Commit 1ceeedb5 authored by Caleb Sander Mateos's avatar Caleb Sander Mateos Committed by Jens Axboe
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ublk: optimize UBLK_IO_UNREGISTER_IO_BUF on daemon task 



ublk_io_release() performs an expensive atomic refcount decrement. This
atomic operation is unnecessary in the common case where the request's
buffer is registered and unregistered on the daemon task before handling
UBLK_IO_COMMIT_AND_FETCH_REQ for the I/O. So if ublk_io_release() is
called on the daemon task and task_registered_buffers is positive, just
decrement task_registered_buffers (nonatomically). ublk_sub_req_ref()
will apply this decrement when it atomically subtracts from io->ref.

Signed-off-by: default avatarCaleb Sander Mateos <csander@purestorage.com>
Reviewed-by: default avatarMing Lei <ming.lei@redhat.com>
Link: https://lore.kernel.org/r/20250620151008.3976463-13-csander@purestorage.com


Signed-off-by: default avatarJens Axboe <axboe@kernel.dk>
parent 8a8fe42d

drivers/block/ublk_drv.c

+8 −1
Original line number Diff line number Diff line
@@ -2030,6 +2030,13 @@ static void ublk_io_release(void *priv) 
	struct ublk_queue *ubq = rq->mq_hctx->driver_data;

	struct ublk_io *io = &ubq->ios[rq->tag];



	/*

	 * task_registered_buffers may be 0 if buffers were registered off task

	 * but unregistered on task. Or after UBLK_IO_COMMIT_AND_FETCH_REQ.

	 */

	if (current == io->task && io->task_registered_buffers)

		io->task_registered_buffers--;

	else

		ublk_put_req_ref(ubq, io, rq);

}