Assembly 翻转EAX寄存器的内容

我正在尝试创建一个程序（使用汇编），允许您输入一个字，然后加密，然后解密

我现在掌握的代码与我所追求的正好相反

目前EAX寄存器为00000037

我希望它是00000073

如何做到这一点

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提前感谢。

以下是翻转EAX内容的代码，由GUI Turbo Assembler x64（）制作，它的作用如下：

为EAX=1234567890分配一个大数字

显示EAX（将EAX转换为字符串并显示消息）

Flip EAX=“0987654321”（逐个提取数字并以相反顺序存储在字符串中）

用消息显示此字符串（EAX翻转）

将字符串转换回EAX=987654321中的数字（这是EAX反转）

显示新的EAX（将EAX转换为字符串并显示消息）

这是：

.model small

.586

.stack 100h

.data

msj1   db 13,10,'Original EAX = $'
msj2   db 13,10,'Flipped  EAX = $'
msj3   db 13,10,'New      EAX = $'

buf    db 11         ;MAX NUMBER OF CHARACTERS ALLOWED (4).
       db ?          ;NUMBER OF CHARACTERS ENTERED BY USER.
       db 11 dup (?) ;CHARACTERS ENTERED BY USER. 

.code          
start:
;INITIALIZE DATA SEGMENT.
  mov  ax, @data
  mov  ds, ax

;CONVERT EAX TO STRING TO DISPLAY IT.
  call dollars  ;NECESSARY TO DISPLAY.
  mov  eax, 1234567890
  call number2string  ;PARAMETER:AX. RETURN:VARIABLE BUF.

;DISPLAY 'ORIGINAL EAX'.
  mov  ah, 9
  mov  dx, offset msj1
  int  21h  

;DISPLAY BUF (EAX CONVERTED TO STRING).
  mov  ah, 9
  mov  dx, offset buf
  int  21h  

;FLIP EAX.
  call dollars  ;NECESSARY TO DISPLAY.
  mov  eax, 1234567890
  call flip_eax  ;PARAMETER:AX. RETURN:VARIABLE BUF.

;DISPLAY 'FLIPPED EAX'.
  mov  ah, 9
  mov  dx, offset msj2
  int  21h  

;DISPLAY BUF (EAX FLIPPED CONVERTED TO STRING).
  mov  ah, 9
  mov  dx, offset buf
  int  21h  

;CONVERT STRING TO NUMBER (FLIPPED EAX TO EAX).
  mov  si, offset buf  ;STRING TO REVERSE.
  call string2number   ;RETURN IN EBX.
  mov  eax, ebx        ;THIS IS THE NEW EAX FLIPPED.

;CONVERT EAX TO STRING TO DISPLAY IT.
  call dollars  ;NECESSARY TO DISPLAY.
  call number2string  ;PARAMETER:EAX. RETURN:VARIABLE BUF.

;DISPLAY 'NEW EAX'.
  mov  ah, 9
  mov  dx, offset msj3
  int  21h  

;DISPLAY BUF (EAX CONVERTED TO STRING).
  mov  ah, 9
  mov  dx, offset buf
  int  21h  

;WAIT UNTIL USER PRESS ANY KEY.
  mov  ah, 7
  int  21h

;FINISH PROGRAM.
  mov  ax, 4c00h
  int  21h           

;------------------------------------------

flip_eax proc
  mov  si, offset buf  ;DIGITS WILL BE STORED IN BUF.
  mov  bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
  mov  cx, 0  ;COUNTER FOR EXTRACTED DIGITS.
extracting:       
;EXTRACT ONE DIGIT.
  mov  edx, 0 ;NECESSARY TO DIVIDE BY EBX.
  div  ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
;INSERT DIGIT IN STRING.
  add  dl, 48 ;CONVERT DIGIT TO CHARACTER.
  mov  [ si ], dl
  inc  si
;NEXT DIGIT.
  cmp  eax, 0     ;IF NUMBER IS
  jne  extracting ;NOT ZERO, REPEAT.

  ret
flip_eax endp  

;------------------------------------------
;CONVERT STRING TO NUMBER IN EBX.
;SI MUST ENTER POINTING TO THE STRING.

string2number proc

;COUNT DIGITS IN STRING.
  mov  cx, 0
find_dollar:                                          
  inc  cx  ;DIGIT COUNTER.
  inc  si  ;NEXT CHARACTER.
  mov  bl, [ si ]
  cmp  bl, '$'
  jne  find_dollar  ;IF BL != '$' JUMP.
  dec  si  ;BECAUSE IT WAS OVER '$', NOT OVER THE LAST DIGIT.

;CONVERT STRING.
  mov  ebx, 0
  mov  ebp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:         
;CONVERT CHARACTER.                    
  mov  eax, 0 ;NOW EAX==AL.
  mov  al, [ si ] ;CHARACTER TO PROCESS.
  sub  al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
  mul  ebp ;EAX*EBP = EDX:EAX.
  add  ebx, eax ;ADD RESULT TO BX. 
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
  mov  eax, ebp
  mov  ebp, 10
  mul  ebp ;AX*10 = EDX:EAX.
  mov  ebp, eax ;NEW MULTIPLE OF 10.  
;CHECK IF WE HAVE FINISHED.
  dec  si ;NEXT DIGIT TO PROCESS.
  loop repeat ;CX-1, IF NOT ZERO, REPEAT.

  ret 
string2number endp

;------------------------------------------
;FILLS VARIABLE STR WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.

dollars proc
  mov  si, offset buf
  mov  cx, 11
six_dollars:      
  mov  bl, '$'
  mov  [ si ], bl
  inc  si
  loop six_dollars

  ret
dollars endp  

;------------------------------------------
;NUMBER TO CONVERT MUST ENTER IN EAX.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (BUF).

number2string proc
  mov  ebx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
  mov  cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:       
  mov  edx, 0 ;NECESSARY TO DIVIDE BY EBX.
  div  ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
  push dx ;PRESERVE DIGIT EXTRACTED (DL) FOR LATER.
  inc  cx  ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
  cmp  eax, 0  ;IF NUMBER IS
  jne  cycle1  ;NOT ZERO, LOOP. 
;NOW RETRIEVE PUSHED DIGITS.
  mov  si, offset buf
cycle2:  
  pop  dx        
  add  dl, 48 ;CONVERT DIGIT TO CHARACTER.
  mov  [ si ], dl
  inc  si
  loop cycle2  

  ret
number2string endp  

end start

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不要链接到您的代码-将其放在您的问题中（作为文本）。我对x86指令不太了解，无法回答，但基本上您希望计算

eax>4

。你需要一个二级reigster来完成这项工作。这很简单，

ror al，4

如果你提到你正在处理的汇编程序和编译器类型，你会得到更好的答案：8086、mips、arm、Visual Studio内联等等。