Re: [PATCH 1/2] io_uring: trigger timeout after any sqe->off CQEs

[Pavel Begunkov <asml.silence@gmail.com>]

On 20/04/2020 23:15, Pavel Begunkov wrote:
> On 20/04/2020 23:12, Pavel Begunkov wrote:
>> On 20/04/2020 22:40, Jens Axboe wrote:
>>> On 4/18/20 11:20 AM, Pavel Begunkov wrote:
>>>> +static void __io_flush_timeouts(struct io_ring_ctx *ctx)
>>>> +{
>>>> +	u32 end, start;
>>>> +
>>>> +	start = end = ctx->cached_cq_tail;
>>>> +	do {
>>>> +		struct io_kiocb *req = list_first_entry(&ctx->timeout_list,
>>>> +							struct io_kiocb, list);
>>>> +
>>>> +		if (req->flags & REQ_F_TIMEOUT_NOSEQ)
>>>> +			break;
>>>> +		/*
>>>> +		 * multiple timeouts may have the same target,
>>>> +		 * check that @req is in [first_tail, cur_tail]
>>>> +		 */
>>>> +		if (!io_check_in_range(req->timeout.target_cq, start, end))
>>>> +			break;
>>>> +
>>>> +		list_del_init(&req->list);
>>>> +		io_kill_timeout(req);
>>>> +		end = ctx->cached_cq_tail;
>>>> +	} while (!list_empty(&ctx->timeout_list));
>>>> +}
>>>> +
>>>>  static void io_commit_cqring(struct io_ring_ctx *ctx)
>>>>  {
>>>>  	struct io_kiocb *req;
>>>>  
>>>> -	while ((req = io_get_timeout_req(ctx)) != NULL)
>>>> -		io_kill_timeout(req);
>>>> +	if (!list_empty(&ctx->timeout_list))
>>>> +		__io_flush_timeouts(ctx);
>>>>  
>>>>  	__io_commit_cqring(ctx);
>>>>  
>>>
>>> Any chance we can do this without having to iterate timeouts on the
>>> completion path?
>>>
>>
>> If you mean the one in __io_flush_timeouts(), then no, unless we forbid timeouts
>> with identical target sequences + some extra constraints. The loop there is not
>> new, it iterates only over timeouts, that need to be completed, and removes
>> them. That's amortised O(1).
> 
> We can think about adding unlock/lock, if that's what you are thinking about.
> 
> 
>> On the other hand, there was a loop in io_timeout_fn() doing in total O(n^2),
>> and it was killed by this patch.
> 

Any thoughts on this?

I'll return fixing the last timeout bug I saw, but I'd prefer to know on top of
what to do that.

-- 
Pavel Begunkov