On Mon, Sep 07, 2020 at 09:52:42AM -0700, Bart Van Assche wrote:
> On 2020-09-07 00:10, Ming Lei wrote:
> > diff --git a/drivers/scsi/scsi_lib.c b/drivers/scsi/scsi_lib.c
> > index 7affaaf8b98e..a05e431ee62a 100644
> > --- a/drivers/scsi/scsi_lib.c
> > +++ b/drivers/scsi/scsi_lib.c
> > @@ -551,8 +551,25 @@ static void scsi_run_queue_async(struct scsi_device *sdev)
> > if (scsi_target(sdev)->single_lun ||
> > !list_empty(&sdev->host->starved_list))
> > kblockd_schedule_work(&sdev->requeue_work);
> > - else
> > - blk_mq_run_hw_queues(sdev->request_queue, true);
> > + else {
>
> Please follow the Linux kernel coding style and balance braces.
Could you provide one document about such style? The patch does pass
checkpatch, or I am happy to follow your suggestion if checkpatch is
updated to this way.
>
> > + /*
> > + * smp_mb() implied in either rq->end_io or blk_mq_free_request
> > + * is for ordering writing .device_busy in scsi_device_unbusy()
> > + * and reading sdev->restarts.
> > + */
> > + int old = atomic_read(&sdev->restarts);
>
> scsi_run_queue_async() has two callers: scsi_end_request() and scsi_queue_rq().
> I don't see how ordering between scsi_device_unbusy() and the above atomic_read()
> could be guaranteed if this function is called from scsi_queue_rq()?
>
> Regarding the I/O completion path, my understanding is that the I/O completion
> path is as follows if rq->end_io == NULL:
>
> scsi_mq_done()
> blk_mq_complete_request()
> rq->q->mq_ops->complete(rq) = scsi_softirq_done
> scsi_finish_command()
> scsi_device_unbusy()
scsi_device_unbusy()
atomic_dec(&sdev->device_busy);
> scsi_cmd_to_driver(cmd)->done(cmd)
> scsi_io_completion()
> scsi_end_request()
> blk_update_request()
> scsi_mq_uninit_cmd()
> __blk_mq_end_request()
> blk_mq_free_request()
> __blk_mq_free_request()
__blk_mq_free_request()
blk_mq_put_tag
smp_mb__after_atomic()
> blk_queue_exit()
> scsi_run_queue_async()
>
> I haven't found any store memory barrier between the .device_busy change in
> scsi_device_unbusy() and the scsi_run_queue_async() call? Did I perhaps overlook
> something?
>
> > + /*
> > + * ->restarts has to be kept as non-zero if there new budget
> > + * contention comes.
>
> Please fix the grammar in the above sentence.
OK.
>
> > + /*
> > + * Order writing .restarts and reading .device_busy. Its pair is
> > + * implied by __blk_mq_end_request() in scsi_end_request() for
> > + * ordering writing .device_busy in scsi_device_unbusy() and
> > + * reading .restarts.
> > + */
> > + smp_mb__after_atomic();
>
> What does "its pair is implied" mean? Please make the above comment
> unambiguous.
See comment in scsi_run_queue_async().
>
> > + /*
> > + * If all in-flight requests originated from this LUN are completed
> > + * before setting .restarts, sdev->device_busy will be observed as
> > + * zero, then blk_mq_delay_run_hw_queues() will dispatch this request
> > + * soon. Otherwise, completion of one of these request will observe
> > + * the .restarts flag, and the request queue will be run for handling
> > + * this request, see scsi_end_request().
> > + */
> > + if (unlikely(atomic_read(&sdev->device_busy) == 0 &&
> > + !scsi_device_blocked(sdev)))
> > + blk_mq_delay_run_hw_queues(sdev->request_queue, SCSI_QUEUE_DELAY);
> > + return false;
>
> What will happen if all in-flight requests complete after
> scsi_run_queue_async() has read .restarts and before it executes
> atomic_cmpxchg()?
One of these completions will run atomic_cmpxchg() successfully, and the
queue is re-run immediately from scsi_run_queue_async().
> Will that cause the queue to be run after a delay
> although it should be run immediately?
Yeah, blk_mq_delay_run_hw_queues() will be called, however:
If scsi_run_queue_async() has scheduled run queue already, this code path
won't queue a dwork successfully. On the other hand, if
blk_mq_delay_run_hw_queues(SCSI_QUEUE_DELAY) has queued a dwork,
scsi_run_queue_async() still can queue the dwork successfully, since the delay
timer can be deactivated easily, see try_to_grab_pending(). In short, the case
you described is an extremely unlikely event. Even though it happens,
forward progress is still guaranteed.
Thanks,
Ming
