On Tue, 2021-06-22 at 19:01 +0100, Pavel Begunkov wrote:
> On 6/22/21 6:54 PM, Pavel Begunkov wrote:
> > On 6/22/21 1:17 PM, Olivier Langlois wrote:
> > >
> >
> > > static bool __io_poll_remove_one(struct io_kiocb *req,
> > > @@ -6437,6 +6445,7 @@ static void __io_queue_sqe(struct io_kiocb
> > > *req)
> > > struct io_kiocb *linked_timeout =
> > > io_prep_linked_timeout(req);
> > > int ret;
> > >
> > > +issue_sqe:
> > > ret = io_issue_sqe(req,
> > > IO_URING_F_NONBLOCK|IO_URING_F_COMPLETE_DEFER);
> > >
> > > /*
> > > @@ -6456,12 +6465,16 @@ static void __io_queue_sqe(struct
> > > io_kiocb *req)
> > > io_put_req(req);
> > > }
> > > } else if (ret == -EAGAIN && !(req->flags &
> > > REQ_F_NOWAIT)) {
> > > - if (!io_arm_poll_handler(req)) {
> > > + switch (io_arm_poll_handler(req)) {
> > > + case IO_APOLL_READY:
> > > + goto issue_sqe;
> > > + case IO_APOLL_ABORTED:
> > > /*
> > > * Queued up for async execution, worker
> > > will release
> > > * submit reference when the iocb is
> > > actually submitted.
> > > */
> > > io_queue_async_work(req);
> > > + break;
> >
> > Hmm, why there is a new break here? It will miscount
> > @linked_timeout
> > if you do that. Every io_prep_linked_timeout() should be matched
> > with
> > io_queue_linked_timeout().
>
> Never mind, I said some nonsense and apparently need some coffee
but this is a pertinant question, imho. I guess that you could get away
without it since it is the last case of the switch statement... I am
not sure what kernel coding standard says about that.
However, I can tell you that there was also a break statement at the
end of the case for IO_APOLL_READY and checkpatch.pl did complain about
it saying that it was useless since it was following a goto statement.
Therefore, I did remove that one.
checkpatch.pl did remain silent about the other remaining break. Hence
this is why I left it there.
Greetings,
