Re: [PATCH 3/4] blk-mq: establish new mapping before cpu starts handling requests
From: Akinobu Mita
Date: Wed Jun 24 2015 - 22:57:04 EST
2015-06-25 1:24 GMT+09:00 Ming Lei <tom.leiming@xxxxxxxxx>:
> On Wed, Jun 24, 2015 at 10:34 PM, Akinobu Mita <akinobu.mita@xxxxxxxxx> wrote:
>> Hi Ming,
>>
>> 2015-06-24 18:46 GMT+09:00 Ming Lei <tom.leiming@xxxxxxxxx>:
>>> On Sun, Jun 21, 2015 at 9:52 PM, Akinobu Mita <akinobu.mita@xxxxxxxxx> wrote:
>>>> ctx->index_hw is zero for the CPUs which have never been onlined since
>>>> the block queue was initialized. If one of those CPUs is hotadded and
>>>> starts handling request before new mappings are established, pending
>>>
>>> Could you explain a bit what the handling request is? The fact is that
>>> blk_mq_queue_reinit() is run after all queues are put into freezing.
>>
>> Notifier callbacks for CPU_ONLINE action can be run on the other CPU
>> than the CPU which was just onlined. So it is possible for the
>> process running on the just onlined CPU to insert request and run
>> hw queue before blk_mq_queue_reinit_notify() is actually called with
>> action=CPU_ONLINE.
>
> You are right because blk_mq_queue_reinit_notify() is alwasy run after
> the CPU becomes UP, so there is a tiny window in which the CPU is up
> but the mapping is updated. Per current design, the CPU just onlined
> is still mapped to hw queue 0 until the mapping is updated by
> blk_mq_queue_reinit_notify().
>
> But I am wondering why it is a problem and why you think flush_busy_ctxs
> can't find the requests on the software queue in this situation?
The problem happens when the CPU has just been onlined first time
since the request queue was initialized. At this time ctx->index_hw
for the CPU is still zero before blk_mq_queue_reinit_notify is called.
The request can be inserted to ctx->rq_list, but blk_mq_hctx_mark_pending()
marks busy for wrong bit position as ctx->index_hw is zero.
flush_busy_ctxs() only retrieves the requests from software queues
which are marked busy. So the request just inserted is ignored as
the corresponding bit position is not busy.
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