> At 02:33 PM 8/7/02 -0500, Adam Smith wrote:
> >The reason I need this routine to be short is because I have to retfie
> >in time for the next interrupt (24 instruction cycles per interrupt, 18
> >after you consider the 4 instructions it takes for the PIC to go to the
> >interrupt vector, etc. and the 2 instructions it takes to retfie).
>
> Sorry.  I read it as "instructions before the [single instr]" rather
> than "before the retfie".
>
> I think that if you have that close of a time constraint, that this
> may not be the way you want to do it.  (I also know you want to see
> if you can do it...and I also know that if it works, then, it works :)
>
>
> >(The code that executes in-between interrupts is just an infinite loop.
> >This is running on a PIC16F872 @ 3.6864 MHz.)
>
> Ok, on the chance that the interrupt must be short, but that
> the time from "timeout" to actually doing that "one instruction"
> is not so critical, use your "infinite loop" time to offload some
> of that decrementing or checking.

Does the foreground infinite loop call any subroutines?  If not, then you
could re-enable interrupts in the interrupt routine.  That causes the
interrupt routine to restart when a new interrupt occurs.  In other words,
an interrupt condition becomes a GOTO 4.  Of course if this happens all the
time, then the foreground loop will never get any cycles (so I guess it
doesn't matter if it can't use the stack).


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