SX Microcontroller Math Method

Divide 24 bits by 16

Andrew David (akdavid at ultronics.co.uk), Software Manager, Ultronics Ltd, Cheltenham says:

This division routine uses the standard binary long-division algorithm. The loop iterates once for each bit in the numerator, so it goes round 24 times. For each loop the numerator (sorry, but I can never remember which is the dividend and the divisor - I guess the divisor is the denominator, but I'd hate to get it wrong) is left shifted 1 bit into the remainder, then the remainder is compared with the denominator. If the remainder is greater than the denominator, the denominator is subtracted from the remainder and a 1 is shifted into the quotient, otherwise a 0 is shifted into the quotient. If you can't see how the routine works, try running through an example on paper, in binary, then think about how you'd write the routine.

;=============================================================================
; DIV24_16u
;
;       Divides a 24bit number by a 16bit number. Unsigned.
;
;       Inputs:
;		24-but numerator in ACCcLO:ACCdHI:ACCdLO
;		16-bit denominator in ACCbHI:ACCbLO
;
;       Outputs:
;		24-bit quotient in ACCcLO:ACCdHI:ACCdLO
;		16-bit rem in ACCaHI:ACCaLO
;
;	Locals used:
;		R5Hi
;
;
; Inputs are not preserved.
;
; No timing analysis performed.
; Andrew David, Software Manager, Ultronics Ltd, Cheltenham
; akdavid at ultronics.co.uk          http://www.ultronics.com
;
;=============================================================================

DIV24_16u:
	mov	W, #24	; for 24 shifts
	mov	R5Hi, W	;

	clr	ACCaHI	; clear remainder.
	clr	ACCaLO	;

d2416lp:rlcf    ACCdLO,f        ; build up remainder.
        rlcf    ACCdHI,f        ;
        rlcf    ACCcLO,f        ;
        rlcf    ACCaLO,f        ;
        rlcf    ACCaHI,f        ;

; remainder is 16-bit, but may have spilled over into carry.

	sb	ALUSTA.C	; check for remainder spill into carry.
	jmp	d2416s	;

        movfp   ACCbLO,WREG     ;
        subwfb  ACCaLO,f        ; Carry bit is the 17th bit of this
subtract!
        movfp   ACCbHI,WREG     ;
        subwfb  ACCaHI,f        ;
	setb	ALUSTA.C	; bit is known to be zero here.
	jmp	d2416ns	;

d2416s: movfp   ACCbLO,WREG     ; Compare remainder with divisor.
	mov	W, ACCaLO-w	;
        movfp   ACCbHI,WREG     ;
        subwfb  ACCaHI,w        ;
	sb	ALUSTA.C	;
	jmp	d2416ns		; (remainder < divisor), shift in a '0'
        movfp   ACCbLO,WREG     ; The remainder is larger, so subtract
the
	sub	ACCaLO, W	; ... divisor FROM the remainder and
        movfp   ACCbHI,WREG     ; ... shift a '1' into the quot.
        subwfb  ACCaHI,f        ;

d2416ns:decfsz  R5Hi,f          ; check all bits
	jmp	d2416lp	;
        rlcf    ACCdLO,f        ; shift in final quotient bit.
        rlcf    ACCdHI,f        ;
        rlcf    ACCcLO,f        ;

	ret