On 12/15/2017 05:54 AM, Peter Zijlstra wrote:
> On Thu, Dec 14, 2017 at 09:42:48PM +0000, Bart Van Assche wrote:
>> On Thu, 2017-12-14 at 21:20 +0100, Peter Zijlstra wrote:
>>> On Thu, Dec 14, 2017 at 06:51:11PM +0000, Bart Van Assche wrote:
>>>> On Tue, 2017-12-12 at 11:01 -0800, Tejun Heo wrote:
>>>>> + write_seqcount_begin(&rq->gstate_seq);
>>>>> + blk_mq_rq_update_state(rq, MQ_RQ_IN_FLIGHT);
>>>>> + blk_add_timer(rq);
>>>>> + write_seqcount_end(&rq->gstate_seq);
>>>>
>>>> My understanding is that both write_seqcount_begin() and
>>>> write_seqcount_end()
>>>> trigger a write memory barrier. Is a seqcount really faster than a
>>>> spinlock?
>>>
>>> Yes lots, no atomic operations and no waiting.
>>>
>>> The only constraint for write_seqlock is that there must not be any
>>> concurrency.
>>>
>>> But now that I look at this again, TJ, why can't the below happen?
>>>
>>> write_seqlock_begin();
>>> blk_mq_rq_update_state(rq, IN_FLIGHT);
>>> blk_add_timer(rq);
>>> <timer-irq>
>>> read_seqcount_begin()
>>> while (seq & 1)
>>> cpurelax();
>>> // life-lock
>>> </timer-irq>
>>> write_seqlock_end();
>>
>> Hello Peter,
>>
>> Some time ago the block layer was changed to handle timeouts in thread
>> context
>> instead of interrupt context. See also commit 287922eb0b18 ("block: defer
>> timeouts to a workqueue").
>
> That only makes it a little better:
>
> Task-A Worker
>
> write_seqcount_begin()
> blk_mq_rw_update_state(rq, IN_FLIGHT)
> blk_add_timer(rq)
> <timer>
> schedule_work()
> </timer>
> <context-switch to worker>
> read_seqcount_begin()
> while(seq & 1)
> cpu_relax();
>
Hi Peter
The current seqcount read side is as below:
do {
start = read_seqcount_begin(&rq->gstate_seq);
gstate = READ_ONCE(rq->gstate);
deadline = rq->deadline;
} while (read_seqcount_retry(&rq->gstate_seq, start));
read_seqcount_retry() doesn't check the bit 0, but whether the saved value from
read_seqcount_begin() is equal to the current value of seqcount.
pls refer:
static inline int __read_seqcount_retry(const seqcount_t *s, unsigned start)
{
return unlikely(s->sequence != start);
}
Thanks
Jianchao
>
> Now normally this isn't fatal because Worker will simply spin its entire
> time slice away and we'll eventually schedule our Task-A back in, which
> will complete the seqcount and things will work.
>
> But if, for some reason, our Worker was to have RT priority higher than
> our Task-A we'd be up some creek without no paddles.
>
> We don't happen to have preemption of IRQs off here? That would fix
> things nicely.
>
