Re: [PATCH 2/2] io_uring: use TWA_SIGNAL for task_work if the task isn't running

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On 8/10/20 2:12 PM, Peter Zijlstra wrote:
> On Mon, Aug 10, 2020 at 01:21:48PM -0600, Jens Axboe wrote:
> 
>>>> Wait.. so the only change here is that you look at tsk->state, _after_
>>>> doing __task_work_add(), but nothing, not the Changelog nor the comment
>>>> explains this.
>>>>
>>>> So you're relying on __task_work_add() being an smp_mb() vs the add, and
>>>> you order this against the smp_mb() in set_current_state() ?
>>>>
>>>> This really needs spelling out.
>>>
>>> I'll update the changelog, it suffers a bit from having been reused from
>>> the earlier versions. Thanks for checking!
>>
>> I failed to convince myself that the existing construct was safe, so
>> here's an incremental on top of that. Basically we re-check the task
>> state _after_ the initial notification, to protect ourselves from the
>> case where we initially find the task running, but between that check
>> and when we do the notification, it's now gone to sleep. Should be
>> pretty slim, but I think it's there.
>>
>> Hence do a loop around it, if we're using TWA_RESUME.
>>
>> diff --git a/fs/io_uring.c b/fs/io_uring.c
>> index 44ac103483b6..a4ecb6c7e2b0 100644
>> --- a/fs/io_uring.c
>> +++ b/fs/io_uring.c
>> @@ -1780,12 +1780,27 @@ static int io_req_task_work_add(struct io_kiocb *req, struct callback_head *cb)
>>  	 * to ensure that the issuing task processes task_work. TWA_SIGNAL
>>  	 * is needed for that.
>>  	 */
>> -	if (ctx->flags & IORING_SETUP_SQPOLL)
>> +	if (ctx->flags & IORING_SETUP_SQPOLL) {
>>  		notify = 0;
>> -	else if (READ_ONCE(tsk->state) != TASK_RUNNING)
>> -		notify = TWA_SIGNAL;
>> +	} else {
>> +		bool notified = false;
>>  
>> -	__task_work_notify(tsk, notify);
>> +		/*
>> +		 * If the task is running, TWA_RESUME notify is enough. Make
>> +		 * sure to re-check after we've sent the notification, as not
> 
> Could we get a clue as to why TWA_RESUME is enough when it's running? I
> presume it is because we'll do task_work_run() somewhere before we
> block, but having an explicit reference here might help someone new to
> this make sense of it all.

Right, it's because we're sure to run task_work in that case. I'll
update the comment.

-- 
Jens Axboe




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