c++ - 程序集解密/逆向？ XOR 是一个问题吗？

Question

我有一个代码函数，我试图扭转它的影响，但没有成功。我原来的功能是:

          ror al,1                      // rotates the al part of the eax register (the Ekey) bitwise by 1 bit, as 1 mod 8 = 1 (al = 2D)
      ror al,1                      // moves the rightmost bit from al (the end of the Ekey) and shifts everything along
      ror al,1                      // rotates al bitwise by 1 bit, as 1 mod 8 = 1 (al = 4B)
      ror al,1                      // rotates the end 8 bits of the Ekey bitwise by 1 bit, as 1 mod 8 = 1 (al = A5)
      push ecx                      // preserves the value of the encrypted character by pushing it on the stack, the stack pointer decrements by 4 to allow this
      not eax                       // completes the ones' complement on the Ekey, toggling the bits
      mov edx,eax                   // copies the current value of the Ekey register and places it in edx, for holding
      pop eax                       // restores original register value from stack
      xor eax,edx                   // completes a bitwise exclusive or on the Ekey, with the previous value of the Ekey that was stored in edx
      ror al,1                      // rotates the last 8 bits of the Ekey bitwise by 1 bit, as 1 mod 8 = 1
      ror al,1                      // rotates al bitwise by 1 bit, as 1 mod 8 = 1
      not eax                       // completes the ones' complement on the Ekey value, 'flipping' eax entirely
      add eax,0x20                  // adds the hex value of 20 (32 in base 10) to the current value in the Ekey

我必须只反转上面代码的效果，而不是每一行。我已经尝试过各种事情...尝试 1(这是错误的):

      sub eax, 0x20
      not eax
      rol al, 2
      xor ecx, eax
      push eax
      mov eax, edx
      not eax
      pop ecx
      rol al, 4

我的第二次尝试如下:

      sub eax, 0x20
      not eax
      rol al, 2 
      not eax
      xor ecx, eax

这有什么问题...异或的效果可以逆转吗？

Answer 1

明显的顺序是这样的:

; inputs:
;     edx: ekey
;     eax: "encrypted" word
; 
not eax
rol al, 1
rol al, 1
not edx
xor eax, edx

在我看来，原始代码也不必要地复杂。我想我会写一些更像这样的东西:

not eax
xchg eax, ecx
xor eax, ecx
rol al, 1
rol al, 1
not eax

我认为还可以进一步简化，但我必须仔细考虑才能确定。