Question

对于我的任务，我需要采取如下所示的程序。

INCLUDE Irvine32.inc

.data
aName BYTE "Abraham Lincoln",0
nameSize = ($ - aName) - 1

.code
main PROC

; Push the name on the stack.
    mov ecx,nameSize
    mov esi,0

L1: movzx eax,aName[esi]    ; get character
    push eax    ; push on stack
    inc esi
    Loop L1

; Pop the name from the stack, in reverse,
; and store in the aName array.
    mov ecx,nameSize
    mov esi,0

L2: pop eax ; get character
    mov aName[esi],al   ; store in string
    inc esi
    Loop L2

; Display the name.
    mov edx,OFFSET aName
    call Writestring
    call Crlf

    exit
main ENDP
END main

并修改程序，以便从用户输入32位整数列表，然后以相反的顺序显示整数。构建失败，说明

上的指令操作数必须相同

mov aName[esi],al ; store in string

这是完整的计划。我没有做什么？

INCLUDE Irvine32.inc
.data
   aName WORD 5 DUP (?)
   nameSize = 5
.code

main PROC

   mov  edx, OFFSET aName
   mov  ecx, 4          ;buffer size - 1


   ; Push the name on the stack.
   mov ecx,nameSize
   mov esi,0
   L1: 
     Call ReadInt
     push eax ; push on stack
     inc esi
   Loop L1
   ; Pop the name from the stack, in reverse,
   ; and store in the aName array.
   mov ecx,nameSize
   mov esi,0
   L2: pop eax ; get character
     mov aName[esi],al ; store in string
     inc esi
   Loop L2
   ; Display the name.
   mov edx,OFFSET aName
   call Writestring
   call Crlf
   exit
   main ENDP
END main

Answer 1

此声明设置为16位值，您的赋值大约为32位整数！
```
aName WORD 5 DUP (?)
```
在汇编程序中，大多数时候操作数必须与大小有关。 AL显然是一个字节，而你将aName定义为word类型！两者都需要成为dword。
```
mov aName[esi],al ; store in string
```
示例程序使用Writestring显示以零结尾的文本。为什么你还希望它还显示一个非零终止的缓冲区，用数字代替字符？
```
mov edx,OFFSET aName
call Writestring
```

您能否看到此代码不需要同时使用ESI？

; Push the name on the stack.
mov ecx,nameSize
mov esi,0
L1: 
 Call ReadInt
 push eax ; push on stack
 inc esi
Loop L1

修改程序，使其从用户输入32位整数列表，然后以相反的顺序显示整数

1 个答案: