On Wed, 21 Nov 2018, Geert Uytterhoeven wrote: > > This suggests that either 0 or N (the latched value) would result from > > a read from the counter immediately following an interrupt. Who can > > say which? Just have to try it. The answer should allow us to avoid > > the risk of a clocksource that jumps forwards and backwards. > > The code in amiga_gettimeoffset() does: > > ticks = hi << 8 | lo; > > if (ticks > jiffy_ticks / 2) > /* check for pending interrupt */ > if (cia_set_irq(&ciab_base, 0) & CIA_ICR_TA) > offset = 10000; >
That _suggests_ that there's no interrupt when ticks == 0. But look what happens next: > ticks = jiffy_ticks - ticks; > > ticks = (10000 * ticks) / jiffy_ticks; > > return (ticks + offset) * 1000; If (hi << 8 | lo) == 0, and you set offset = 10000, then the return value would be maximal. Let's immediately call this function again. This time (hi << 8 | lo) == N. Let's add the offset again. I'm afraid the clock just jumped backwards. So the logic you quoted has a rationale which is unrelated to the question. --
