From 1ceeedb5974937a6d881fd0d92f143ca16f55577 Mon Sep 17 00:00:00 2001 From: Caleb Sander Mateos Date: Fri, 20 Jun 2025 09:10:06 -0600 Subject: [PATCH] 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: Caleb Sander Mateos Reviewed-by: Ming Lei Link: https://lore.kernel.org/r/20250620151008.3976463-13-csander@purestorage.com Signed-off-by: Jens Axboe --- drivers/block/ublk_drv.c | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/drivers/block/ublk_drv.c b/drivers/block/ublk_drv.c index e38d4fe5e8333..c3c0601db4871 100644 --- a/drivers/block/ublk_drv.c +++ b/drivers/block/ublk_drv.c @@ -2030,7 +2030,14 @@ static void ublk_io_release(void *priv) struct ublk_queue *ubq = rq->mq_hctx->driver_data; struct ublk_io *io = &ubq->ios[rq->tag]; - ublk_put_req_ref(ubq, io, rq); + /* + * 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); } static int ublk_register_io_buf(struct io_uring_cmd *cmd, -- 2.47.2