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Date:   Thu, 11 Nov 2021 16:55:12 -0500
From:   Waiman Long <longman@...hat.com>
To:     Peter Zijlstra <peterz@...radead.org>
Cc:     Hillf Danton <hdanton@...a.com>,
        马振华 <mazhenhua@...omi.com>,
        mingo <mingo@...hat.com>, will <will@...nel.org>,
        "boqun.feng" <boqun.feng@...il.com>,
        linux-kernel <linux-kernel@...r.kernel.org>
Subject: Re: [BUG]locking/rwsem: only clean RWSEM_FLAG_HANDOFF when already
 set


On 11/11/21 16:53, Peter Zijlstra wrote:
> On Thu, Nov 11, 2021 at 04:25:56PM -0500, Waiman Long wrote:
>> On 11/11/21 16:01, Waiman Long wrote:
>>> On 11/11/21 15:26, Peter Zijlstra wrote:
>>>> On Thu, Nov 11, 2021 at 02:36:52PM -0500, Waiman Long wrote:
>>>>
>>>>> @@ -434,6 +430,7 @@ static void rwsem_mark_wake(struct
>>>>> rw_semaphore *sem,
>>>>>                if (!(oldcount & RWSEM_FLAG_HANDOFF) &&
>>>>>                    time_after(jiffies, waiter->timeout)) {
>>>>>                    adjustment -= RWSEM_FLAG_HANDOFF;
>>>>> +                waiter->handoff_set = true;
>>>>>                    lockevent_inc(rwsem_rlock_handoff);
>>>>>                }
>>>> Do we really need this flag? Wouldn't it be the same as waiter-is-first
>>>> AND sem-has-handoff ?
>>> That is true. The only downside is that we have to read the count first
>>> in rwsem_out_nolock_clear_flags(). Since this is not a fast path, it
>>> should be OK to do that.
>> I just realize that I may still need this flag for writer to determine if it
>> should spin after failing to acquire the lock. Or I will have to do extra
>> read of count value in the loop. I don't need to use it for writer now.
> Maybe it's too late here, but afaict this is right after failing
> try_write_lock(), which will have done at least that load you're
> interested in, no?
>
> Simply have try_write_lock() update &count or something.

You are right. I have actually decided to do an extra read after second 
thought.

Cheers,
Longman

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