2017-03-17 121 views
1

我的问题是,我无法从下面的C代码中获得AES 256 CTR输出以匹配下面的OpenSSL命令的输出。CTR-AES256加密不匹配OpenSSL -aes-256-ctr

C代码产生这样的:

5f b7 18 d1 28 62 7f 50 35 ba e9 67 a7 17 ab 22 
f9 e4 09 ce 23 26 7b 93 82 02 d3 87 eb 01 26 ac 
96 2c 01 8c c8 af f3 de a4 18 7f 29 46 00 2e 00 

OpenSSL的命令行产生这样的:

5f b7 18 d1 28 62 7f 50 35 ba e9 67 a7 17 ab 22 
3c 01 11 bd 39 14 74 76 31 57 a6 53 f9 00 09 b4 
6f a9 49 bc 6d 00 77 24 2d ef b9 c4 

通知前16个字节是相同的,因为nonceIV是一样的,但是,当nonceIV在下一次迭代中更新,然后与明文XOR,接下来的16个字节不同,等等......?

我不明白为什么会发生这种情况?任何人都知道为什么十六进制代码在第一个16字节块之后有所不同?

声明:我不是C专家。

谢谢!

Fox.txt

The quick brown fox jumped over the lazy dog 

然后运行下面的OpenSSL命令创建foxy.exe

openssl enc -aes-256-ctr -in fox.txt -out foxy.exe -K 603deb1015ca71be2b73aef0857d77811f352c073b6108d72d9810a30914dff4 -iv f0f1f2f3f4f5f6f7f8f9fafbfcfdfeff -nosalt -nopad -p 

这里是foxy.exe包含:

5f b7 18 d1 28 62 7f 50 35 ba e9 67 a7 17 ab 22 
3c 01 11 bd 39 14 74 76 31 57 a6 53 f9 00 09 b4 
6f a9 49 bc 6d 00 77 24 2d ef b9 c4 

他重新的代码。

#include <Windows.h> 

    // What is AES CTR 
    // 
    // AES - CTR (counter) mode is another popular symmetric encryption algorithm. 
    // 
    // It is advantageous because of a few features : 
    // 1. The data size does not have to be multiple of 16 bytes. 
    // 2. The encryption or decryption for all blocks of the data can happen in parallel, allowing faster implementation. 
    // 3. Encryption and decryption use identical implementation. 
    // 
    // Very important note : choice of initial counter is critical to the security of CTR mode. 
    // The requirement is that the same counter and AES key combination can never to used to encrypt more than more one 16 - byte block. 

    // Notes 
    // ----- 
    // * CTR mode does not require padding to block boundaries. 
    // 
    // * The IV size of AES is 16 bytes. 
    // 
    // * CTR mode doesn't need separate encrypt and decrypt method. Encryption key can be set once. 
    // 
    // * AES is a block cipher : it takes as input a 16 byte plaintext block, 
    // a secret key (16, 24 or 32 bytes) and outputs another 16 byte ciphertext block. 
    // 
    // References 
    // ---------- 
    // https://en.wikipedia.org/wiki/Block_cipher_mode_of_operation#Counter_.28CTR.29 
    // https://www.cryptopp.com/wiki/CTR_Mode#Counter_Increment 
    // https://modexp.wordpress.com/2016/03/10/windows-ctr-mode-with-crypto-api/ 
    // https://msdn.microsoft.com/en-us/library/windows/desktop/jj650836(v=vs.85).aspx 
    // http://www.cryptogrium.com/aes-ctr.html 
    // http://www.bierkandt.org/encryption/symmetric_encryption.php 


    #define IV_SIZE 16 
    #define AES_BLOCK_SIZE 16 

    typedef struct _key_hdr_t { 
     PUBLICKEYSTRUC hdr;   // Indicates the type of BLOB and the algorithm that the key uses. 
     DWORD   len;   // The size, in bytes, of the key material. 
     char   key[32];  // The key material. 
    } key_hdr; 


