装配程序中的错误很少。对于业余爱好者

Question

我必须制作一个可以作为计算器使用的程序。我写了这个：

push ds
push 0000H


data segment
    imsg1 db 13,10,'enter 1st number:$'
    imsg2 db 13,10,'enter 2nd number:$'
    imsg3 db 13,10,'sum:$'
    imsg4 db 13,10,'diff:$'
    imsg5 db 13,10,'div:$'
    imsg6 db 13,10,'product:$'
    num dw ?
    num2 dw ?
data ends
   MOV ax,data
   MOV ds,ax
      mov ah,09h
      mov dx,offset imsg1
      int 21h
   mov cx,0
   mov bx,10
   mov num,0            ; Using num instead of dx, as we usually do, because we have to store another number too.

   r1:mov ah,01h        ; For storing first number
      int 21h          
      cmp al,13
      je yy
      sub al,48
      mov cl,al
      mov ax,num
      mul bx
      add ax,cx
      mov num,ax
      jmp r1  

   yy:
      ;to enter 2nd number  
      mov ah,09h
      mov dx,offset imsg2
      int 21h

       mov cx,0
       mov bx,10
       mov num2,0

       r2:mov ah,01h
       int 21h
       cmp al,13
       je yy1
       sub al,48
       mov cl,al
       mov ax,num2
       mul bx
       add ax,cx
       mov num2,ax
       jmp r2


    yy1:
    ;TO PRINT PROMPT MSG FOR SUM
      mov ah,09h
      mov dx,offset imsg3
      int 21h

      mov ax,num                ; Move the numbers to registers before doing any operation
      mov bx,num2               ; Because we will also need the numbers for other operations
      add ax,bx                 ; Adding and storing in ax
      mov cx,0
      mov bx,10
      mov dx,0
    r3:                                 ; To print the sum
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r3
       mov ah,02h
       print:
        pop dx
        int 21h
       loop print

       ;TO PRINT SUBTRACTION PROMPT
       mov ah,09h
      mov dx,offset imsg4
      int 21h
       mov ax,num
       mov bx,num2
       sub ax,bx                ; Subtracting and storing in ax

       mov cx,0
      mov bx,10
      mov dx,0
    r4:                                 ; Printing the subtracted value
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r4
       mov ah,02h
       print1:
        pop dx
        int 21h
       loop print1

       ;TO PRINT DIVISION PROMPT
       mov ah,09h
       mov dx,offset imsg5
       int 21h
       mov dx,0
       mov ax,num
       mov bx,num2
       div bx                   ; Quotient stored in AX

       mov cx,0
       mov bx,10
       mov dx,0
    r5:                                 ; Printing the Quotient
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r5
       mov ah,02h
       print2:
        pop dx
        int 21h
       loop print2

       ;TO PRINT MULTIPLICATION PROMPT
       mov ah,09h
       mov dx,offset imsg6
       int 21h
       mov dx,0
       mov ax,num
       mov bx,num2
       mul bx                   ; Solution in AX

       mov cx,0
       mov bx,10
       mov dx,0
    r6:                                 ; Printing the solution
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r6
       mov ah,02h
       print3:
        pop dx
        int 21h
       loop print3
ret
main endp
code ends
end main

我有错误：

  

第5行（数据段）解析器：预期指令
  第18行（mov dx，offset imsg1）：逗号或预期行尾   第39行（mov dx，offset imsg2）：逗号或预期行尾   第62行（mov dx，offset imsg3）：逗号或预期行结束
  第87行（mov dx，offset imsg4）：逗号或预期结束的行   第112行（mov dx，offset imsg5）：逗号或预期行尾   第138行（mov dx，offset imsg6）：逗号或预期行尾   第162行（主要endp）：错误：解析器：指令预期
  第163行（主要目的）：错误：解析器：指令预期
  第164行（结束主要）：错误：解析器：指令预期

我试图解决这些问题，因为一周没有成功。对不起，我已经在这里推出了完整的代码，但它不是那么长的片段，所以也许有人可以帮我这个.. 谢谢！

Answer 1

最简单的事情可能是使用预期的汇编程序......但我曾经做过很多M / Tasm翻译。这是没有问题的！...但是没有抱怨的组装...我的偏好是将它组装到.com文件，但原始代码是用于MZ文件（需要链接器）。

; nasm -f bin -o myprog.com myprog.asm
; or
; nasm -f obj myprog.asm
; link ???

; for -f obj, remove this line
org 100h

segment data
    imsg1 db 13,10,'enter 1st number:$'
    imsg2 db 13,10,'enter 2nd number:$'
    imsg3 db 13,10,'sum:$'
    imsg4 db 13,10,'diff:$'
    imsg5 db 13,10,'div:$'
    imsg6 db 13,10,'product:$'

segment .bss
    num resw 1
    num2 resw 1

segment .text
; if -f obj, put these lines back
;   MOV ax,data
;   MOV ds,ax

      mov ah,09h
      mov dx, imsg1
      int 21h
   mov cx,0
   mov bx,10
   mov word [num],0            ; Using num instead     of dx, as we usually do, because we have to store another number too.

   r1:mov ah,01h        ; For storing first number
      int 21h          
      cmp al,13
      je yy
      sub al,48
      mov cl,al
      mov ax,[num]
      mul bx
      add ax,cx
      mov [num],ax
      jmp r1  

   yy:
      ;to enter 2nd number  
      mov ah,09h
      mov dx, imsg2
      int 21h

       mov cx,0
       mov bx,10
       mov word [num2],0

       r2:mov ah,01h
       int 21h
       cmp al,13
       je yy1
       sub al,48
       mov cl,al
       mov ax,[num2]
       mul bx
       add ax,cx
       mov [num2],ax
       jmp r2


    yy1:
    ;TO PRINT PROMPT MSG FOR SUM
      mov ah,09h
      mov dx, imsg3
      int 21h

      mov ax,[num]              ; Move the numbers to registers before doing any operation
      mov bx,[num2]             ; Because we will also need the numbers for other operations
      add ax,bx                 ; Adding and storing in ax
      mov cx,0
      mov bx,10
      mov dx,0
    r3:                                 ; To print the sum
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r3
       mov ah,02h
       print:
        pop dx
        int 21h
       loop print

