Re: [PATCH] blk-mq: Fix blk_mq_tagset_busy_iter() for shared tags

[John Garry]

> > > blk_mq_queue_tag_busy_iter() needn't such change? >> I didn't think so.>>>> blk_mq_queue_tag_busy_iter() will indeed 
re-iter the tags per hctx. However
> > in bt_iter(), we check rq->mq_hctx == hctx for calling the iter callback:
> >
> > static bool bt_iter(struct sbitmap *bitmap, unsigned int bitnr, void *data)
> > {
> > 	...
> >
> > 	if (rq->q == hctx->queue && rq->mq_hctx == hctx)
> > 		ret = iter_data->fn(hctx, rq, iter_data->data, reserved);
> >
> > And this would only pass for the correct hctx which we're iter'ing for.
> It is true for both shared and non-shared sbitmap since we don't share
> hctx, so what does matter?

It matters that we are doing the right thing for shared tags. My point 
is we iter but don't call the callback unless the correct hctx.

As I see, this has not changed in transitioning from shared sbitmap to 
shared tags.

> With single shared tags, you can iterate over
> all requests originated from all hw queues, right?
Right, for the same request queue, we should do that.

> > Indeed, it would be nice not to iter excessive times, but I didn't see a
> > straightforward way to change that.


> In Kashyap's report, the lock contention is actually from
> blk_mq_queue_tag_busy_iter(), see:
>
> https://lore.kernel.org/linux-block/8867352d-2107-1f8a-0f1c-ef73450bf256@xxxxxxxxxx/

As I understand, Kashyap mentioned no throughput regression with my 
series, but just higher cpu usage in blk_mq_find_and_get_req().

I'll see if I can see such a thing in my setup.

But could it be that since we only have a single sets of requests per 
tagset, and not a set of requests per HW queue, there is more contention 
on the common set of requests in the refcount_inc_not_zero() call ***, 
below:

static struct request *blk_mq_find_and_get_req(struct blk_mq_tags *tags,
unsigned int bitnr)
{
	...

	rq = tags->rqs[bitnr];
	if (... || !refcount_inc_not_zero(&rq->ref)) ***
	...
}

But I wonder why this function is even called often...

> > There is also blk_mq_all_tag_iter():
> >
> > void blk_mq_all_tag_iter(struct blk_mq_tags *tags, busy_tag_iter_fn *fn,
> > 		void *priv)
> > {
> > 	__blk_mq_all_tag_iter(tags, fn, priv, BT_TAG_ITER_STATIC_RQS);
> > }
> >
> > But then the only user is blk_mq_hctx_has_requests():
> >
> > static bool blk_mq_hctx_has_requests(struct blk_mq_hw_ctx *hctx)
> > {
> > 	struct blk_mq_tags *tags = hctx->sched_tags ?
> > 			hctx->sched_tags : hctx->tags;
> > 	struct rq_iter_data data = {
> > 		.hctx	= hctx,
> > 	};
> >
> > 	blk_mq_all_tag_iter(tags, blk_mq_has_request, &data);
> > 	return data.has_rq;
> > }
> This above one only iterates over the specified hctx/tags, it won't be
> affected.
>
> > But, again like bt_iter(), blk_mq_has_request() will check the hctx matches:
> Not see what matters wrt. checking hctx.

I'm just saying that something like the following would be broken for 
shared tags:

static bool blk_mq_has_request(struct request *rq, void *data, bool 
reserved)
{
	struct rq_iter_data *iter_data = data;

	iter_data->has_rq = true;
	return true;
}

static bool blk_mq_hctx_has_requests(struct blk_mq_hw_ctx *hctx)
{
	struct rq_iter_data data = {
	};

	blk_mq_all_tag_iter(tags, blk_mq_has_request, &data);
	return data.has_rq;
}

As it ignores that we want to check for a specific hctx.

Thanks,
John