Incorrect value of the variable ans that stores the LCM of two numbers (8086 Program)

问题

Following is the code I wrote to find LCM of two numbers in EMU8086. When I ran it, I am getting value 0 in the Ans variable.

.MODEL SMALL 
.DATA 
Num1 DW 250 
Num2 DW 100
Ans DW ? 
.CODE 
MOV AX,@DATA 
MOV DS, AX 
MOV AX, Num1 
MOV BX, Num2 
MOV DX, 0000h 
NEXT: PUSH AX 
PUSH DX 
DIV BX 
CMP DX, 0000h 
JZ LAST 
POP DX 
POP AX 
ADD AX, Num1 
JNC NEXT 
INC DX 
JMP NEXT 
LAST: POP Ans+2 
POP Ans 
MOV AH, 4Ch 
INT 21h 
END

回答1:

LCM(a, b) = a * b / GCD(a, b)

Due to this equation, you can find GCD using Euclid's algorithm and then calculate LCM. Assuming numbers a and b are in al and dl, this code calculate LCM.

; Save multiplication value
MOV AL, DL ; This 2 lines is AL * DH
MUL DH
PUSH AX ; Store result in stack

FINDBCD: ; General idea is : LCM(a, b) = a*b/BCD(a,b)
    ; We calculate BCD with euclidean algorithm
    CMP DH, DL
    JNS CALCULATE ; Jump if DL < DH else swap them
    MOV CL, DL ; This 3 line swap DL and DH
    MOV DL, DH
    MOV DH, CL
    CALCULATE:
    MOV AL, DH ; Move greater number in AL
    XOR AH, AH ; Zero AX 8 second bits
    DIV DL ; This is AL/DL
    CMP AH, 0 ; If remainder is zero so we found BCD
    JE AFTERFINDBCD
    SUB DH, DL ; Else substract smaller number from greater number
    JMP FINDBCD ; Do this until find BCD

AFTERFINDBCD:
    CMP DH, DL 
    JNS FINDLCM ; Jump if DL < DH
    MOV CL, DL ; This 3 line swap DL and DH
    MOV DL, DH
    MOV DH, CL

FINDLCM:
    POP AX ; Retreive multiplication result
    DIV DL ; This is AX/DL
    AND AX, 0000000011111111B ; Ignore remainder

来源：https://stackoverflow.com/questions/64729112/incorrect-value-of-the-variable-ans-that-stores-the-lcm-of-two-numbers-8086-pro

标签

x86-16

emu8086

lcm