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Message-ID: <CAAmzW4N5HxN+Ha_kwwKSf9na-g6bnro1UumQ+ZiQEmgS4kacrA@mail.gmail.com>
Date: Sat, 28 Jul 2012 22:56:03 +0900
From: JoonSoo Kim <js1304@...il.com>
To: Christoph Lameter <cl@...ux.com>
Cc: Pekka Enberg <penberg@...nel.org>, linux-kernel@...r.kernel.org,
linux-mm@...ck.org
Subject: Re: [PATCH] slub: remove one code path and reduce lock contention in __slab_free()
2012/7/28 Christoph Lameter <cl@...ux.com>:
> On Sat, 28 Jul 2012, Joonsoo Kim wrote:
>
>> Subject and commit log are changed from v1.
>
> That looks a bit better. But the changelog could use more cleanup and
> clearer expression.
>
>> @@ -2490,25 +2492,17 @@ static void __slab_free(struct kmem_cache *s, struct page *page,
>> return;
>> }
>>
>> + if (unlikely(!new.inuse && n->nr_partial > s->min_partial))
>> + goto slab_empty;
>> +
>
> So we can never encounter a empty slab that was frozen before? Really?
In my suggestion, 'was_frozen = 1' is "always" handled without taking a lock.
Then, never hit following code.
+ if (unlikely(!new.inuse && n->nr_partial > s->min_partial))
+ goto slab_empty;
+
Instead, hit following code.
if (likely(!n)) {
/*
* If we just froze the page then put it onto the
* per cpu partial list.
*/
if (new.frozen && !was_frozen) {
put_cpu_partial(s, page, 1);
stat(s, CPU_PARTIAL_FREE);
}
/*
* The list lock was not taken therefore no list
* activity can be necessary.
*/
if (was_frozen)
stat(s, FREE_FROZEN);
return;
}
So, even if we encounter a empty slab that was frozen before, we just
do "stat(s, FREE_FROZEN)".
Please let me know my answer is sufficient.
Thanks!!
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