User's cron job creates zombie process on 5.3

[Kevin Oberman]

> Date: Thu, 20 Jan 2005 02:06:38 -0800 (PST)
> From: spam maps <spamrefuse at yahoo.com>
> Sender: owner-freebsd-stable at freebsd.org
> 
> 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: oberman at es.net			Phone: +1 510 486-8634