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Commit 442b06bc ("slub: Remove node check in slab_free") added a call to deactivate_slab() in the debug case in __slab_alloc(), which unlocks the current slab used for allocation. Going to the label 'unlock_out' then does it again. Also, in the debug case we do not need all the other processing that the 'unlock_out' path does. We always fall back to the slow path in the debug case. So the tid update is useless. Similarly, ALLOC_SLOWPATH would just be incremented for all allocations. Also a pretty useless thing. So simply restore irq flags and return the object. Signed-off-by:Christoph Lameter <cl@linux.com> Reported-and-bisected-by:
James Morris <jmorris@namei.org> Reported-by:
Ingo Molnar <mingo@elte.hu> Reported-by:
Jens Axboe <jaxboe@fusionio.com> Cc: Pekka Enberg <penberg@kernel.org> Signed-off-by:
Linus Torvalds <torvalds@linux-foundation.org>
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