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:) 我正在努力使用一种汇编语言，主要是我的字符串的打印顺序。这就是我所拥有的！！

.model small
.stack 100h
.DATA


;DISPLAYS

msgquestion db 13,10, "Addition or Subtraction?: $ "  
msgfirst db 13,10, "Enter first number : $"     ;MESSAGE first   
msgsecond db  "Enter second Number : $"         ;MESSAGE second
msgSum db 13,10, "The result is : $"
msgDif db 13,10, "The result is : $"
msgterminator db 13,10, "X to END: $"
.code 

;START OF CODE

start:
    mov ax,@data 
    mov ds, ax

    lea dx, msgfirst  ;Displays first
    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,msgsecond  ;DISPLAYS MESSAGE 'second'
        mov ah,09h
        int 21h

        mov bx,0

start2:
        mov ah,01h
        int 21h
        cmp al,0dh
        je QUESTION
        mov ah,0
        sub al,30h
        push ax
        mov ax,10d
        mul bx
        pop bx
        add bx,ax 
        jmp start2


    QUESTION:

        lea dx, msgquestion
        mov ah, 09h
        int 21h

        mov ah, 01h
        int 21h

        cmp al,'+'  
        je adding

        cmp al,'-'
        je subtracting

        cmp al,'x'
        mov ah, 4ch
        int 21h
        jmp break


TERMINATOR:

        lea dx, msgterminator
        mov ah, 09h
        int 21h

        mov ah, 01h
        int 21h

        cmp al,'+'  
        je adding

        cmp al,'-'
        je subtracting

        cmp al,'x'
        mov ah, 4ch
        int 21h
        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

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 TERMINATOR
    end start

;END OF THE PROGRAM

我想先显示加法和减法问题。还请提出我犯的任何其他错误！我真的很感激伙计们，谢谢一千！<3

4

1 回答 1

0

我从您的问题中了解到，您希望首先显示Addition or subtraction:，然后输入您的数字并执行计算。那么解决方案很简单。一旦您的代码开始，您可以询问用户添加或减去的问题，然后将保存的值（来自al）存储到变量中。稍后您可以简单地使用该保存的值进行比较。

这是新代码

    .model small
.stack 100h
.DATA


;DISPLAYS

msgquestion db 13,10, "Addition or Subtraction?: $ "  
msgfirst db 13,10, "Enter first number : $"     ;MESSAGE first   
msgsecond db  "Enter second Number : $"         ;MESSAGE second
msgSum db 13,10, "The result is : $"
msgDif db 13,10, "The result is : $"
msgterminator db 13,10, "X to END: $"
v1 db ?
.code 

;START OF CODE

main proc
    mov ax,@data 
    mov ds, ax

    ;//CHANGES MADE HERE
    ;//add or sub
    lea dx, msgquestion
    mov ah, 09h
    int 21h

    mov ah, 01h
    int 21h

    mov v1,al
    ;//END CHANGES

    lea dx, msgfirst  ;Displays first
    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,msgsecond  ;DISPLAYS MESSAGE 'second'
        mov ah,09h
        int 21h

        mov bx,0

start2:
        mov ah,01h
        int 21h
        cmp al,0dh
        je QUESTION
        mov ah,0
        sub al,30h
        push ax
        mov ax,10d
        mul bx
        pop bx
        add bx,ax 
        jmp start2


    QUESTION:

        mov al,v1     ;will now move the value stored in v1 earlier into al for comparision

        cmp al,'+'  
        je adding

        cmp al,'-'
        je subtracting

        cmp al,'x'
        mov ah, 4ch
        int 21h
        jmp break


TERMINATOR:

        lea dx, msgterminator
        mov ah, 09h
        int 21h

        mov ah, 01h
        int 21h

        cmp al,'+'  
        je adding

        cmp al,'-'
        je subtracting

        cmp al,'x'
        mov ah, 4ch
        int 21h
        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

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 TERMINATOR
main endp
end main
于 2013-09-12T18:35:55.353 回答