以下是16位除16位的程序.大家来谈谈它是怎么样实现的.
org 0000h
goto main
;*******************************************************************;
; Division : ACCb(16 bits) / ACCa(16 bits) -> ACCb(16 bits) with
; Remainder in ACCc (16 bits)
; (a) Load the Denominator in location ACCaHI & ACCaLO ( 16 bits )
; (b) Load the Numerator in location ACCbHI & ACCbLO ( 16 bits )
; (c) CALL D_div
; (d) The 16 bit result is in location ACCbHI & ACCbLO
; (e) The 16 bit Remainder is in locations ACCcHI & ACCcLO
;
; NOTE :
; The performance specs are for Unsigned arithmetic ;
;*******************************************************************
D_divF
setup
movlw D'16' ; for 16 shifts
movwf temp
movf ACCbHI,w ; ACCb = ACCd
movwf ACCdHI
movf ACCbLO,w
movwf ACCdLO
clrf ACCbHI
clrf ACCbLO
clrf ACCcHI
clrf ACCcLO
divv bcf STATUS,C
rlf ACCdLO
rlf ACCdHI
rlf ACCcLO
rlf ACCcHI
movf ACCaHI,w
subwf ACCcHI,w ;check if a>c
btfss STATUS,Z
goto NOCHK
movf ACCaLO,w
subwf ACCcLO,w ;if msb equal then check lsb
NOCHK
btfss STATUS,C ;carry set if c>a
goto NOGO
movf ACCaLO,w ;c-a into c
subwf ACCcLO
btfss STATUS,C
decf ACCcHI
movf ACCaHI,w
subwf ACCcHI
bsf STATUS,C ;shift a 1 into b (result)
NOGO
rlf ACCbLO
rlf ACCbHI
decfsz temp,1
goto divv
return
;
;*******************************************************************
; 测试程序(注意用法,ACCb/ACCa,结果放在ACCb里,余数放在ACCc里)
;*******************************************************************
; Load constant values to ACCa & ACCb for testing
;
main
movlw 04
movwf ACCaHI
movlw 0e2 ; loads ACCa = 01FF
movwf ACCaLO
;
movlw 0f4
movwf ACCbHI
movlw 024 ; loads ACCb = 7FFF
movwf ACCbLO
call D_divF ; remainder in ACCc. Here ACCb =0040 & ACCc=003F
;
self goto self ;如不是测试程序,本句无用
END
;****************************************************************
