Jan Kiszka wrote:
> Philippe Gerum wrote:
> > On Sat, 2006-07-29 at 16:20 +0200, Jan Kiszka wrote:
> >>>> :|func 6 xnintr_clock_handler (__ipipe_dispatch_wired)
> >>>> :|func 6 xnintr_irq_handler (xnintr_clock_handler)
> >>>> :|func 7 xnpod_announce_tick (xnintr_irq_handler)
> >>>> :|func 8+ xntimer_do_tick_aperiodic (xnpod_announce_tick)
> >>>> :|func 9 xnthread_periodic_handler (xntimer_do_tick_aperiodic)
> >>>> :|func 10 xnpod_resume_thread (xnthread_periodic_handler)
> >>>> :|[21559] 11+ xnpod_resume_thread (xnthread_periodic_handler)
> >>>> :|func 13+ xnthread_periodic_handler (xntimer_do_tick_aperiodic)
> >> ...
> >>
> >>>> :|func 363+ xnthread_periodic_handler (xntimer_do_tick_aperiodic)
> >> That are a lot of overruns. Haven't counted, but it should be one
> >> xnthread_periodic_handler per missed 100 us period (20000 / 100 = 200!).
> >> [BTW, I think we should handle even this failure scenario without
> >> looping.
> >
> > We need to loop in the aperiodic handler in order to catch timers that
> > could have elapsed while processing the current tick. However,
>
> No, that was not what I meant. I know that we need the timer loop. But I
> was thinking of something like this for the tick handler's error path:
>
> if (unlikely((timer.date += timer.interval) < now))
> timer.date = now + timer.interval -
> (now - timer.date) % timer.interval;
Actually,
while (timer.date < now)
timer.date += timer.interval
cost much less cycles in the normal/fast case than going through a
division...
--
Gilles Chanteperdrix.
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