    // NIST specifies two types of counters. 
    // 
    // First is a counter which is made up of a nonce and counter. 
    // The nonce is random, and the remaining bytes are counter bytes (which are incremented). 
    // For example, a 16 byte block cipher might use the high 8 bytes as a nonce, and the low 8 bytes as a counter. 
    // 
    // Second is a counter block, where all bytes are counter bytes and can be incremented as carries are generated. 
    // For example, in a 16 byte block cipher, all 16 bytes are counter bytes. 
    // 
    // This uses the second method, which means the entire byte block is treated as counter bytes. 

    void IncrementCounterByOne(char *inout) 
    { 
     int i; 

     for (i = 16 - 1; i >= 0; i--) { 
      inout[i]++; 
      if (inout[i]) { 
       break; 
      } 
     } 
    } 


    void XOR(char *plaintext, char *ciphertext, int plaintext_len) 
    { 
     int i; 

     for (i = 0; i < plaintext_len; i++) 
     { 
      plaintext[i] ^= ciphertext[i]; 
     } 
    } 


    unsigned int GetAlgorithmIdentifier(unsigned int aeskeylenbits) 
    { 
     switch (aeskeylenbits) 
     { 
     case 128: 
      return CALG_AES_128; 
     case 192: 
      return CALG_AES_192; 
     case 256: 
      return CALG_AES_256; 
     default: 
      return 0; 
     } 
    } 


    unsigned int GetKeyLengthBytes(unsigned int aeskeylenbits) 
    { 
     return aeskeylenbits/8; 
    } 


    void SetKeyData(key_hdr *key, unsigned int aeskeylenbits, char *pKey) 
    { 
     key->hdr.bType = PLAINTEXTKEYBLOB; 
     key->hdr.bVersion = CUR_BLOB_VERSION; 
     key->hdr.reserved = 0; 
     key->hdr.aiKeyAlg = GetAlgorithmIdentifier(aeskeylenbits); 
     key->len = GetKeyLengthBytes(aeskeylenbits); 
     memmove(key->key, pKey, key->len); 
    } 

    // point = pointer to the start of the plaintext, extent is the size (44 bytes) 
    void __stdcall AESCTR(char *point, int extent, char *pKey, char *pIV, unsigned int aeskeylenbits, char *bufOut) 
    { 
     HCRYPTPROV hProv; 
     HCRYPTKEY hSession; 
     key_hdr key; 
     DWORD  IV_len; 
     div_t  aesblocks; 
     char  nonceIV[64]; 
     char  tIV[64]; 
     char  *bufIn; 

     bufIn = point; 

     memmove(nonceIV, pIV, IV_SIZE); 

     SetKeyData(&key, aeskeylenbits, pKey); 

     CryptAcquireContext(&hProv, NULL, NULL, PROV_RSA_AES, CRYPT_VERIFYCONTEXT | CRYPT_SILENT); 

     CryptImportKey(hProv, (PBYTE)&key, sizeof(key), 0, CRYPT_NO_SALT, &hSession); 

     aesblocks = div(extent, AES_BLOCK_SIZE); 

     while (aesblocks.quot != 0) 
     { 
      IV_len = IV_SIZE; 
      memmove(tIV, nonceIV, IV_SIZE); 
      CryptEncrypt(hSession, 0, FALSE, 0, (BYTE *)tIV, &IV_len, sizeof(tIV)); 
      XOR(bufIn, tIV, AES_BLOCK_SIZE); 
      IncrementCounterByOne(nonceIV); 
      bufIn += AES_BLOCK_SIZE; 
      aesblocks.quot--; 
     } 

     if (aesblocks.rem != 0) 
     { 
      memmove(tIV, nonceIV, IV_SIZE); 
      CryptEncrypt(hSession, 0, TRUE, 0, (BYTE *)tIV, &IV_len, sizeof(tIV)); 
      XOR(bufIn, tIV, aesblocks.rem); 
     } 

     memmove(bufOut, point, extent); 

