这组VB6加密功能可以被破解吗?

时间:2012-05-24 02:59:35

标签: encryption vb6 cryptography

 Public Function EncryptString(theString As String, TheKey As String) As String
    Dim X As Long
    Dim eKey As Byte, eChr As Byte, oChr As Byte, tmp$
    For i = 1 To Len(TheKey)
         'generate a key
          eKey = Asc(Mid$(TheKey, i, 1)) Xor eKey
    Next

    'reset random function
    Rnd -1
    'initilize our key as the random seed
    Randomize eKey
    'generate a pseudo old char
    oChr = Int(Rnd * 256)
    'start encryption
    For X = 1 To Len(theString)
        pp = pp + 1
        If pp > Len(TheKey) Then pp = 1
        eChr = Asc(Mid$(theString, X, 1)) Xor _
                   Int(Rnd * 256) Xor Asc(Mid$(TheKey, pp, 1)) Xor oChr
        tmp$ = tmp$ & Chr(eChr)
        oChr = eChr
    Next
    EncryptString = AsctoHex(tmp$)    

End Function


Public Function DecryptString(theString As String, TheKey As String) As String

Dim X As Long
Dim eKey As Byte, eChr As Byte, oChr As Byte, tmp$
For i = 1 To Len(TheKey)
     'generate a key
     eKey = Asc(Mid$(TheKey, i, 1)) Xor eKey
Next
'reset random function
Rnd -1
'initilize our key as the random seed
Randomize eKey
'generate a pseudo old char
oChr = Int(Rnd * 256)
'start decryption
tmp$ = HexToAsc(theString)
    DecryptString = ""
    For X = 1 To Len(tmp$)
    pp = pp + 1
    If pp > Len(TheKey) Then pp = 1
    If X > 1 Then oChr = Asc(Mid$(tmp$, X - 1, 1))
    eChr = Asc(Mid$(tmp$, X, 1)) Xor Int(Rnd * 256) Xor _
           Asc(Mid$(TheKey, pp, 1)) Xor oChr
        DecryptString = DecryptString & Chr$(eChr)
Next

End Function


Private Function AsctoHex(ByVal astr As String)

For X = 1 To Len(astr)
hc = Hex$(Asc(Mid$(astr, X, 1)))
nstr = nstr & String(2 - Len(hc), "0") & hc
Next
AsctoHex = nstr

End Function

2 个答案:

答案 0 :(得分:6)

您不应该尝试自己实现这样的加密。很难做到正确,并且很容易意外地构建漏洞。

找到一个已被证明可行且经过大量测试的现有解决方案会更容易,更安全。 This可能是更好的解决方案。

答案 1 :(得分:4)

这有几个缺陷:

如果你想要在与加密的系统不同的系统上解密(或者如果你更新你的系统......),你将遇到麻烦:Randomize eKey以及{{1}的一组调用不保证在重启后返回相同的序列。它肯定不会在不同的系统上返回相同的序列。

您将密码减少为单个8位值(eKey),因此加密的事实密钥长度为8位。

简而言之:无论谁有权访问类似于你的系统(即系统,您的解密实际上会生成明文),只需克隆您的Rnd函数并使用eKey = 0运行它。 255。

忘记它,使用有效的东西。阅读Schneier on homebrew encryption