       ;TO PRINT SUBTRACTION PROMPT
       mov ah,09h
      mov dx, imsg4
      int 21h
       mov ax,[num]
       mov bx,[num2]
       sub ax,bx                ; Subtracting and storing in ax

       mov cx,0
      mov bx,10
      mov dx,0
    r4:                                 ; Printing the subtracted value
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r4
       mov ah,02h
       print1:
        pop dx
        int 21h
       loop print1

       ;TO PRINT DIVISION PROMPT
       mov ah,09h
       mov dx, imsg5
       int 21h
       mov dx,0
       mov ax,[num]
       mov bx,[num2]
       div bx                   ; Quotient stored in AX

       mov cx,0
       mov bx,10
       mov dx,0
    r5:                                 ; Printing the Quotient
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r5
       mov ah,02h
       print2:
        pop dx
        int 21h
       loop print2

       ;TO PRINT MULTIPLICATION PROMPT
       mov ah,09h
       mov dx, imsg6
       int 21h
       mov dx,0
       mov ax,[num]
       mov bx,[num2]
       mul bx                   ; Solution in AX

       mov cx,0
       mov bx,10
       mov dx,0
    r6:                                 ; Printing the solution
       mov dx,0
       div bx
       add dx,48
       push dx
       inc cx
       cmp ax,0
       jg r6
       mov ah,02h
       print3:
        pop dx
        int 21h
       loop print3
ret

; nasm -f bin -o myprog.com myprog.asm ; or ; nasm -f obj myprog.asm ; link ??? ; for -f obj, remove this line org 100h segment data imsg1 db 13,10,'enter 1st number:$' imsg2 db 13,10,'enter 2nd number:$' imsg3 db 13,10,'sum:$' imsg4 db 13,10,'diff:$' imsg5 db 13,10,'div:$' imsg6 db 13,10,'product:$' segment .bss num resw 1 num2 resw 1 segment .text ; if -f obj, put these lines back ; MOV ax,data ; MOV ds,ax mov ah,09h mov dx, imsg1 int 21h mov cx,0 mov bx,10 mov word [num],0 ; Using num instead of dx, as we usually do, because we have to store another number too. r1:mov ah,01h ; For storing first number int 21h cmp al,13 je yy sub al,48 mov cl,al mov ax,[num] mul bx add ax,cx mov [num],ax jmp r1 yy: ;to enter 2nd number mov ah,09h mov dx, imsg2 int 21h mov cx,0 mov bx,10 mov word [num2],0 r2:mov ah,01h int 21h cmp al,13 je yy1 sub al,48 mov cl,al mov ax,[num2] mul bx add ax,cx mov [num2],ax jmp r2 yy1: ;TO PRINT PROMPT MSG FOR SUM mov ah,09h mov dx, imsg3 int 21h mov ax,[num] ; Move the numbers to registers before doing any operation mov bx,[num2] ; Because we will also need the numbers for other operations add ax,bx ; Adding and storing in ax mov cx,0 mov bx,10 mov dx,0 r3: ; To print the sum mov dx,0 div bx add dx,48 push dx inc cx cmp ax,0 jg r3 mov ah,02h print: pop dx int 21h loop print ;TO PRINT SUBTRACTION PROMPT mov ah,09h mov dx, imsg4 int 21h mov ax,[num] mov bx,[num2] sub ax,bx ; Subtracting and storing in ax mov cx,0 mov bx,10 mov dx,0 r4: ; Printing the subtracted value mov dx,0 div bx add dx,48 push dx inc cx cmp ax,0 jg r4 mov ah,02h print1: pop dx int 21h loop print1 ;TO PRINT DIVISION PROMPT mov ah,09h mov dx, imsg5 int 21h mov dx,0 mov ax,[num] mov bx,[num2] div bx ; Quotient stored in AX mov cx,0 mov bx,10 mov dx,0 r5: ; Printing the Quotient mov dx,0 div bx add dx,48 push dx inc cx cmp ax,0 jg r5 mov ah,02h print2: pop dx int 21h loop print2 ;TO PRINT MULTIPLICATION PROMPT mov ah,09h mov dx, imsg6 int 21h mov dx,0 mov ax,[num] mov bx,[num2] mul bx ; Solution in AX mov cx,0 mov bx,10 mov dx,0 r6: ; Printing the solution mov dx,0 div bx add dx,48 push dx inc cx cmp ax,0 jg r6 mov ah,02h print3: pop dx int 21h loop print3 ret

Answer 2

在评论部分，您提到要使用NASM汇编代码，但代码是使用TASM / MASM语法编写的。这两种风格之间存在一些重要的差异：

在TASM / MASM语法中：

mov ax,num

将使用 ax中存储的值加载num。在NASM语法中，它将使用地址 ax加载num。要在NASM中获取TASM行为，您需要在地址周围加上括号：

mov ax,[num]

---

在TASM / MASM语法中：

mov dx,offset imsg1

使用dx的地址加载imsg1（实际上是段内的偏移量）。在NASM语法中，它将简单地变为：

mov dx,imsg1

---

某些指令（如ENDP）特定于TASM / MASM，并且在NASM中没有真正的对应物。