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Date:      Thu, 2 Oct 2014 13:13:36 -0500
From:      Paul Albrecht <palbrecht@glccom.com>
To:        Adrian Chadd <adrian@freebsd.org>
Cc:        "freebsd-hackers@freebsd.org" <freebsd-hackers@freebsd.org>
Subject:   Re: freebsd 10 kqueue timer regression
Message-ID:  <8587D819-AA2F-4387-A4E9-523014384672@glccom.com>
In-Reply-To: <CAJ-VmonJQKWeW7K6%2BjY6=FpmZrm%2B6HQOuBmhhjJEapyVpwNFdQ@mail.gmail.com>
References:  <8ABC0977-FB8F-45E7-ACCC-BFA92EE22E1C@glccom.com> <CAJ-VmonJQKWeW7K6%2BjY6=FpmZrm%2B6HQOuBmhhjJEapyVpwNFdQ@mail.gmail.com>

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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:
>>=20
>> Hi,
>>=20
>> What=92s up with freebsd 10? I=92m testing some code that uses the =
kqueue timer for timing and it doesn=92t work because the precision of =
the timer is off.
>=20
> Can you provide a test case for it?

Here=92s 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.


>=20
> I just chased down one of those recently; maybe it's the same thing
> (callout() API changes.)
>=20
>=20
> -a




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