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Subject: Re: [PATCH net-next 2/3] net: sched: implement TCQ_F_CAN_BYPASS for
 lockless qdisc
To:     Jakub Kicinski <kuba@kernel.org>,
        Yunsheng Lin <yunshenglin0825@gmail.com>
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From:   Yunsheng Lin <linyunsheng@huawei.com>
Message-ID: <3c2fbc70-841f-d90b-ca13-1f058169be50@huawei.com>
Date:   Mon, 31 May 2021 08:40:14 +0800
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On 2021/5/31 4:21, Jakub Kicinski wrote:
> On Sun, 30 May 2021 09:37:09 +0800 Yunsheng Lin wrote:
>> On 2021/5/30 2:49, Jakub Kicinski wrote:
>>> The fact that MISSED is only cleared under q->seqlock does not matter,
>>> because setting it and ->enqueue() are not under any lock. If the thread
>>> gets interrupted between:
>>>
>>> 	if (q->flags & TCQ_F_CAN_BYPASS && nolock_qdisc_is_empty(q) &&
>>> 	    qdisc_run_begin(q)) {
>>>
>>> and ->enqueue() we can't guarantee that something else won't come in,
>>> take q->seqlock and clear MISSED.
>>>
>>> thread1                thread2             thread3
>>> # holds seqlock
>>>                        qdisc_run_begin(q)
>>>                        set(MISSED)
>>> pfifo_fast_dequeue
>>>   clear(MISSED)
>>>   # recheck the queue
>>> qdisc_run_end()  
>>>                        ->enqueue()  
>>>                                             q->flags & TCQ_F_CAN_BYPASS..
>>>                                             qdisc_run_begin() # true
>>>                                             sch_direct_xmit()
>>>                        qdisc_run_begin()
>>>                        set(MISSED)
>>>
>>> Or am I missing something?
>>>
>>> Re-checking nolock_qdisc_is_empty() may or may not help.
>>> But it doesn't really matter because there is no ordering
>>> requirement between thread2 and thread3 here.  
>>
>> I were more focued on explaining that using MISSED is reliable
>> as sch_may_need_requeuing() checking in RFCv3 [1] to indicate a
>> empty qdisc, and forgot to mention the data race described in
>> RFCv3, which is kind of like the one described above:
>>
>> "There is a data race as below:
>>
>>       CPU1                                   CPU2
>> qdisc_run_begin(q)                            .
>>         .                                q->enqueue()
>> sch_may_need_requeuing()                      .
>>     return true                               .
>>         .                                     .
>>         .                                     .
>>     q->enqueue()                              .
>>
>> When above happen, the skb enqueued by CPU1 is dequeued after the
>> skb enqueued by CPU2 because sch_may_need_requeuing() return true.
>> If there is not qdisc bypass, the CPU1 has better chance to queue
>> the skb quicker than CPU2.
>>
>> This patch does not take care of the above data race, because I
>> view this as similar as below:
>>
>> Even at the same time CPU1 and CPU2 write the skb to two socket
>> which both heading to the same qdisc, there is no guarantee that
>> which skb will hit the qdisc first, becuase there is a lot of
>> factor like interrupt/softirq/cache miss/scheduling afffecting
>> that."
>>
>> Does above make sense? Or any idea to avoid it?
>>
>> 1. https://patchwork.kernel.org/project/netdevbpf/patch/1616404156-11772-1-git-send-email-linyunsheng@huawei.com/
> 
> We agree on this one.
> 
> Could you draw a sequence diagram of different CPUs (like the one
> above) for the case where removing re-checking nolock_qdisc_is_empty()
> under q->seqlock leads to incorrect behavior? 

When nolock_qdisc_is_empty() is not re-checking under q->seqlock, we
may have:


        CPU1                                   CPU2
  qdisc_run_begin(q)                            .
          .                                enqueue skb1
deuqueue skb1 and clear MISSED                  .
          .                        nolock_qdisc_is_empty() return true
    requeue skb                                 .
   q->enqueue()                                 .
    set MISSED                                  .
        .                                       .
 qdisc_run_end(q)                               .
        .                              qdisc_run_begin(q)
        .                             transmit skb2 directly
        .                           transmit the requeued skb1

The problem here is that skb1 and skb2  are from the same CPU, which
means they are likely from the same flow, so we need to avoid this,
right?

> 
> If there is no such case would you be willing to repeat the benchmark
> with and without this test?
> 
> Sorry for dragging the review out..
> 
> .
>