Hello John .


> > I would, most of all, appreciate an add with carry and subtract with
> > borrow.  This would __greatly__ simplify multi byte arithmetic!  But,
> > alas...
>
> True, but even without those a 4-byte add isn't too bad...
>
> ; D<-D+S
>         movf    S0,w
>         addwf   D0,f
Ok.

>         movf    S1,w
>         btfsc   C
>          incfsz S1,w
>          addwf  D1,f

Stop. For example S1=0xFF , D1=0x00 , Carry=1 . After executing your
code result will
be equal 0x00 with Carry=0 not =1 . It's error . In my code to correct
this situation
I save previous Carry's in S1 .

        movfw   S1
        skpnc
        addlw   1
        rlf     S1,F
        addwf   D1,F
        btfsc   S1,0
        setc

... etc


>         movf    S2,w
>         btfsc   C
>          incfsz S2,w
>          addwf  D2,w
>         rlf     KZ,w    ; KZ == known-zero address
>         addwf   S3,w
>         addwf   D3,f
>
> Total of 13 instructions, 13 cycles.  Note that if the destination is not
> one of the sources, you should copy the middle bytes from source to
> destination and then do the add.


WBR Dmitry.