On Wed, 13 Nov 1996, Robert Lunn wrote:

>         This is just unrolling the loop once.  The number of
>         instructions is the same (you save a test on "overflow"
>         but add a "lo = timer").

That was what I thought when I first rewrote it, but it's not really just
an unrolling (though you could arrive at that by unrolling and then
deducing which parts were unnecessary, I suppose), and if the PIC had a
conditional branch I think the instruction counts would be the same.  My
assumption is that lo can held in W after it has been captured and then
assigned once "past the end" of the pseudo code.

>         I prefer the looped version because it handles the
>         situation where the timer overflows immediately after
>         you've cleared overflow.

Both version handle that case, in that they leave a coherent sample of the
timer's state in hi:lo.  Neither version will work well if the code isn't
executed at least often enought to catch every overflow, of course.

>         This seems a trifle more robust to me.  It also more
>         closely matches what I intuitively expect the code to
>         do, which is to tell me the current 'time'.

I disagree that either version is more robust.  They do give results
different by one count for a right-near-the-edge timing coincidence, but
that's a half-full/half-empty choice.  I would suggest that if this choice
makes any significant difference then you really needed a higher
resolution in the timer anyway.


code originally at top of message:

> >>                 repeat
> >>                     if overflow then
> >>                         begin
> >>                         hi := hi + 1;
> >>                         clear overflow
> >>                         end{if};
> >>                     lo := timer
> >>                 until not overflow
> >
> >Or just
> >
> >        lo = timer
> >        if (overflow) {
> >            lo = timer
> >            ++hi
> >            clear overflow
> >        }