I assume you have an 8 bit signed value and would like to add it to a 16 bit
signed, right? The 2's complient had already been explained to you, so you
know you do not have to worry about whenever your value is negative or
positive as long as both numbers represented in 8 bits. But you have to add
that 8 bits to 16 so you have to extend it first like this:

        clrf    tmp       ; high value of input is now 0x00
        btfsc   WREG, 7
        comf    tmp,f     ; oh, that's a negative one? put 0xFF instead

so if WREG is positive tmp holds 0 but if negative holds 0xFF. So now all
you have to do is to add the two 16 bit numbers as normal:

        addwf   blahL,f
        movf    tmp, W
        addwfc  blahH,f

If you would like to extend it to add two 16 signed numbers then all you
have to do is to forget the first 3 instruction which just expands the 8 bit
signed to 16 one...

Do not forget that there are always some tricks to avoid using extra memory
space and save some cycles :-)

        addwf   blahL,f     ; that's standard
        btfsc   STATUS, C
        incf    blahH       ; add carry to the high byte
        btfsc   WREG, 7
        decf    blahH       ; erm, negative number, adjust high byte

With this one you saved 1 cycle and no more RAM needed -- but you can't
extend this to add two 16 signed.

Regards,
Tamas






On 14/07/06, Bob J. <rocketbob@gmail.com> wrote:
>
> Sorry guys I fat fingered my previous message.  Here's the rest of it.
>
> Guys, I have a value in W that's signed.  Adding this value to a set of
> two
> variables is easy enough:
> addwf     blahL,f
> clrf         wreg
> addwfc   blahH,f
>
> W contains a signed value.  If the value of W is a negative, doing the
> following doesn't work (to subtract it instead of adding it.)
>
>     addwf    blahL,f
>      clrf        wreg
>      bnc       subtract
>      addwfc blahH,f
> subtract
>      subwfb, blahH,F
>
> I see why this is incorrect, but I'm not sure how you test for a 2's
> complement value in order to properly skip the subtraction if the value in
> W
> is not setting the carry or borrow flag when adding the lower byte.  Can
> someone please steer me down the correct path?
>
> Regards,
> Bob
>
>
>
>
>
> On 7/14/06, Bob J. < rocketbob@gmail.com> wrote:
> >
> > Guys, I have a value in W that's signed.  Adding this value to a set of
> > two variables is easy enough:
> >
> > addwf     blah,f
> > clrf         wreg
> > addwfc
> >
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