--- Nikolai Golovchenko <golovchenko@MAIL.RU> wrote:

> The topic was about how to divide a 16 bit number by
> 255 and round the
> result. I only now clarified for myself the way to
> do this. Better later
> than never..
>
> There was a suggestion to take higher byte and add
> one if MSB of lower byte
> is set. This method is not correct for every number.
> For example 383/255
> (~=2) will give 1.
> 1/255 In binary form  is 0.0000 0001 0000 0001 0000
> 0001....
> This means that for 16 bit divident and rounded
> result two values should be
> added:
> b15b14b13b12b11b10b9b8.b7b6b5b4b3b2b1b0 and
> 0.b15b14b13b12b11b10b9b8.
> You can see that the division and rounding result is
> b15b14b13b12b11b10b9b8
> + e, where e=0, 1, or 2 depending on sum of
> fractional parts.
>
> The correct way is:
>     cblock
>     RegH    ;higher byte
>     RegL    ;lower byte
>     endc
> ;*******Divide by 255****************
> ;RegL, RegH - input
> ;RegL, RegH - output
> ;***************************************
>     movf RegH, w
>     addwf RegL, f
>     clrw
>     btfsc _C
>     movlw 1
>     btfsc RegL, 7
>     addlw 1
>     addwf RegH, w
>     movwf RegL

 Are you sure???:

>     clrf RegH
>     rlf RegH, f
> ;***************************************


And the fast way:

        movf    RegH,w
        addwf   RegL,f
        skpnc
         incf   RegH,f
        btfsc   RegL,7
         incf   RegH,f

.lo

=====

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