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SubjectRe: [PATCH] blk-mq: put the driver tag of nxt rq before first one is requeued
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On 09/12/2017 07:39 PM, jianchao.wang wrote:
>
>
> On 09/13/2017 09:24 AM, Ming Lei wrote:
>> On Wed, Sep 13, 2017 at 09:01:25AM +0800, jianchao.wang wrote:
>>> Hi ming
>>>
>>> On 09/12/2017 06:23 PM, Ming Lei wrote:
>>>>> @@ -1029,14 +1029,20 @@ bool blk_mq_dispatch_rq_list(struct request_queue *q, struct list_head *list)
>>>>> if (list_empty(list))
>>>>> bd.last = true;
>>>>> else {
>>>>> - struct request *nxt;
>>>>> -
>>>>> nxt = list_first_entry(list, struct request, queuelist);
>>>>> bd.last = !blk_mq_get_driver_tag(nxt, NULL, false);
>>>>> }
>>>>>
>>>>> ret = q->mq_ops->queue_rq(hctx, &bd);
>>>>> if (ret == BLK_STS_RESOURCE) {
>>>>> + /*
>>>>> + * If an I/O scheduler has been configured and we got a
>>>>> + * driver tag for the next request already, free it again.
>>>>> + */
>>>>> + if (!list_empty(list)) {
>>>>> + nxt = list_first_entry(list, struct request, queuelist);
>>>>> + blk_mq_put_driver_tag(nxt);
>>>>> + }
>>>> The following way might be more simple and clean:
>>>>
>>>> if (nxt)
>>>> blk_mq_put_driver_tag(nxt);
>>>>
>>>> meantime 'nxt' need to be cleared inside the 'if (list_empty(list))'
>>>> before .queue_rq().
>>>
>>> I had ever thought about that, but to avoid add extra command in the
>>> fast path, I made the patch above.
>>
>> Got it, so how about changing to the following way simply:
>>
>> if (nxt && !list_empty(list))
>> blk_mq_put_driver_tag(nxt);
>>
> It seems that we even could change it as following:
> if (!list_empty(list))
> blk_mq_put_driver_tag(nxt);

This is starting to get too clever for its own good, I generally don't
like to sacrifice readability for performance. In reality, the compiler
probably figures it out anyway...

So either make it explicit, or add a nice comment as to why it is the
way that it is.

--
Jens Axboe

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