Re: [PATCH RFC 3/5] blk-mq: Facilitate a shared tags per tagset

[John Garry <john.garry@huawei.com>]

On 15/11/2019 17:57, Bart Van Assche wrote:
> On 11/15/19 2:24 AM, John Garry wrote:
>> Bart Van Assche wrote:
>> > How about sharing tag sets across hardware
>> > queues, e.g. like in the (totally untested) patch below?
>>
>> So this is similar in principle what Ming Lei came up with here:
>> https://lore.kernel.org/linux-block/20190531022801.10003-1-ming.lei@redhat.com/ 
>>
>> However your implementation looks neater, which is good.
>>
>> My concern with this approach is that we can't differentiate which 
>> tags are allocated for which hctx, and sometimes we need to know that.
>>

Hi Bart,

>> An example here was blk_mq_queue_tag_busy_iter(), which iterates the 
>> bits for each hctx. This would just be broken by that change, unless 
>> we record which bits are associated with each hctx.
> 
> I disagree. In bt_iter() I added " && rq->mq_hctx == hctx" such that 
> blk_mq_queue_tag_busy_iter() only calls the callback function for 
> matching (hctx, rq) pairs.

OK, I see. I assumed that rq->mq_hctx was statically set when we 
initially allocate the static requests per hctx; but that doesn’t appear 
so - it's set in blk_mq_get_request()->blk_mq_rq_ctx_init().

> 
>> Another example was __blk_mq_tag_idle(), which looks problematic.
> 
> Please elaborate.

Again, this was for the same reason being that I thought we could not 
differentiate which rqs were associated with which hctx.

> 
>> For debugfs, when we examine 
>> /sys/kernel/debug/block/.../hctxX/tags_bitmap, wouldn't that be the 
>> tags for all hctx (hctx0)?
>>
>> For debugging reasons, I would say we want to know which tags are 
>> allocated for a specific hctx, as this is tightly related to the 
>> requests for that hctx.
> 
> That is an open issue in the patch I posted and something that needs to 
> be addressed. One way to address this is to change the 
> sbitmap_bitmap_show() calls into calls to a function that only shows 
> those bits for which rq->mq_hctx == hctx.

Yeah, understood.

> 
>>> @@ -341,8 +341,11 @@ void blk_mq_tagset_busy_iter(struct 
>>> blk_mq_tag_set *tagset,
>>>       int i;
>>>
>>>       for (i = 0; i < tagset->nr_hw_queues; i++) {
>>> -        if (tagset->tags && tagset->tags[i])
>>> +        if (tagset->tags && tagset->tags[i]) {
>>>               blk_mq_all_tag_busy_iter(tagset->tags[i], fn, priv);
>>
>> As I mentioned earlier, wouldn't this iterate over all tags for all 
>> hctx's, when we just want the tags for hctx[i]?
>>
>> Thanks,
>> John
>>
>> [Not trimming reply for future reference]
>>
>>> +            if (tagset->share_tags)
>>> +                break;
>>> +        }
>>>       }
>>>   }
>>>   EXPORT_SYMBOL(blk_mq_tagset_busy_iter);
> 
> Since blk_mq_tagset_busy_iter() loops over all hardware queues all what 
> is changed is the order in which requests are examined. I am not aware 
> of any block driver that calls blk_mq_tagset_busy_iter() and that 
> depends on the order of the requests passed to the callback function.
> 

OK, fine.

So, to me, this approach also seems viable then.

I am however not so happy with how we use blk_mq_tag_set.tags[0] for the 
shared tags; I would like to use blk_mq_tag_set.shared_tags and make 
blk_mq_tag_set.tags[] point at blk_mq_tag_set.shared_tags or maybe not 
blk_mq_tag_set.tags[] at all. However maybe that change may be more 
intrusive.

And another more real concern is that we miss a check somewhere for 
rq->mq_hctx == hctx when examining the bits on the shared tags.

Thanks,
John