Re: [PATCH -next v2 2/5] block, bfq: add fake weight_counter for weight-raised queue

From: yukuai (C)
Date: Mon Apr 25 2022 - 21:49:17 EST


在 2022/04/26 0:16, Jan Kara 写道:
Hello!

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.
Hi,

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
service.

2) bfq_release_process_ref(), user thread is gone / moved, or old bfqq
is gone due to merge / ioprio change.

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?

Thanks,
Kuai

Honza