> Date: Thu, 20 Jan 2005 02:06:38 -0800 (PST)
> From: spam maps <[EMAIL PROTECTED]>
> Sender: [EMAIL PROTECTED]
>
> Peter Jeremy wrote:
> > On Wed, 2005-Jan-19 21:14:26 -0800, spam maps wrote:
> >
> >>> ( /usr/bin/ssh -n -f ${tunnel} & )
> >>
> >>Alas, no success. Still get the <defunct> zombie
> >>process.
> >>
> >>I actually wonder if this is an odd or buggy
> >>behaviour of ssh, or is cron making a mistake here?
> >
> >
> > The cron daemon (which will have a PPID of 1) forks
> > a copy of itself to actually handle the cron job
> > (I suspect this is the parent of the zombie that
> > you are seeing). This child process runs
> > "/bin/sh -c CRONJOB" (where CRONJOB is the line in
> > your crontab) and I suspect this is the zombie you
> > are seeing.
> >
> > My guess is that your ssh process is holding open
> > file descriptors and the cron child process is
> > waiting for these descriptors to close before
> > wait()ing for the child. If this is true, then you
> > should avoid it with something like:
> > ( /usr/bin/ssh -n -f ${tunnel} >/dev/null 2>&1 & )
> >
>
> BINGO!
> That works. Zombie has gone. Thank you.
>
> >>Leaving a zombie process around, means there's a
> kind
> >>of bug/mistake somewhere, right?
> >
> > Yes. But it's not necessarily a bug in FreeBSD :-).
>
> So, after you've given me a complicated solution to
> avoid the zombie, can you tell which program is at
> error here? Cron, ssh, or FreeBSD?
FreeBSD is an operating system, not a kernel like Linux. Both cron and
ssh are part of FreeBSD. That said, the problem could be in sh or in the
program executed, too.
I trivial note: -f implies -n. There is no need to specify both, but no
harm in doing so.
--
R. Kevin Oberman, Network Engineer
Energy Sciences Network (ESnet)
Ernest O. Lawrence Berkeley National Laboratory (Berkeley Lab)
E-mail: [EMAIL PROTECTED] Phone: +1 510 486-8634
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