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我正在做一个代码，用汇编语言（英特尔8086）添加2个4位数的数字。在代码中的某一点，我想通过BX寄存器（使用div BX）来分割AX，例如AX = 2AB3（十进制为10931）和BX = 2710（十进制为10000）。当我用汇编语言划分时出现溢出
正常情况下，结果我应该有AX = 1（商）和DX = 3A3（余数），问题是模拟器显示溢出消息。
下面是代码：
DATA SEGMENT
MSG1 DB 0DH,0AH, "first number : $" ,0DH
MSG2 DB 0DH,0AH, "second number : $" ,0DH
RST DB 0DH,0AH, "result : $" ,0DH
DATA ENDS
PILE SEGMENT PARA STACK
DB 128 DUP (?)
PILE ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA,SS:PILE
DEBUT:
mov ax,data
mov ds,ax
mov dx,offset MSG1 ; first msg
mov ah,9
int 21h
mov ah,1 ; multipling first number by 1000 and store it
int 21h
sub al,30h
mov ah,0
mov bx,1000
mul bx
push ax
mov ah,1 ; multipling second number by 100 and store it
int 21h
sub al,30h
mov ah,0
mov bx,100
mul bx
push ax
mov ah,1 ; multipling third number by 10 and store it
int 21h
sub al,30h
mov ah,0
mov bx,10
mul bx
push ax
mov ah,1 ; last number
int 21h
sub al,30h
mov ah,0
push ax
mov dx,offset MSG2 ;same thing for the second 4 digits number
mov ah,9
int 21h
mov ah,1
int 21h
sub al,30h
mov ah,0
mov bx,1000
mul bx
push ax
mov ah,1
int 21h
sub al,30h
mov ah,0
mov bx,100
mul bx
push ax
mov ah,1
int 21h
sub al,30h
mov ah,0
mov bx,10
mul bx
push ax
mov ah,1
int 21h
sub al,30h
mov ah,0
push ax
;-------------------------------
;here I'm doing the addition to all the stored numbers to have the result number
;in the DX register
pop dx
pop bx
add dx,bx
pop bx
add dx,bx
pop bx
add dx,bx
pop bx
add dx,bx
pop bx
add dx,bx
pop bx
add dx,bx
pop bx
add dx,bx
push dx ; I've stored the result number because I'm going to use the DX register to show the
; third message
mov dx,offset RST
mov ah,9
int 21h
pop dx ;restore the result number
mov ax,dx
mov bx,10000
div bx ;I'm dividing the result number by 10000 to have the first number of the result number
;in AX register (quotient) , the error message show up here
push dx
push ax
pop dx
add dl,30h
mov ah,2
int 21h
pop ax
mov bx,1000 ;deviding by 1000 to have the second number
div bx ;same error
push dx
push ax
pop dx
add dl,30h
mov ah,2
int 21h
pop ax
mov bx,100
div bx
push dx
push ax
pop dx
add dl,30h
mov ah,2
int 21h
pop ax
mov bx,10
div bx
push dx
push ax
pop dx
add dl,30h
mov ah,2
int 21h
pop dx
add dl,30h
mov ah,2
int 21h
愿我们看到你的代码'回覆工作或其相关部分？ – halfer
您是否咨询了指令集参考？你有没有看到股息是在'DX：AX'？您是否正确设置了“DX”？你为什么没有显示你的代码？你看过SO上的gazillion重复吗？ – Jester
（你的英语对我来说似乎很好，我们更喜欢这里的问题不要包含[关于语言的道歉]（https://stackoverflow.com/search?tab=newest&q=sorry%20for%20my%20English） - 这只是另一件事编辑出来）。 – halfer