linux-kernel - Re: [PATCH] oom killer: break from infinite loop

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Message-ID: <20100329120904.GB11838@desktop>
Date:	Mon, 29 Mar 2010 20:09:04 +0800
From:	anfei <anfei.zhou@...il.com>
To:	Oleg Nesterov <oleg@...hat.com>
Cc:	Andrew Morton <akpm@...ux-foundation.org>, rientjes@...gle.com,
	kosaki.motohiro@...fujitsu.com, nishimura@....nes.nec.co.jp,
	kamezawa.hiroyu@...fujitsu.com, linux-mm@...ck.org,
	linux-kernel@...r.kernel.org
Subject: Re: [PATCH] oom killer: break from infinite loop

On Mon, Mar 29, 2010 at 01:46:30PM +0200, Oleg Nesterov wrote:
> On 03/29, anfei wrote:
> >
> > On Sun, Mar 28, 2010 at 06:28:21PM +0200, Oleg Nesterov wrote:
> > > On 03/28, anfei wrote:
> > > >
> > > > Assume thread A and B are in the same group.  If A runs into the oom,
> > > > and selects B as the victim, B won't exit because at least in exit_mm(),
> > > > it can not get the mm->mmap_sem semaphore which A has already got.
> > >
> > > I see. But still I can't understand. To me, the problem is not that
> > > B can't exit, the problem is that A doesn't know it should exit. All
> >
> > If B can exit, its memory will be freed,
> 
> Which memory? I thought, we are talking about the memory used by ->mm ?
> 
There is also a little kernel struct related to the task can be freed,
but I think you are correct, the memory used by ->mm takes more effect,
and it won't be freed even B exits. So I agree you on:
"
the problem is not that B can't exit, the problem is that A doesn't know
it should exit. All threads should exit and free ->mm. Even if B could
exit, this is not enough. And, to some extent, it doesn't matter if it
holds mmap_sem or not.
"

Thanks,
Anfei.

> Oleg.
> 
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