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

[Ming Lei <ming.lei@redhat.com>]

On Fri, Jun 20, 2025 at 09:10:06AM -0600, Caleb Sander Mateos wrote:
> 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 <csander@purestorage.com>
> ---
>  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 <ming.lei@redhat.com>

Thanks,
Ming