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Message-ID: <20091005122436.GA5626@elte.hu>
Date: Mon, 5 Oct 2009 14:24:36 +0200
From: Ingo Molnar <mingo@...e.hu>
To: Peter Zijlstra <peterz@...radead.org>,
Oleg Nesterov <oleg@...hat.com>,
Andrew Morton <akpm@...ux-foundation.org>,
Linus Torvalds <torvalds@...ux-foundation.org>
Cc: Thomas Gleixner <tglx@...utronix.de>,
Anirban Sinha <ani@...rban.org>, linux-kernel@...r.kernel.org,
Darren Hart <dvhltc@...ibm.com>,
Kaz Kylheku <kaz@...gmasystems.com>,
Anirban Sinha <asinha@...gmasystems.com>
Subject: Re: futex question
* Peter Zijlstra <peterz@...radead.org> wrote:
> On Mon, 2009-10-05 at 13:59 +0200, Thomas Gleixner wrote:
>
> > Stared at the same place a minute ago :) But still I wonder if it's
> > a good idea to silently release locks and set the state to OWNERDEAD
> > instead of hitting the app programmer with a big clue stick in case
> > the app holds locks when calling execve().
>
> Agreed, I rather like the feedback. With regular exit like things
> there's just not much we can do to avoid the mess, but here we can
> actually avoid it, seems a waste not to do so.
Well, exec() has been a 'exit() + boot-strap next process' kind of thing
from the get go - with little state carried over into the new task. This
has security and robustness reasons as well.
So i think exec() should release all existing state, unless told
otherwise. Making it behave differently for robust futexes sounds
assymetric to me.
It might make sense though - a 'prevent exec because you are holding
locks!' thing. Dunno.
Cc:-ed a few execve() semantics experts who might want to chime in.
If a (buggy) app calls execve() with a (robust) futex still held should
we auto-force-release robust locks held, or fail the exec with an error
code? I think the forced release is a 'anomalous exit' thing mostly,
while calling exec() is not anomalous at all.
Ingo
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