Regulus Berdin wrote:
>
> Considering 2digit bcd is in one byte, here is my solution (may have some
> bugs):
>
>         swapf   BCD,w
>         andlw   0x0f                    ;w = bcdh
>         movwf   RESULT                  ;result = bcdh
>         addwf   RESULT,w                ;w = bcdh*2
>         addwf   RESULT,f                ;result = bcdh + w
>         addwf   RESULT,f                ;result = (bcdh + w) + w = bcdh*5
>         rlf     RESULT,f                ;result = result * 2 = bcdh*10
>                                         ;no need to clear carry flag due
>                                         ;  previous adds which cannot have
 overflow
>         movf    BCD,w
>         andlw   0x0f                    ;w = bcdl
>         addwf   RESULT,f                ;result = bcdh*10 + bcdl

Good Reggie, but how about this


    rrf   bcd,W
    andlw 01111000b  ;W = tens*8
    movwf temp
    clrc
    rrf   temp,f     ;W = tens*4
    rrf   temp,f     ;W = tens*2
    subwf bcd,W      ;W = tens*16 + ones - tens*8
                     ;W = tens*8 + ones
    addwf temp,W     ;W = tens*10 + ones


Or if you have the BCD digits in the lower nibbles of the
two bytes bcdl and bcdh (per Keitz's naming convention) then
consider something Andy posted a long time ago:

    clrc
    rlf   bcdh,W     ;W = tens*2
    swapf bcdh,F     ;tens = tens*16
    rrf   bcdh,F     ;tens = tens*8
    addwf bcdh,W     ;W = tens *10

    addwf bcdl,W     ;W = tens*10 + ones

or if you wish to preserve the bcd digits:

    swapf bcdh,W     ;W = tens*16
    movwf temp
    clrc
    rrf   temp,F     ;temp = tens*8
    rlf   bcdh,W     ;W = tens*2
    addwf bcdl,W     ;W = tens*2 + ones
    addwf temp,W     ;W = tens*10 + ones

None of these routines have been tested. But all of my free
code is either untested or uncommented. :)

Scott