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我正在做一個代碼，用匯編語言（英特爾8086）添加2個4位數的數字。在代碼中的某一點，我想通過BX寄存器（使用div BX）來分割AX，例如AX = 2AB3（十進制爲10931）和BX = 2710（十進制爲10000）。當我用匯編語言劃分時出現溢出
正常情況下，結果我應該有AX = 1（商）和DX = 3A3（餘數），問題是模擬器顯示溢出消息。
下面是代碼：
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
願我們看到你的代碼'回覆工作或其相關部分？ – halfer
您是否諮詢了指令集參考？你有沒有看到股息是在'DX：AX'？您是否正確設置了「DX」？你爲什麼沒有顯示你的代碼？你看過SO上的gazillion重複嗎？ – Jester
（你的英語對我來說似乎很好，我們更喜歡這裏的問題不要包含[關於語言的道歉]（https://stackoverflow.com/search?tab=newest&q=sorry%20for%20my%20English） - 這只是另一件事編輯出來）。 – halfer