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Date:   Wed, 18 Mar 2020 12:07:37 +0100
From:   Paolo Valente <paolo.valente@...aro.org>
To:     Zhiqiang Liu <liuzhiqiang26@...wei.com>
Cc:     Jens Axboe <axboe@...nel.dk>,
        linux-block <linux-block@...r.kernel.org>,
        "linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
        Mingfangsen <mingfangsen@...wei.com>,
        Yanxiaodan <yanxiaodan@...wei.com>,
        "wubo (T)" <wubo40@...wei.com>, renxudong <renxudong1@...wei.com>,
        Louhongxiang <louhongxiang@...wei.com>
Subject: Re: [PATCH] block, bfq: fix use-after-free in
 bfq_idle_slice_timer_body



> Il giorno 18 mar 2020, alle ore 10:52, Zhiqiang Liu <liuzhiqiang26@...wei.com> ha scritto:
> 
> 
> 
> On 2020/3/18 16:45, Paolo Valente wrote:
>> 
>> 
>>>>> 	spin_lock_irqsave(&bfqd->lock, flags);
>>>>> -	bfq_clear_bfqq_wait_request(bfqq);
>>>>> -
>>>>> 	if (bfqq != bfqd->in_service_queue) {
>>>>> 		spin_unlock_irqrestore(&bfqd->lock, flags);
>>>>> 		return;
>>>>> 	}
>>>>> 
>>>>> +	bfq_clear_bfqq_wait_request(bfqq);
>>>>> +
>>>> 
>>>> Please add a comment on why you (correctly) clear this flag only if bfqq is in service.
>>>> 
>>>> For the rest, seems ok to me.
>>>> 
>>>> Thank you very much for spotting and fixing this bug,
>>>> Paolo
>>>> 
>>> Thanks for your reply.
>>> Considering that the bfqq may be in race, we should firstly check whether bfqq is in service before
>>> doing something on it.
>>> 
>> 
>> The comment you propose is correct, but the correctness issue I raised
>> is essentially the opposite.  Sorry for not being clear.
>> 
>> Let me put it the other way round: why is it still correct that, if
>> bfqq is not the queue in service, then that flag is not cleared at all?
>> IOW, why is it not a problem that that flag remains untouched is bfqq
>> is not in service?
>> 
>> Thanks,
>> Paolo
>> 
> Thanks for your patient.
> As you comment in bfq_idle_slice_timer, there are two race situations as follows,
> a) bfqq is null
>   bfq_idle_slice_timer will not call bfq_idle_slice_timer_body ->no problem
> b) bfqq are not in service
>   1) bfqq is freed
>      it will cause use-after-free problem before calling bfq_clear_bfqq_wait_request
>      in bfq_idle_slice_timer_body. -> use-after-free problem as analyzed in the patch.
>   2) bfqq is not freed
>      it means in_service_queue has been set to a new bfqq. The old bfqq has been expired
>      through __bfq_bfqq_expire func. Then the wait_request flags of old bfqq will be cleared
>      in __bfq_bfqd_reset_in_service func. -> it is no a problem to re-clear the wait_request
>      flags before checking whether bfqq is in service.

Great, this item 2 is exactly what I meant.  We need a comment
because, even if now this stuff is clear to you, imagine somebody
else getting to your modified piece of code after reading hundreds of
lines of code, about a non-trivial state machine as BFQ ...  :)

Thanks,
Paolo

> 
> In one word, the old bfqq in race has already cleared the wait_request flag when switching in_service_queue.
> 
> Thanks,
> Zhiqiang Liu
> 
>>>>> 
>>>> 
>>>> 
>>>> .
>> 
>> 
>> .
>> 
> 

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