> "Richard B. Johnson" wrote:
> >
> > On Fri, 2 Mar 2001, Christopher Friesen wrote:
> >
> > > John Being wrote:
> > >
> > > > gives following result on box in question
> > > > root@******:# ./clo
> > > > Leap found: -1687 msec
> > > > and prints nothing on all other my boxes.
> > > > This gives me bunch of troubles with occasional hang ups and I found nothing
> > > > in kernel archives at
> > > > http://www.uwsg.indiana.edu/hypermail/linux/kernel/index.html
> > > > just some notes about smth like this for SMP boxes with ntp. Is this issue
> > > > known, and how can I fix it?
> > >
> > > I've run into non-monotonic gettimeofday() on a PPC system with 2.2.17, but it
> > > always seemed to be almost exactly a jiffy out, as though it was getting
> > > hundredths of a second from the old tick, and microseconds from the new tick.
> > > Your leap seems to be more unusual, and the first one I've seen on an x86 box.
> > >
> > > Have you considered storing the results to see what happens on the next call?
> > > Does it make up the difference, or do you just lose that time?
> > >
> > > Chris
> >
> > I think it's a math problem in the test code. Try this:
> >
> > #include <stdio.h>
> > #include <sys/time.h>
> >
> > #define DEB(f)
> >
> > int main()
> > {
> > struct timeval t;
> > double start_us;
> > double stop_us;
> > for(;;)
> > {
> > gettimeofday(&t, NULL);
> > start_us = (double) t.tv_sec * 1e6;
> > start_us += (double) t.tv_usec;
> > gettimeofday(&t, NULL);
> > stop_us = (double) t.tv_sec * 1e6;
> > stop_us += (double) t.tv_usec;
> > if(stop_us <= start_us)
> > break;
> > DEB(fprintf(stdout, "Start = %f, Stop = %f\n", start_us, stop_us));
> > }
> > fprintf(stderr, "Start = %f, Stop = %f\n", start_us, stop_us);
> > return 0;
> > }
> >
> > Note that two subsequent calls to gettimeofday() must not return the
> > same time even if your CPU runs infinitely fast. I haven't seen any
> > kernel in the past few years that fails this test.
>
> Oh! With only micro second resolution how is this avoided? The only
> "legal" thing to do to avoid this is for the fast boxes to loop until
> the requirement is satisfied. Is this really done?
>
> George
>
Yes and no. It takes microseconds to call the kernel for anything (time
getpid() ), so it seldom loops. All the kernel has to do is remember
the last value returned. If the time isn't past that time yet, bump
that value and return it instead of waiting.
Cheers,
Dick Johnson
Penguin : Linux version 2.4.1 on an i686 machine (799.53 BogoMips).
"Memory is like gasoline. You use it up when you are running. Of
course you get it all back when you reboot..."; Actual explanation
obtained from the Micro$oft help desk.
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