Hi all.

I need a small delay in my program, which bitbangs an i2c interface connected to the parallel port. This is mainly for educational purposes, so please don't question the wiseness of this decision.

Is there any simple way (short of writing a kernel module) of delaying less than the HZ value? Weirdly, fpNanosleep seems to delay 8.3 ms minimally (weird, 'cause that's less than the HZ value).

I'm using a 2.6 kernel.

John

Is there any simple way (short of writing a kernel module) of delaying less than the HZ value? Weirdly, fpNanosleep seems to delay 8.3 ms minimally (weird, 'cause that's less than the HZ value).

Just an extra note: delay(1) seems to delay 4.1 ms, so it _does_ seem to be possible to do less than 8.3?

John

Is there any simple way (short of writing a kernel module) of delaying less than the HZ value?

Very unelegant, but it works:

procedure usleep(time: longint);
var
  tv: timeval;
  tz: timezone;
  diff,
  ref: clong;
begin
  fpGetTimeOfDay(@tv, @tz);
  ref := tv.tv_usec;
  repeat
    fpGetTimeOfDay(@tv, @tz);
    diff := tv.tv_usec - ref;
    if diff < 0 then
      diff := diff + 1000000;
  until diff > time;
end;

in case someone is interested.

John

If you know the Processor Speed (MHz) then there is an assembler command which reads the cpu cycles (for example a 1MHz Processor has 1.000.000 cycles/second) so you can do it in this way.

http://en.wikipedia.org/wiki/RDTSC

Rainer

http://en.wikipedia.org/wiki/RDTSC

Interesting... I'm guessing gettimeofday would use this register to obtain the hi-res time it provides - does it?

Thanks
John

John Coppens wrote

Just an extra note: delay(1) seems to delay 4.1 ms, so it _does_ seem to be possible to do less than 8.3?

You wait till the next tick that is after the requested delay. So it's never a HZ multiple or so.

Micha