Date: Thu, 2 Oct 2014 12:42:38 -0700 From: Adrian Chadd <adrian@freebsd.org> To: Paul Albrecht <palbrecht@glccom.com> Cc: "freebsd-hackers@freebsd.org" <freebsd-hackers@freebsd.org> Subject: Re: freebsd 10 kqueue timer regression Message-ID: <CAJ-VmomKFd_oTXp1bVhU22MgHG6U1V7mr6iwWmoyUGKpSqPy1Q@mail.gmail.com> In-Reply-To: <8587D819-AA2F-4387-A4E9-523014384672@glccom.com> References: <8ABC0977-FB8F-45E7-ACCC-BFA92EE22E1C@glccom.com> <CAJ-VmonJQKWeW7K6%2BjY6=FpmZrm%2B6HQOuBmhhjJEapyVpwNFdQ@mail.gmail.com> <8587D819-AA2F-4387-A4E9-523014384672@glccom.com>
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Right, and jhb@ mentioned callout() and related stuff.
Let me take a look. I bet it's not doing things "right".
-a
On 2 October 2014 11:13, Paul Albrecht <palbrecht@glccom.com> wrote:
>
> On Oct 2, 2014, at 12:18 PM, Adrian Chadd <adrian@freebsd.org> wrote:
>
> On 2 October 2014 08:07, Paul Albrecht <palbrecht@glccom.com> wrote:
>
>
> Hi,
>
> What=E2=80=99s up with freebsd 10? I=E2=80=99m testing some code that use=
s the kqueue timer
> for timing and it doesn=E2=80=99t work because the precision of the timer=
is off.
>
>
> Can you provide a test case for it?
>
>
> Here=E2=80=99s the code:
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <string.h>
> #include <unistd.h>
> #include <errno.h>
> #include <sys/types.h>
> #include <sys/event.h>
> #include <sys/time.h>
>
> int
> main(void)
> {
> int i,msec;
> int kq,nev;
> struct kevent inqueue;
> struct kevent outqueue;
> struct timeval start,end;
>
> if ((kq =3D kqueue()) =3D=3D -1) {
> fprintf(stderr, "kqueue error!? errno =3D %s",
> strerror(errno));
> exit(EXIT_FAILURE);
> }
> EV_SET(&inqueue, 1, EVFILT_TIMER, EV_ADD | EV_ENABLE, 0, 20, 0);
>
> gettimeofday(&start, 0);
> for (i =3D 0; i < 50; i++) {
> if ((nev =3D kevent(kq, &inqueue, 1, &outqueue, 1, NULL))=
=3D=3D
> -1) {
> fprintf(stderr, "kevent error!? errno =3D %s",
> strerror(errno));
> exit(EXIT_FAILURE);
> } else if (outqueue.flags & EV_ERROR) {
> fprintf(stderr, "EV_ERROR: %s\n",
> strerror(outqueue.data));
> exit(EXIT_FAILURE);
> }
> }
> gettimeofday(&end, 0);
>
> msec =3D ((end.tv_sec - start.tv_sec) * 1000) + (((1000000 +
> end.tv_usec - start.tv_usec) / 1000) - 1000);
>
> printf("msec =3D %d\n", msec);
>
> close(kq);
> return EXIT_SUCCESS;
> }
>
> When I run it on my system I get these results:
>
> ./a.out
> msec =3D 1072
> ./a.out
> msec =3D 1071
> ./a.out
> msec =3D 1071
>
> Which is over about 3.5 times the wait time per second.
>
>
>
> I just chased down one of those recently; maybe it's the same thing
> (callout() API changes.)
>
>
> -a
>
>
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