On Tue, 20 Feb 2001, Roman Black wrote:

> Drew Vassallo wrote:
> >
> > (don't mind the "." in front of each line... my email client seems not to
> > recognize tabs or spaces in front of text)
> >
> > Hmm, how about this.  6 cycles regardless of which bit changed.  7 cycles if
> > neither changed 0->1.
> >
> > .   movf PORTB, w
> > .   andlw 0x30  ; mask out 4,5
> > .   xorwf last_input, f
> > .   btfss last_input, 5
> > .   btfss last_input, 4
> > .   goto error_path
> > ;; Either 4 or 5 changed 0->1
> > .   movf PORTB, w
> > .   movwf last_input
> > .
> > .   etc...
> > error_path
> > ;; Otherwise, neither changed
> > ;   do error stuff here
> > ;
>
>
> Hi Andrew (and thanks everyone else re this challenge!).
>
> But now I'm getting confused. Doesn't the xorwf
> produce a 1 for every bit that has changed, and a 0
> for every unchanged bit??
>
> So this is just a test for change?? How does it
> just detect a 0-1 change but not a 1-0 change?
>
> I already had a fast routine to detect any change:
>
> subwf last_input,w
> skpz
>
> so did I miss something? :o)
>
> I forgot to mention before too that after it detects
> a 0-1 going input I need to know which input it was
> to get to that routine asap. :o)


Another way to approach this problem is with state machines. As I see it you
have these conditions which are of interest:

    bit5    bit4
SA:  low     low
SB:  low    high
SC: high     low
SD: high    high


Then the low to high transitions may be checked slightly faster:

while in state SA:

    movf   PORTB,W
    andlw  MASK
    skpnz
  ...


However, this assumes your in a tight polling loop. If you're really
multitasking, then you have the overhead of determining which state to switch
to.
Just a thought.

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