     CryptDestroyKey(hSession); 
     CryptReleaseContext(hProv, 0); 
    } 

我能够通过在M $ CryptEncrypt()建议的伪代码得到这个工作说明部分https://msdn.microsoft.com/en-us/library/windows/desktop/aa379924(v=vs.85).aspx

// Set the IV for the original key. Do not use the original key for 
// encryption or decryption after doing this because the key's 
// feedback register will get modified and you cannot change it. 
CryptSetKeyParam(hOriginalKey, KP_IV, newIV) 

while(block = NextBlock()) 
{ 
    // Create a duplicate of the original key. This causes the 
    // original key's IV to be copied into the duplicate key's 
    // feedback register. 
    hDuplicateKey = CryptDuplicateKey(hOriginalKey) 

    // Encrypt the block with the duplicate key. 
    CryptEncrypt(hDuplicateKey, block) 

    // Destroy the duplicate key. Its feedback register has been 
    // modified by the CryptEncrypt function, so it cannot be used 
    // again. It will be re-duplicated in the next iteration of the 
    // loop. 
    CryptDestroyKey(hDuplicateKey) 
} 

下面是用两条新线的更新程式码:

HCRYPTKEY hDuplicateKey; 
boolean final; 

while (aesblocks.quot != 0) 
{ 
    CryptDuplicateKey(hOriginalKey, NULL, 0, &hDuplicateKey); 
    IV_len = IV_SIZE; 
    memmove(tIV, nonceIV, IV_len); 
    final = (aesblocks.quot == 1 && aesblocks.rem == 0) ? TRUE : FALSE; 
    CryptEncrypt(hDuplicateKey, 0, final, 0, (BYTE *)tIV, &IV_len, sizeof(tIV)); 
    XOR(bufIn, tIV, AES_BLOCK_SIZE); 
    IncrementCounterByOne(nonceIV); 
    bufIn += AES_BLOCK_SIZE; 
    aesblocks.quot--; 
    CryptDestroyKey(hDuplicateKey); 
} 

if (aesblocks.rem != 0) 
{ 
    CryptDuplicateKey(hOriginalKey, NULL, 0, &hDuplicateKey); 
    final = TRUE; 
    memmove(tIV, nonceIV, IV_SIZE); 
    CryptEncrypt(hDuplicateKey, 0, final, 0, (BYTE *)tIV, &IV_len, sizeof(tIV)); 
    XOR(bufIn, tIV, aesblocks.rem); 
    CryptDestroyKey(hDuplicateKey); 
} 
+0

您可能需要[openssl c EVP_BytesToKey](https://www.google.com/search?q=openssl+c+EVP_BytesToKey)。我不记得在WinCrypt(尚未)的上下文中看到关于“EVP_BytesToKey”的问题。 – jww

回答

2

我对Microsoft API不熟悉,但我相信CryptEncrypt()默认使用CBC模式 - 所以第一块加密的输出会自动变为f编入第二个块的输入。您正在构建CTR模式,自己形成草稿(顺便说一句,通常不是一个明智的做法 - 您应该使用加密库的功能,而不是“推出自己的”加密)。要获得预期的输出,您可能需要使CryptEncrypt在ECB模式下使用AES - 我相信可以使用CryptptSetKeyParam(https://msdn.microsoft.com/en-us/library/aa380272.aspx)并将KP_MODE设置为CRYPT_MODE_ECB来完成。

+0

感谢您的提示。再次访问CryptEncrypt()文档,发现M $暗示了这一点: “设置原始密钥的IV。这样做后,请勿使用原始密钥进行加密或解密,因为密钥的 反馈寄存器将被修改,您无法更改它。“ – vengy

0

确保您的输入文件不包含任何额外的字符,如新行等Openssl将包括这些额外的字符,同时加密。