linux-kernel - Re: [PATCH 3/3] slub: release a lock if freeing object with a lock is failed in __slab

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Message-ID: <CAAmzW4NJyX9e_dMyJBA5zDiVYVmL1vbUkaRHNoSbbhDZWW7iMg@mail.gmail.com>
Date:	Fri, 6 Jul 2012 23:19:27 +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 3/3] slub: release a lock if freeing object with a lock is
 failed in __slab_free()

2012/7/5 Christoph Lameter <cl@...ux.com>:
> On Sat, 23 Jun 2012, Joonsoo Kim wrote:
>
>> In some case of __slab_free(), we need a lock for manipulating partial list.
>> If freeing object with a lock is failed, a lock doesn't needed anymore
>> for some reasons.
>>
>> Case 1. prior is NULL, kmem_cache_debug(s) is true
>>
>> In this case, another free is occured before our free is succeed.
>> When slab is full(prior is NULL), only possible operation is slab_free().
>> So in this case, we guess another free is occured.
>> It may make a slab frozen, so lock is not needed anymore.
>
> A free cannot freeze the slab without taking the lock. The taken lock
> makes sure that the thread that first enters slab_free() will be able to
> hold back the thread that wants to freeze the slab.

I don't mean we can freeze the slab without taking the lock.
We can fail cmpxchg_double_slab with taking the lock.
And in this case, we don't need lock anymore, so let's release lock.

For example,
When we try to free object A at cpu 1, another process try to free
object B at cpu 2 at the same time.
object A, B is in same slab, and this slab is in full list.

CPU 1                           CPU 2
prior = page->freelist;    prior = page->freelist
....                                  ...
new.inuse--;                   new.inuse--;
taking lock                      try to take the lock, but failed, so
spinning...
free success                   spinning...
add_partial
release lock                    taking lock
                                       fail cmpxchg_double_slab
                                       retry
                                       currently, we don't need lock

At CPU2, we don't need lock anymore, because this slab already in partial list.

Case 2 is similar as case 1.
So skip explain.

Case 1, 2 in commit message explain almost retry case with taking lock.
So, below is reasonable.

@@ -2450,13 +2449,17 @@ static void __slab_free(struct kmem_cache *s,
struct page *page,
                return;

        do {
+               if (unlikely(n)) {
+                       spin_unlock_irqrestore(&n->list_lock, flags);
+                       n = NULL;
+               }
                prior = page->freelist;
                counters = page->counters;
                set_freepointer(s, object, prior);


>> Case 2. inuse is NULL
>>
>> In this case, acquire_slab() is occured before out free is succeed.
>> We have a last object for slab, so other operation for this slab is
>> not possible except acquire_slab().
>> Acquire_slab() makes a slab frozen, so lock is not needed anymore.
>
> acquire_slab() also requires lock acquisition and would be held of by
> slab_free holding the lock.

See above explain.

>> This also make logic somehow simple that 'was_frozen with a lock' case
>> is never occured. Remove it.
>
> That is actually interesting and would be a good optimization.
>

So, I think patch is valid.
Thanks for comments.
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