assembly - 汇编中的程序求解表达式

Question

我在汇编中的简单程序有问题。我正在使用 DOSBox 和 TASM。问题是第 76、78 和 80 行中的操作数类型不匹配。这是在乘法之后。我尝试通过使用不同的可变大小来进行一些更改。

; --------------------------------------------
; Equation=(a+c*b)/d-2*c,
; --------------------------------------------
.model small
.stack 100h
.data
        a       db 0                           
        b       db 0
        c       db 0
        d       db 0
        result1 db ?
        result2 db ?
       
 
       
        message1 db "Equation: (a+c*b)/d-2*c   a=$"
        message2 db "b=$"
        message3 db "c=$"
        message4 db "d=$"
        message5 db "Result=$"
.code
 
start:  mov ax,@data
                mov ds,ax                      

                mov ax, seg message1   ;get a and save to a variable
                mov ds,ax      
                mov dx,offset message1
                mov ah, 9h
                int 21h
                mov ah, 1h
                int 21h
                sub al,30h    ;converting to real number
                mov a,al
               
                mov ax, seg message2 ;get b and save to a variable
                mov ds,ax      
                mov dx,offset message2
                mov ah, 9h
                int 21h
                mov ah, 1h
                int 21h
                sub al,30h    ;converting to real number
                mov b,al
               
               
                mov ax, seg message3    ;get c and save to a variable
                mov ds,ax      
                mov dx,offset message3
                mov ah, 9h
                int 21h
                mov ah, 1h
                int 21h
                sub al,30h    ;converting to real number
                mov c,al
               
            
                mov ax, seg message4   ;get d and save to a variable
                mov ds,ax      
                mov dx,offset message4
                mov ah, 9h
                int 21h
                mov ah, 1h
                int 21h
                sub al,30h   ;converting to real number
                mov d,al
               
               
                mov al,b         ; (a+c*b) ------------------------error
                mul c                          
                add ax,a       ; ------------------------error
       
                push ax     ;save current ax
       
                mov ax,c     ;d-2*c------------------------error
                shl ax,2
                sub d,ax
               
               
                pop bx     ;get previous ax  to bx
               
                div bx     ; div ax:bx
               
                mov result1,al
                mov result2,ah
       
                add result1,30h   ;converting to string
                add result2,30h    ;converting to string
               
                mov al,result1
                mov bl,result2
               
                mov ax, seg message5
                mov ds,ax      
                mov dx,offset message5
                mov ah, 9h
                int 21h
                mov al,result1
                mov bl,result2
                mov dl, al
                mov ah , 2h
                int 21h
                mov dl, bl
                mov ah , 2h
                int 21h
               
                mov ax,4C00h           
                int 21h
 
end             start

Answer 1

你的程序几乎很好，你只有操作数大小的一些问题，这是正常的。所以我拿了你的代码并做了一些小改动，这些改动用箭头 (<========) 注释和指出，它们是:

修复了操作数大小问题。我仍然使用 DB，因为我注意到您将数字捕获为单个字符。

(d-2*c) 的结果存储在 BX 中。这是因为我们需要除以 (a+c*b)/(d-2*c)，并且您在 BX 中弹出 (a+c*b)，因此，当您执行 div bx 时你在做 (d-2*c)/(a+c*b) 。

将商和余数分开显示。

已添加 13,10消息的换行符。

已修复 shl ax,2来自 shl ax,1 .一 shl是 x2，两个 shl是 x2x2。

余数从 dl 获得因为当 div用一个词做除数，余数留在dx .

