Re: io_uring_prep_timeout_update on linked timeouts

[Jens Axboe <axboe@kernel.dk>]

On 8/26/21 7:40 PM, Victor Stewart wrote:
> On Wed, Aug 25, 2021 at 2:27 AM Victor Stewart <v@nametag.social> wrote:
>>
>> On Tue, Aug 24, 2021 at 11:43 PM Victor Stewart <v@nametag.social> wrote:
>>>
>>> we're able to update timeouts with io_uring_prep_timeout_update
>>> without having to cancel
>>> and resubmit, has it ever been considered adding this ability to
>>> linked timeouts?
>>
>> whoops turns out this does work. just tested it.
> 
> doesn't work actually. missed that because of a bit of misdirection.
> returns -ENOENT.
> 
> the problem with the current way of cancelling then resubmitting
> a new a timeout linked op (let's use poll here) is you have 3 situations:
> 
> 1) the poll triggers and you get some positive value. all good.
> 
> 2) the linked timeout triggers and cancels the poll, so the poll
> operation returns -ECANCELED.
> 
> 3) you cancel the existing poll op, and submit a new one with
> the updated linked timeout. now the original poll op returns
> -ECANCELED.
> 
> so solely from looking at the return value of the poll op in 2) and 3)
> there is no way to disambiguate them. of course the linked timeout
> operation result will allow you to do so, but you'd have to persist state
> across cqe processings. you can also track the cancellations and know
> to skip the explicitly cancelled ops' cqes (which is what i chose).
> 
> there's also the problem of efficiency. you can imagine in a QUIC
> server where you're constantly updating that poll timeout in response
> to idle timeout and ACK scheduling, this extra work mounts.
> 
> so i think the ability to update linked timeouts via
> io_uring_prep_timeout_update would be fantastic.

Hmm, I'll need to dig a bit, but whether it's a linked timeout or not
should not matter. It's a timeout, it's queued and updated the same way.
And we even check this in some of the liburing tests.

Do you have a test case that doesn't work for you? Always easier to
reason about a test case.

-- 
Jens Axboe