c++ - アセンブリの復号化/反転? XORは問題ですか？

Question

運が悪かったので、効果を元に戻そうとしたコード関数があります。私が持っている元の機能は次のとおりです。

          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

特定の行ごとではなく、上記のコードの効果のみを元に戻す必要があります。私はさまざまなことを試しました...試行1（これは間違っています）：

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

私の2番目の試みは以下の通りです：

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

これの何が問題なのですか... xor の効果を元に戻すことはできますか?