我在下面有这个计算器代码，它工作正常，但它不会跳到“分裂”..我仍然在学习汇编编程的过程...需要帮助解决这个问题...
.model small
.stack 100h
.data
msg1 db 13,10,13,10, "Enter 1st Number : $"
msg2 db 13,10, "Enter 2nd Number : $"
msgEr db 13,10, "Error $"
msgCh db 13,10, "Press A to ADD , S to SUBTRACT ,D to MULTIPLY, F to DIVIDE, X to EXIT : $ "
msgSum db 13,10,13,10, "Sum is : $"
msgDif db 13,10,13,10, "Difference is : $"
msgDiv db 13,10,13,10, "Quotient is : $"
msgMul db 13,10,13,10, "Product is : $"
tmp db ?
.code
start:
mov ax, @data
mov ds, ax
lea dx, msg1
mov ah, 09h
int 21h
mov bx, 0
start1:
mov ah, 01h
int 21h
cmp al,0dh
je next1
mov ah,0
sub al,30h
push ax
mov ax,10d
mul bx
pop bx
add bx,ax
jmp start1
next1:
push bx
lea dx,msg2
mov ah,09h
int 21h
mov bx,0
start2:
mov ah,01h
int 21h
cmp al,0dh
je choice
mov ah,0
sub al,30h
push ax
mov ax,10d
mul bx
pop bx
add bx,ax
jmp start2
choice:
lea dx, msgCh
mov ah, 09h
int 21h
mov ah, 01h
mov answer, al
int 21h
cmp al,'f'
je dividing
cmp al,'a'
je adding
cmp al,'s'
je subtracting
cmp al,'d'
je multiplying
cmp al,'x'
mov ah, 4ch
int 21h
error:
lea dx,msgEr
mov ah,09h
int 21h
jmp start
dividing:
pop ax
div bx
push ax
lea dx,msgDiv
mov ah,09h
int 21h
pop ax
mov cx,0
mov dx,0
mov bx,10d
jmp break
adding:
pop ax
add ax,bx
push ax
lea dx,msgSum
mov ah,09h
int 21h
pop ax
mov cx,0
mov dx,0
mov bx,10d
jmp break
multiplying:
pop ax
mul bx
push ax
lea dx,msgMul
mov ah,09h
int 21h
pop ax
mov cx,0
mov dx,0
mov bx,10d
jmp break
subtracting:
pop ax
sub ax,bx
push ax
lea dx,msgDif
mov ah,09h
int 21h
pop ax
mov cx,0
mov dx,0
mov bx,10d
break:
div bx
push dx
mov dx,0
inc cx
or ax,ax
jne break
ans:
pop dx
add dl,30h
mov ah,02h
int 21h
loop ans
jmp start
end start
1 回答
在执行操作之前，您必须清除DX . x86中的部门将在
DX:AX上运行 . 如果您没有先清除它并且它有一些内容，您可能最终会出现溢出情况 . 显然，这很糟糕 . :)所以：
应成为：