Thread (26 messages) 26 messages, 3 authors, 2025-06-27

Re: [PATCH v2 12/14] ublk: optimize UBLK_IO_UNREGISTER_IO_BUF on daemon task

From: Ming Lei <hidden>
Date: 2025-06-23 09:45:22

On Fri, Jun 20, 2025 at 09:10:06AM -0600, Caleb Sander Mateos wrote:
quoted hunk ↗ jump to hunk
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 <redacted>
---
 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 b2925e15279a..199028f36ec8 100644
--- a/drivers/block/ublk_drv.c
+++ b/drivers/block/ublk_drv.c
@@ -2012,11 +2012,18 @@ static void ublk_io_release(void *priv)
 {
 	struct request *rq = 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);
 }
Reviewed-by: Ming Lei <redacted>

Thanks,
Ming
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