linux-kernel - Re: [PATCH] slub: remove one code path and reduce lock contention in __slab

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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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