On Tue 26-04-22 16:27:46, yukuai (C) wrote:
在 2022/04/26 15:40, Jan Kara 写道:
On Tue 26-04-22 09:49:04, yukuai (C) wrote:
在 2022/04/26 0:16, Jan Kara 写道:
On Mon 25-04-22 21:34:16, yukuai (C) wrote:
在 2022/04/25 17:48, Jan Kara 写道:
On Sat 16-04-22 17:37:50, Yu Kuai wrote:
Weight-raised queue is not inserted to weights_tree, which makes it
impossible to track how many queues have pending requests through
weights_tree insertion and removel. This patch add fake weight_counter
for weight-raised queue to do that.
Signed-off-by: Yu Kuai <yukuai3@xxxxxxxxxx>
This is a bit hacky. I was looking into a better place where to hook to
count entities in a bfq_group with requests and I think bfq_add_bfqq_busy()
and bfq_del_bfqq_busy() are ideal for this. It also makes better sense
conceptually than hooking into weights tree handling.
bfq_del_bfqq_busy() will be called when all the reqs in the bfqq are
dispatched, however there might still some reqs are't completed yet.
Here what we want to track is how many bfqqs have pending reqs,
specifically if the bfqq have reqs are't complted.
Thus I think bfq_del_bfqq_busy() is not the right place to do that.
Yes, I'm aware there will be a difference. But note that bfqq can stay busy
with only dispatched requests because the logic in __bfq_bfqq_expire() will
not call bfq_del_bfqq_busy() if idling is needed for service guarantees. So
I think using bfq_add/del_bfqq_busy() would work OK.
I didn't think of that before. If bfqq stay busy after dispathing all
the requests, there are two other places that bfqq can clear busy:
1) bfq_remove_request(), bfqq has to insert a new req while it's not in
Yes and the request then would have to be dispatched or merged. Which
generally means another bfqq from the same bfqg is currently active and
thus this should have no impact on service guarantees we are interested in.
2) bfq_release_process_ref(), user thread is gone / moved, or old bfqq
is gone due to merge / ioprio change.
Yes, here there's no new IO for the bfqq so no point in maintaining any
service guarantees to it.
I wonder, will bfq_del_bfqq_busy() be called immediately when requests
are completed? (It seems not to me...). For example, a user thread
issue a sync io just once, and it keep running without issuing new io,
then when does the bfqq clears the busy state?
No, when bfqq is kept busy, it will get scheduled as in-service queue in
the future. Then what happens depends on whether it will get more requests
or not. But generally its busy state will get cleared once it is expired
for other reason than preemption.
Thanks for your explanation.
I think in normal case using bfq_add/del_bfqq_busy() if fine.
There is one last situation that I'm worried: If some disk are very
slow that the dispatched reqs are not completed when the bfqq is
rescheduled as in-service queue, and thus busy state can be cleared
while reqs are not completed.
Using bfq_del_bfqq_busy() will change behaviour in this specail case,
do you think service guarantees will be broken?
Well, I don't think so. Because slow disks don't tend to do a lot of
internal scheduling (or have deep IO queues for that matter). Also note
that generally bfq_select_queue() will not even expire a queue (despite it
not having any requests to dispatch) when we should not dispatch other
requests to maintain service guarantees. So I think service guarantees will
be generally preserved. Obviously I could be wrong, we we will not know
until we try it :).