On Mon, Jan 30, 2023 at 12:31 PM Andrei Gherzan
<andrei.gherzan@xxxxxxxxxxxxx> wrote:
>
> On 23/01/30 11:29AM, Willem de Bruijn wrote:
> > On Mon, Jan 30, 2023 at 11:23 AM Andrei Gherzan
> > <andrei.gherzan@xxxxxxxxxxxxx> wrote:
> > >
> > > On 23/01/30 04:15PM, Andrei Gherzan wrote:
> > > > On 23/01/30 11:03AM, Willem de Bruijn wrote:
> > > > > On Mon, Jan 30, 2023 at 9:28 AM Andrei Gherzan
> > > > > <andrei.gherzan@xxxxxxxxxxxxx> wrote:
> > > > > >
> > > > > > On 23/01/30 08:35AM, Willem de Bruijn wrote:
> > > > > > > On Mon, Jan 30, 2023 at 7:51 AM Andrei Gherzan
> > > > > > > <andrei.gherzan@xxxxxxxxxxxxx> wrote:
> > > > > > > >
> > > > > > > > On 23/01/30 09:26AM, Paolo Abeni wrote:
> > > > > > > > > On Fri, 2023-01-27 at 17:03 -0500, Willem de Bruijn wrote:
> > > > > > > > > > On Fri, Jan 27, 2023 at 1:16 PM Andrei Gherzan
> > > > > > > > > > <andrei.gherzan@xxxxxxxxxxxxx> wrote:
> > > > > > > > > > >
> > > > > > > > > > > The tx and rx test programs are used in a couple of test scripts including
> > > > > > > > > > > "udpgro_bench.sh". Taking this as an example, when the rx/tx programs
> > > > > > > > > > > are invoked subsequently, there is a chance that the rx one is not ready to
> > > > > > > > > > > accept socket connections. This racing bug could fail the test with at
> > > > > > > > > > > least one of the following:
> > > > > > > > > > >
> > > > > > > > > > > ./udpgso_bench_tx: connect: Connection refused
> > > > > > > > > > > ./udpgso_bench_tx: sendmsg: Connection refused
> > > > > > > > > > > ./udpgso_bench_tx: write: Connection refused
> > > > > > > > > > >
> > > > > > > > > > > This change addresses this by adding routines that retry the socket
> > > > > > > > > > > operations with an exponential back off algorithm from 100ms to 2s.
> > > > > > > > > > >
> > > > > > > > > > > Fixes: 3a687bef148d ("selftests: udp gso benchmark")
> > > > > > > > > > > Signed-off-by: Andrei Gherzan <andrei.gherzan@xxxxxxxxxxxxx>
> > > > > > > > > >
> > > > > > > > > > Synchronizing the two processes is indeed tricky.
> > > > > > > > > >
> > > > > > > > > > Perhaps more robust is opening an initial TCP connection, with
> > > > > > > > > > SO_RCVTIMEO to bound the waiting time. That covers all tests in one
> > > > > > > > > > go.
> > > > > > > > >
> > > > > > > > > Another option would be waiting for the listener(tcp)/receiver(udp)
> > > > > > > > > socket to show up in 'ss' output before firing-up the client - quite
> > > > > > > > > alike what mptcp self-tests are doing.
> > > > > > > >
> > > > > > > > I like this idea. I have tested it and it works as expected with the
> > > > > > > > exeception of:
> > > > > > > >
> > > > > > > > ./udpgso_bench_tx: sendmsg: No buffer space available
> > > > > > > >
> > > > > > > > Any ideas on how to handle this? I could retry and that works.
> > > > > > >
> > > > > > > This happens (also) without the zerocopy flag, right? That
> > > > > > >
> > > > > > > It might mean reaching the sndbuf limit, which can be adjusted with
> > > > > > > SO_SNDBUF (or SO_SNDBUFFORCE if CAP_NET_ADMIN). Though I would not
> > > > > > > expect this test to bump up against that limit.
> > > > > > >
> > > > > > > A few zerocopy specific reasons are captured in
> > > > > > > https://www.kernel.org/doc/html/latest/networking/msg_zerocopy.html#transmission.
> > > > > >
> > > > > > I have dug a bit more into this, and it does look like your hint was in
> > > > > > the right direction. The fails I'm seeing are only with the zerocopy
> > > > > > flag.
> > > > > >
> > > > > > From the reasons (doc) above I can only assume optmem limit as I've
> > > > > > reproduced it with unlimited locked pages and the fails are transient.
> > > > > > That leaves optmem limit. Bumping the value I have by default (20480) to
> > > > > > (2048000) made the sendmsg succeed as expected. On the other hand, the
> > > > > > tests started to fail with something like:
> > > > > >
> > > > > > ./udpgso_bench_tx: Unexpected number of Zerocopy completions: 774783
> > > > > > expected 773707 received
> > > > >
> > > > > More zerocopy completions than number of sends. I have not seen this before.
> > > > >
> > > > > The completions are ranges of IDs, one per send call for datagram sockets.
> > > > >
> > > > > Even with segmentation offload, the counter increases per call, not per segment.
> > > > >
> > > > > Do you experience this without any other changes to udpgso_bench_tx.c.
> > > > > Or are there perhaps additional sendmsg calls somewhere (during
> > > > > initial sync) that are not accounted to num_sends?
> > > >
> > > > Indeed, that looks off. No, I have run into this without any changes in
> > > > the tests (besides the retry routine in the shell script that waits for
> > > > rx to come up). Also, as a data point.
> > >
> > > Actually wait. I don't think that is the case here. "expected" is the
> > > number of sends. In this case we sent 1076 more messages than
> > > completions. Am I missing something obvious?
> >
> > Oh indeed.
> >
> > Receiving fewer completions than transmission is more likely.
>
> Exactly, yes.
>
> > This should be the result of datagrams still being somewhere in the
> > system. In a qdisc, or waiting for the network interface to return a
> > completion notification, say.
> >
> > Does this remain if adding a longer wait before the final flush_errqueue?
>
> Yes and no. But not realiably unless I go overboard.
>
> > Or, really, the right fix is to keep polling there until the two are
> > equal, up to some timeout. Currently flush_errqueue calls poll only
> > once.
>
> That makes sense. I have implemented a retry and this ran for a good
> while now.
>
> - flush_errqueue(fd, true);
> + while (true) {
> + flush_errqueue(fd, true);
> + if ((stat_zcopies == num_sends) || (delay >= MAX_DELAY))
> + break;
> + usleep(delay);
> + delay *= 2;
> + }
>
> What do you think?
Thanks for running experiments.
We can avoid the unconditional sleep, as the poll() inside
flush_errqueue already takes a timeout.
One option is to use start_time = clock_gettime(..) or gettimeofday
before poll, and restart poll until either the exit condition or
timeout is reached, with timeout = orig_time - elapsed_time.
