;双字节无符号除法程序 (R5R2R3R4/R6R7)=(R2)R3R4 余数R6R7
;入口: R5,R2,R3,R4,R6,R7
;占用资源: ACC,B,F0
;堆栈需求: 4字节
;出口: (R2),R3,R4,R6,R7,OV
NDIV42 : MOV A,R1
PUSH A
MOV B,#00H
NDV421 : MOV A,R2
CLR C
SUBB A,R7
MOV R1,A
MOV A,R5
SUBB A,R6
JC NDV422
MOV R5,A
MOV A,R1
MOV R2,A
INC B
SJMP NDV421
NDV422 : PUSH B
MOV B,#10H
NDV423 : CLR C
MOV A,R4
RLC A
MOV R4,A
MOV A,R3
RLC A
MOV R3,A
MOV A,R2
RLC A
MOV R2,A
XCH A,R5
RLC A
XCH A,R5
MOV F0,C
CLR C
SUBB A,R7
MOV R1,A
MOV A,R5
SUBB A,R6
JB F0,NCV424
JC NDV425
NCV424 : MOV R5,A
MOV A,R1
MOV R2,A
INC R4
NDV425 : DJNZ B,NDV423
POP A
CLR OV
JNZ NDV426
SETB OV
NDV426 : XCH A,R2
MOV R7,A
MOV A,R5
MOV R6,A
POP A
MOV R1,A
RET
这是一个比较经典的编程思想.我这里程序就不提供了,我只提供一个经典的思路:
移位相减,在作"商"处理