linux-kernel - Re: OOPS in perf_mmap

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Message-ID: <20130523163901.GG23650@twins.programming.kicks-ass.net>
Date:	Thu, 23 May 2013 18:39:01 +0200
From:	Peter Zijlstra <peterz@...radead.org>
To:	Christoph Lameter <cl@...ux.com>
Cc:	Al Viro <viro@...IV.linux.org.uk>,
	Vince Weaver <vincent.weaver@...ne.edu>,
	linux-kernel@...r.kernel.org, Paul Mackerras <paulus@...ba.org>,
	Ingo Molnar <mingo@...hat.com>,
	Arnaldo Carvalho de Melo <acme@...stprotocols.net>,
	trinity@...r.kernel.org
Subject: Re: OOPS in perf_mmap_close()

On Thu, May 23, 2013 at 04:12:14PM +0000, Christoph Lameter wrote:
> On Thu, 23 May 2013, Peter Zijlstra wrote:
> 
> > The patch completely fails to explain how RLIMIT_LOCKED is supposed to
> > deal with pinned vs locked. Perf used to account its pages against
> > RLIMIT_LOCKED, with the patch it compares pinned against RLIMIT_LOCKED
> > but completely discards any possible locked pages.
> 
> Pinned pages are different from mlock. Mlock semantics means that the
> pages are kept in memory but the pages are movable (subject to page
> migration f.e.).
> 
> Pinned pages have to stay where they are since the physical addresses may
> be used for device I/O or other stuff.
> 
> Both pinned and mlocked pages cannot be evicted from memory. If one wants
> to account for unevictable pages then both are contributing. However,
> since a pinned page may be mlocked simply adding up the counter may cause
> problems. The sum could be used as a worst case estimate though.
> 
> We could mlock all pinned pages but then the issue arises on how to track
> that properly in order to unpin when the I/O action is done since the app
> may have also mlocked pages.

I know all that, and its completely irrelevant to the discussion.

You cannot simply take away pinned pages from the RLIMIT_MEMLOCK
accounting without mention nor replacement limits.

You now have double the amount of memory you can loose, once to actual
mlock() and once through whatever generates pinned -- if it bothers with
checking limits at all.

Where we had the guarantee that x < y; you did x := x1 + x2; which then
should result in: x1 + x2 < y, instead you did: x1 < y && x2 < y, not
the same and completely wrong.
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