kqueue periodic timer confusion

[Paul Albrecht]

Hi,

Sorry about this repost but I'm confused about the responses I received
in my last post so I'm looking for some clarification.

Specifically, I though I could use the kqueue timer as essentially a
"drop in" replacement for linuxfd_create/read, but was surprised that
the accuracy of the kqueue timer is much less than what I need for my
application.

So my confusion at this point is whether this is consider to be a bug or
"feature"?

Here's some test code if you want to verify the problem:

#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 = kqueue()) == -1) {
                fprintf(stderr, "kqueue error!? errno = %s", strerror(errno));
                exit(EXIT_FAILURE);
        }
        EV_SET(&inqueue, 1, EVFILT_TIMER, EV_ADD | EV_ENABLE, 0, 20, 0);

        gettimeofday(&start, 0);
        for (i = 0; i < 50; i++) {
                if ((nev = kevent(kq, &inqueue, 1, &outqueue, 1, NULL)) == -1) {
                        fprintf(stderr, "kevent error!? errno = %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 = ((end.tv_sec - start.tv_sec) * 1000) + (((1000000 + end.tv_usec - start.tv_usec) / 1000) - 1000);

        printf("msec = %d\n", msec);

        close(kq);
        return EXIT_SUCCESS;
}


-- 
Paul Albrecht