linux-kernel - Re: [PATCH v2 07/11] mm: Allocate kernel pages to the right memcg

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Date:	Wed, 15 Aug 2012 17:51:40 +0400
From:	Glauber Costa <glommer@...allels.com>
To:	Mel Gorman <mgorman@...e.de>
CC:	<linux-kernel@...r.kernel.org>, <linux-mm@...ck.org>,
	<cgroups@...r.kernel.org>, <devel@...nvz.org>,
	Michal Hocko <mhocko@...e.cz>,
	Johannes Weiner <hannes@...xchg.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	<kamezawa.hiroyu@...fujitsu.com>, Christoph Lameter <cl@...ux.com>,
	David Rientjes <rientjes@...gle.com>,
	Pekka Enberg <penberg@...nel.org>,
	Pekka Enberg <penberg@...helsinki.fi>,
	Suleiman Souhlal <suleiman@...gle.com>
Subject: Re: [PATCH v2 07/11] mm: Allocate kernel pages to the right memcg

On 08/15/2012 05:22 PM, Mel Gorman wrote:
>> I believe it
>> > to be a better and less complicated approach then letting a page appear
>> > and then charging it. Besides being consistent with the rest of memcg,
>> > it won't create unnecessary disturbance in the page allocator
>> > when the allocation is to fail.
>> > 
> I still don't get why you did not just return a mem_cgroup instead of a
> handle.
> 

Forgot this one, sorry:

The reason is to keep the semantics simple.

What should we return if the code is not compiled in? If we return NULL
for failure, the test becomes

memcg = memcg_kmem_charge_page(gfp, order);
if (!memcg)
  exit;

If we're not compiled in, we'd either return positive garbage or we need
to wrap it inside an ifdef

I personally believe to be a lot more clear to standardize on true
to mean "allocation can proceed".

the compiled out case becomes:

if (!true)
   exit;

which is easily compiled away altogether. Now of course, using struct
mem_cgroup makes sense, and I have already changed that here.

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