Gilles Chanteperdrix wrote: > 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... >
Yeah, most probably the right way for the timer IRQ. If we have to loop here significantly often, then we already had a much too long IRQ-off period and are toasted anyway. Jan
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