-----Mensagem Original-----
De: Andy David <akdavid@ultronics.co.uk>
Para: <anbarsystem>
Enviada em: Quinta-feira, 28 de Outubro de 1999 09:40
Assunto: FW: fw06@pop.dial.pipex.com r. unknown " 16-BIT DIVISION"


Luis,

I can't post directly to the PIClist, can you forward this to the list
for me, pls?

Best regards,


- Andy.
---------------------------------------------------------
Andrew David, Software Manager, Ultronics Ltd, Cheltenham
akdavid@ultronics.co.uk          http://www.ultronics.com

> -----Original Message-----
> From: Andy David
> Sent: Thursday, October 28, 1999 12:11 PM
> To: 'pic microcontroller discussion list'
> Subject: RE: fw06@pop.dial.pipex.com r. unknown "      16-BIT
> DIVISION"
>
> >I am looking for a routine in ASM that make a division of a
> 16-bit binary
> >number for 5 .
> >For example.
> >015Eh / 05 = 046h
> >
> >So my 16-bit binary number will never be larger than 03DEH,
> with this I will
> >need only 1 byte for the result.
>
> Luis,
>
> instead of dividing by 5, how about multiplying by 0.2 (1/5)?
>
> i.e., 0x015e* 0x00cd >> 0x000a
>
> which will probably execute more quickly (unless someone else like
> Dmitry or Bob Fehrenbach) knows a good algorithm that specifically
> gives x/5) - but beware that 0x00cd >> 0x000a (205/1024) gives an
> approximation to 0x015e/5, but it's pretty close - it gives the
> correct answer for 0 to 0x03de, but can be 1 out for larger numbers
> (I've just used a spreadsheet to test the results).
>
> Now I come to think of it, I'm sure Andy Warren posted a divide-by-5
> algorithm a few years ago (1996, I think) that was quick, it should be
> in the archives at: http://www.iversoft.com/piclist/
>
>
> - Andy.
>
> ---------------------------------------------------------
> Andrew David, Software Manager, Ultronics Ltd, Cheltenham
> akdavid@ultronics.co.uk          http://www.ultronics.com
>
> I don't have a 16 by 8 division handy, but here's a 16 by 8 multiply
> and a right shifter, both for 14-bit core PICs. I've just thrown this
> together and I haven't assembled it, but it shows use of multiplying
> by 1/5.
>
> ;=====================================================================
> =
>
> CBLOCK
> ACCaHI
> ACCaLO
> ACCbHI
> ACCbLO
> ACCcHI
> ACCcLO
> ACCdHI
> ACCdLO
> ENDC
>
TestNumber equ 0x0153 ; Original test
number = 0x0153
DivGain equ 0xcd ; Div by 5 = multiply by
1/5
DivScale equ 0x0a ; ... which is
close to 205/2^10


> movlw HIGH TestNumber ; Test number into ACCaHI:ACCaLO
> movwf ACCaHI ;
> movlw LOW TestNumber ;
> movwf ACCaLO ;
> movlw DivGain ; x/5 is approximated by:
> movwf ACCbHI ; ... x * 205 / 2^10
> call mpy8_16 ;
>
> movlw DivScale ; Right shift result of
> multiplication
> movwf ACCdLO ; ... 10 bits to
> complete calculation.
> call scldn24 ; Final result in ACCcLO:ACCdHI
>
> ;=====================================================================
> =
>
>
> ;=====================================================================
> =
> ; mpy8_16
> ;
> ;       Multiplies an 8 bit number by a 16 bit number.
> ;
> ;       Ins:     8 bit input in ACCbHI
> ;                16 bit input in ACCaHI:ACCaLO
> ;
> ;       Outs:    24 bit result in ACCcHI:ACCcLO:ACCdHI
> ;
> ;      Uses:    ACCdLO
> ;=====================================================================
> =
>
> mpy8_16:movlw   8               ; Temp1h used as a loop counter, set
> to
>         movwf   ACCdLO          ; ... 8 loops.
>         clrf    ACCcHI          ; ACCcHI:ACCcLO:ACCdHI is the result
>         clrf    ACCcLO          ; ... accumulator, this must be set to
>
>         clrf    ACCdHI          ; ... 0 at the start.
>
> m8_16loop:rrf   ACCbHI, F       ;
>         btfss   STATUS,C        ;
>         goto    noadd           ;
>
>         movf    ACCaLO,w        ; Add the multiplier to the partial
> product.
>         addwf   ACCcLO,f        ; ... Note that this addition can be
> made
>         movf    ACCaHI,w        ; ... 1 cycle shorter if the carry out
> of
>         btfsc   STATUS,C        ; ... the high byte is not important.
>         incfsz  ACCcHI,f        ;
>         addwf   ACCcHI,w        ;
>         movwf   ACCcHI          ;
>
> noadd:  bcf     STATUS,C
>         rrf     ACCcHI,f        ; shift partial result right
>         rrf     ACCcLO,f        ;
>         rrf     ACCdHI,f        ;
>
>         decfsz  ACCdLO, F       ;
>         goto    m8_16loop
>
>         return
>
> ;=====================================================================
> ========
> ; scldn24
> ;
> ;      24 bit right shift routine - right shifts ACCcHI:ACCcLO:ACCdHI
> by
> ; number of bits given in ACCdLO.
> ;
> ;=====================================================================
> ========
>
> scldn24:incf    ACCdLO,f        ; add 1 - to cope with sc of 0
> s24strt:decfsz  ACCdLO,f        ;
>         goto    shft24          ;
>
>         return
>
> shft24: bcf     STATUS,C
>         rrf     ACCcHI,f        ; rotate bytes in turn through carry.
>         rrf     ACCcLO,f        ;
>         rrf     ACCdHI,f        ;
>
>         goto    s24strt         ;