如何在C＃中使用三重DES执行ISO 9797-1 MAC？ [英] How can I do an ISO 9797-1 MAC with triple DES in C#?

问题描述

我有一个项目，它为24字节的数据块规定了以下加密规则。

1）加密应该使用9797-1中定义的完全三重DES MAC算法作为MAC
算法3，其中输出变换3没有截断，并且在CBC模式中具有DES，而块C
具有ICV设置为零的加密。最后8个字节的加密数据构成了我们需要的值。

程序说加密完成是错误的。是否还有其他事情需要做匹配上面的规范？

数据是一个24字节的值，加密的输出应该是8字节，我猜根据规格）。我得到整个24字节作为输出：（

我写了以下代码实现所述规范：

  des.KeySize = 128; 
 des.Key = ParseHex（key）; 
 des.Mode = CipherMode.CBC; 
 des.Padding = PaddingMode.None; 
 
 ICryptoTransform ic = des.CreateEncryptor（）; 
 
 CryptoOutput = ic.TransformFinalBlock（CryptoOutput，0，24）; 
  

我也尝试过：

  MACTripleDES des = new MACTripleDES（ParseHex（key））; 
 byte [] CDCryptp = des.ComputeHash（CryptoOutput）; 
  

ISO 9797-1 MAC算法3包括使用第一个DES密钥执行CBC MAC，然后只对最后一个块执行完整的3-DES操作

尝试以下操作：

  byte [] keybytes = ParseHex key; 
 byte [] key1 = new byte [8]; 
 Array.Copy（keybytes，0，key1，0，8）; 
 byte [] key2 = new byte [8 ]; 
 Array.Copy（keybytes，8，key2，0，8）; 
 
 DES des1 = DES.Create（）; 
 des1Key = key1; 
 des1.Mode = CipherMode.CBC; 
 des1.Padding = PaddingMode.None; 
 des1.IV = new byte [8]; 
 
 DES DES2 = DES.Create（）; 
 des2.Key = key2; 
 des2.Mode = CipherMode.CBC; 
 des2.Padding = PaddingMode.None; 
 des2.IV = new byte [8]; 
 
 // MAC算法3 
 byte [] intermediate = des1.CreateEncryptor（）。TransformFinalBlock（data，0，data.Length）; 
 
 //输出转换3 
 byte [] intermediate2 = des2.CreateDecryptor（）。TransformFinalBlock（intermediate，intermediate.Length  -  8，8）; 
 byte [] result = des1.CreateEncryptor（）。TransformFinalBlock（intermediate2，0，8）; 
  

I've got a project which stipulates the following encryption rules for a 24 byte block of data.

1) Cryptography should be done using full triple DES MAC algorithm as defined in 9797-1 as MAC algorithm 3 with output transformation 3 without truncation and with DES in CBC mode as block cipher with ICV set to zeros. Last 8 bytes of encrypted data constitute the value we need.

The program is saying the encryption done is wrong. Are there any other things I need to do to match the above spec?

The data is a 24 byte value and output of the encryption should be 8 bytes, I guess (as per the spec). I am getting the whole 24 bytes as output :(

I wrote the following code to achieve the said specification:

des.KeySize = 128;
des.Key = ParseHex(key);
des.Mode = CipherMode.CBC;
des.Padding = PaddingMode.None;

ICryptoTransform ic = des.CreateEncryptor();

CryptoOutput = ic.TransformFinalBlock(CryptoOutput, 0, 24);

I tried this also:

MACTripleDES des = new MACTripleDES(ParseHex(key));
byte[] CDCryptp = des.ComputeHash(CryptoOutput);

解决方案

ISO 9797-1 MAC Algorithm 3 consists of using the first DES key to perform a CBC MAC and then only for the final block perform a full 3-DES operation.

Try this:

byte[] keybytes = ParseHex(key);
byte[] key1 = new byte[8];
Array.Copy(keybytes, 0, key1, 0, 8);
byte[] key2 = new byte[8];
Array.Copy(keybytes, 8, key2, 0, 8);

DES des1 = DES.Create();
des1.Key = key1;
des1.Mode = CipherMode.CBC;
des1.Padding = PaddingMode.None;
des1.IV = new byte[8];

DES des2 = DES.Create();
des2.Key = key2;
des2.Mode = CipherMode.CBC;
des2.Padding = PaddingMode.None;
des2.IV = new byte[8];

// MAC Algorithm 3
byte[] intermediate = des1.CreateEncryptor().TransformFinalBlock(data, 0, data.Length);

// Output Transformation 3
byte[] intermediate2 = des2.CreateDecryptor().TransformFinalBlock(intermediate, intermediate.Length - 8, 8);
byte[] result = des1.CreateEncryptor().TransformFinalBlock(intermediate2, 0, 8);

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