c – 生成的加密字符串在PyCrypto和Crypto中的大小不同

我最近使用Python脚本来加密字符串.但无法使用Crypto在C中解密它.我只是比较生成的加密字符串,发现它们不一样.有人可以帮忙吗?

这是我的Python代码:

key  = "0123456789abcdef"
data = "ccccccccccccccccdddddddddddddddd"
iv = "aaaaaaaaaaaaaaaa"
encryptor = AES.new(key, AES.MODE_CBC, iv)
enc = encryptor.encrypt(data)
print enc

这是我的C代码:

std::string key = "0123456789abcdef";
std::string iv  = "aaaaaaaaaaaaaaaa";


std::string plaintext = "ccccccccccccccccdddddddddddddddd";
std::string ciphertext;
std::string decryptedtext;

std::cout << "Plain Text (" << plaintext.size() << " bytes)" << std::endl;
std::cout << plaintext;
std::cout << std::endl << std::endl;

CryptoPP::AES::Encryption aesEncryption((byte *)key.c_str(), CryptoPP::AES::DEFAULT_KEYLENGTH);
CryptoPP::CBC_Mode_ExternalCipher::Encryption cbcEncryption( aesEncryption, (byte *)iv.c_str() );

CryptoPP::StreamTransformationFilter stfEncryptor(cbcEncryption, new CryptoPP::StringSink( ciphertext ) );
stfEncryptor.Put( reinterpret_cast<const unsigned char*>( plaintext.c_str() ), plaintext.length() + 1 );
stfEncryptor.MessageEnd();

std::cout << "Cipher Text (" << ciphertext.size() << " bytes)" << std::endl;

for( int i = 0; i < ciphertext.size(); i++ ) {

    std::cout << "0x" << std::hex << (0xFF & static_cast<byte>(ciphertext[i])) << " ";
}

CryptoPP::AES::Decryption aesDecryption((byte *)key.c_str(), CryptoPP::AES::DEFAULT_KEYLENGTH);
CryptoPP::CBC_Mode_ExternalCipher::Decryption cbcDecryption( aesDecryption, (byte *)iv.c_str() );

CryptoPP::StreamTransformationFilter stfDecryptor(cbcDecryption, new CryptoPP::StringSink( decryptedtext ) );
stfDecryptor.Put( reinterpret_cast<const unsigned char*>( ciphertext.c_str() ), ciphertext.size() );
stfDecryptor.MessageEnd();

std::cout << "Decrypted Text: " << std::endl;
std::cout << decryptedtext;
std::cout << std::endl << std::endl;

解决方法:

默认情况下,Crypto使用PKCS5填充到明文字符串.这会在字符串末尾添加填充,以确保字符串是块大小的倍数,对于AES,它是16字节.如果明文已经是16的倍数,那么Crypto会在明文中添加另外16个字节的填充,然后对整个内容进行加密. PyCrypto不会添加这个额外的填充,而是由用户决定以确保正确的块大小.请注意,解密加密加密密文时,会自动删除额外填充.

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