Re: [BUG]locking/rwsem: only clean RWSEM_FLAG_HANDOFF when already set

From: Waiman Long <longman@redhat.com>
To: Peter Zijlstra <peterz@infradead.org>
Cc: "Hillf Danton" <hdanton@sina.com>, 马振华 <mazhenhua@xiaomi.com>,
	mingo <mingo@redhat.com>, will <will@kernel.org>,
	"boqun.feng" <boqun.feng@gmail.com>,
	linux-kernel <linux-kernel@vger.kernel.org>
Subject: Re: [BUG]locking/rwsem: only clean RWSEM_FLAG_HANDOFF when already set
Date: Thu, 11 Nov 2021 16:25:56 -0500	[thread overview]
Message-ID: <436bcf39-297a-f5a6-ac58-a82e77cb3b83@redhat.com> (raw)
In-Reply-To: <be3dc705-494a-913e-230f-9533c7404ac2@redhat.com>

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.

Cheers,
Longman