gnomerc=Zombie

[Francois Ouellette]

By resources I meant CPU... but you are right, the process might be using
memory and other stuff as well as the process entry itself. I have seen
zombies in all UNIX systems I worked with.

  François Ouellette
<fouellet-cpI+UMyWUv9BDgjK7y7TUQ at public.gmane.org>


----- Original Message -----
From: "Taavi Burns" <jaaaarel-Re5JQEeQqe8AvxtiuMwx3w at public.gmane.org>
To: <tlug-lxSQFCZeNF4 at public.gmane.org>
Sent: Friday, 12 November, 2004 16:35
Subject: Re: [TLUG]: Re: gnomerc=Zombie


> On Fri, 12 Nov 2004 16:15:58 -0500 (EST), Francois Ouellette
> <fouellet-cpI+UMyWUv9BDgjK7y7TUQ at public.gmane.org> wrote:
> > A zombie process exists solely as a process table entry and consumes no
> > other resources. This entry is retained to hold the child's exit status
> > until the parent process wants to retrieve it. The parent can also be
> > notified asynchronously via a signal of the child's termination.
> >
> > --------------
> >
> > According to these definitions (and my experience) a zombie does not use
> > resources! So the parent process of your gnomerc (probably init?) is
still
> > waiting for a signal to terminate it for good.
>
> But it IS consuming resources: an entry in the process table.  I'm
> quite sure that
> I've heard of machines which required a reboot because the process table
> became full (the process table size on UNIX has historically been a boot-
or
> compile-time parameter).  I don't know if modern Linux kernels have
improved
> on this or not, but last time I checked 32-bit Linux still only used
> 16-bit PIDs,
> meaning that when you reach around 65534 zombies (since you need init, and
> I don't think PID 0 is valid), you still can't fork anything new.
> Now, that would be a
> pathological case.  And yet it could happen (Yet Another way to implement
a
> local DoS).
>
> --
> taa


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