slub: remove one code path and reduce lock contention in __slab_free()

When we try to free object, there is some of case that we need
to take a node lock. This is the necessary step for preventing a race.
After taking a lock, then we try to cmpxchg_double_slab().
But, there is a possible scenario that cmpxchg_double_slab() is failed
with taking a lock. Following example explains it.

CPU A               CPU B
need lock
...                 need lock
...                 lock!!
lock..but spin      free success
spin...             unlock
lock!!
free fail

In this case, retry with taking a lock is occured in CPU A.
I think that in this case for CPU A,
"release a lock first, and re-take a lock if necessary" is preferable way.

There are two reasons for this.

First, this makes __slab_free()'s logic somehow simple.
With this patch, 'was_frozen = 1' is "always" handled without taking a lock.
So we can remove one code path.

Second, it may reduce lock contention.
When we do retrying, status of slab is already changed,
so we don't need a lock anymore in almost every case.
"release a lock first, and re-take a lock if necessary" policy is
helpful to this.

Change-Id: I8d85a296343c61b0a20c6357aac1f9497a1cd2df
Signed-off-by: Joonsoo Kim <js1304@gmail.com>
Acked-by: Christoph Lameter <cl@linux.com>
Signed-off-by: Pekka Enberg <penberg@kernel.org>