere?" So I set $SIG{CHLD} to 'IGNORE' to prevent zombies. It went OK for awhile but then the same problem appeared again even with no zombies!

Am still looking to see whats truly causing the command to not work. I'll give your method a shot in a bit.

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Re: Re: Re: timeout process problems
by shrubbery (Acolyte) on Nov 08, 2001 at 00:42 UTC
    Ahh, I found out that $! is returning "Too many files open" during the 57th try. Any ideas?
[reply]
      I'm guessing, but perhaps the alarm is leaving a pipe open to each of your children and 56 is your limit of open pipes/file descriptors?

      I'd try the non-alarm solution.

      bluto

[reply]

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