我花了一些时间(〜2天)来解决同样的问题。将“des.exe”重写到C#中。最后,我得到了libdes资源,并对逻辑进行了逆向设计。
初始化向量都是(8)个零。即新的字节[8]应该做。
但是,诀窍可能是如何将字符串密码转换为8个字节长的密钥。如果您使用简单的DES(未三重DES),这个代码可以为你做的伎俩:
public class LibDesPasswordConvertor
{
public byte[] PasswordToKey(string password)
{
if (password == null)
throw new ArgumentNullException("password");
if (password == "")
throw new ArgumentException("password");
var key = new byte[8];
for (int i = 0; i < password.Length; i++)
{
var c = (int)password[i];
if ((i % 16) < 8)
{
key[i % 8] ^= (byte)(c << 1);
}
else
{
// reverse bits e.g. 11010010 -> 01001011
c = (((c << 4) & 0xf0) | ((c >> 4) & 0x0f));
c = (((c << 2) & 0xcc) | ((c >> 2) & 0x33));
c = (((c << 1) & 0xaa) | ((c >> 1) & 0x55));
key[7 - (i % 8)] ^= (byte)c;
}
}
AddOddParity(key);
var target = new byte[8];
var passwordBuffer =
Encoding.ASCII.GetBytes(password).Concat(new byte[8]).Take(password.Length + (8 - (password.Length % 8)) % 8).ToArray();
var des = DES.Create();
var encryptor = des.CreateEncryptor(key, key);
for (int x = 0; x < passwordBuffer.Length/8; ++x)
{
encryptor.TransformBlock(passwordBuffer, 8 * x, 8, target, 0);
}
AddOddParity(target);
return target;
}
private void AddOddParity(byte[] buffer)
{
for (int i = 0; i < buffer.Length; ++i)
{
buffer[i] = _oddParityTable[buffer[i]];
}
}
private static byte[] _oddParityTable = {
1, 1, 2, 2, 4, 4, 7, 7, 8, 8, 11, 11, 13, 13, 14, 14,
16, 16, 19, 19, 21, 21, 22, 22, 25, 25, 26, 26, 28, 28, 31, 31,
32, 32, 35, 35, 37, 37, 38, 38, 41, 41, 42, 42, 44, 44, 47, 47,
49, 49, 50, 50, 52, 52, 55, 55, 56, 56, 59, 59, 61, 61, 62, 62,
64, 64, 67, 67, 69, 69, 70, 70, 73, 73, 74, 74, 76, 76, 79, 79,
81, 81, 82, 82, 84, 84, 87, 87, 88, 88, 91, 91, 93, 93, 94, 94,
97, 97, 98, 98,100,100,103,103,104,104,107,107,109,109,110,110,
112,112,115,115,117,117,118,118,121,121,122,122,124,124,127,127,
128,128,131,131,133,133,134,134,137,137,138,138,140,140,143,143,
145,145,146,146,148,148,151,151,152,152,155,155,157,157,158,158,
161,161,162,162,164,164,167,167,168,168,171,171,173,173,174,174,
176,176,179,179,181,181,182,182,185,185,186,186,188,188,191,191,
193,193,194,194,196,196,199,199,200,200,203,203,205,205,206,206,
208,208,211,211,213,213,214,214,217,217,218,218,220,220,223,223,
224,224,227,227,229,229,230,230,233,233,234,234,236,236,239,239,
241,241,242,242,244,244,247,247,248,248,251,251,253,253,254,254};
}
(重复使用的libdes一些代码块,虽然我不得不找出DES“校验”的一部分我自己)。
最后一个缺点是,libdes使用非标准的填充机制。它几乎和ISO一样,但最后一个字节不是添加的字节数,而是8 - 这个数字。我将Padding属性设置为None并自己处理填充。
在回答下面的答案时,初始化向量是必需的。我相信第三方使用CBC模式。 – 2009-10-24 13:37:35
好吧,正如我所说,如果他们使用IV,他们必须把它给你,因为你无法猜测它。 – svens 2009-10-24 21:26:06
不知何故,它们不需要它们提供des.exe的命令行工具。你只需输入密钥。这就是为什么我相信这是可以推导出来的。感谢您的意见。 – 2009-10-24 23:24:21