这是您的代码有一些小的变化(在 EMU8086 上测试):

; --------------------------------------------
; Equation=(a+c*b)/d-2*c,
; --------------------------------------------.model small
.stack 100h
.data
    a   db 0                
    b   db 0
    c   db 0
    d   db 0
    result1 db ?
    result2 db ?



    message1 db 13,10,"Equation: (a+c*b)/d-2*c",13,10,"a=$"
    message2 db 13,10,"b=$"         ;<================= 13,10 IS
    message3 db 13,10,"c=$"         ;<================= LINEBREAK.
    message4 db 13,10,"d=$"         ;<=================
    message5 db 13,10,"Quotient=$"  ;<=================
    message6 db 13,10,"Remainder=$" ;<=================
.code

start:  mov ax,@data
        mov ds,ax           



        mov ax, seg message1   ;get a and save to a variable
        mov ds,ax   
        mov dx,offset message1
        mov ah, 9h
        int 21h
        mov ah, 1h 
        int 21h
        sub al,30h    ;converting to real number
        mov a,al

        mov ax, seg message2 ;get b and save to a variable
        mov ds,ax   
        mov dx,offset message2
        mov ah, 9h
        int 21h
        mov ah, 1h
        int 21h
        sub al,30h    ;converting to real number
        mov b,al


        mov ax, seg message3    ;get c and save to a variable
        mov ds,ax   
        mov dx,offset message3
        mov ah, 9h
        int 21h
        mov ah, 1h 
        int 21h
        sub al,30h    ;converting to real number
        mov c,al


        mov ax, seg message4   ;get d and save to a variable
        mov ds,ax   
        mov dx,offset message4
        mov ah, 9h
        int 21h
        mov ah, 1h 
        int 21h
        sub al,30h   ;converting to real number
        mov d,al


        mov al,b          ; (a+c*b)
        mul c
        mov cl,A    ;<======== MOV A TO CX TO
        mov ch,0    ;<======== ADD IT TO AX.
        add ax,CX   ;<======== C*B + A.

       ;push ax     ;<======== NO LONGER NECESSARY BECAUSE
                    ;<======== IN NEXT BLOCK WE USE BX.

        mov bl,C    ;<======== MOV C TO BL AND CLEAR
        mov bh,0    ;<======== BH. NOW C IS IN BX.
        shl bx,1    ;<======== 2*c. ONE SHIFT IS x2, TWO SHIFTS ARE x2x2.
        sub d,bl          ;d - 2c
        mov bl,d    ;<======== MOV D TO BL AND CLEAR BH. NOW
        mov bh,0    ;<======== D IS IN BX. BX = (D-2C).

       ;pop ax      ;<======== NO LONGER NECESSARY. AX CONTAINS (A+C*B).

        mov dx,0    ;<======== CLEAR DX, BECAUSE DIVISOR IS A WORD.
                    ;<======== WHEN DIVISOR IS A WORD, DIV USES DX:AX. 
        div bx      ;<======== dx:ax / bx == DX:(A+C*B) / (D-2C). THIS
                    ;<======== DIVISION IS UNSIGNED, FOR SIGNED USE IDIV.

        mov result1,al ;<===== QUOTIENT.
        mov result2,dl ;<===== REMAINDER, BECAUSE DIVISOR IS WORD.

        add result1,30h   ;converting to string
        add result2,30h    ;converting to string

        mov al,result1
        mov bl,result2

      ;DISPLAY QUOTIENT  <=============
        mov ax, seg message5
        mov ds,ax   
        mov dx,offset message5
        mov ah, 9h
        int 21h
        mov al,result1
        mov dl, al
        mov ah , 2h
        int 21h       
      ;DISPLAY REMAINDER  <=============
        mov ax, seg message6
        mov ds,ax   
        mov dx,offset message6
        mov ah, 9h
        int 21h
        mov dl, bl
        mov ah , 2h
        int 21h

        mov ax,4C00h        
        int 21h

end     start

接下来是您的“待办事项” list :

将操作数的大小从 DB 更改为 DW，以允许您的程序处理更大的数字。

将 DIV 更改为 IDIV，因为 DIV 未签名，而 IDIV 已签名。 IDIV 会让您处理负面结果。

使用 int=21h ah=0Ah 捕获数字作为字符串(而不是单个字符)。稍后，您将字符串转换为数字。接下来的两个链接将带您到将字符串转换为数字的过程:

Assembly x86 Date to Number - Breaking a string into smaller sections

32 bit Calculator in 8086 Assembly

最后，测试数据:

(a+c*b) / (d-2*c)

a=1
b=2
c=3
d=8

a+c*b = 1+3*2 = 7
d-2*c = 8-2*3 = 2

7 / 2 = quotient 3, remainder 1