C#RFC2898DeriveBytes正在工作，但Python PBKDF2生成的密钥和IV不能用于Python AES解密

Question

我手头有一个解密AES加密密码文本的问题，该密码文本由以下规范组成：·256字节的RFC2898派生salt，然后是使用密码“password”和派生IV进行AES加密的消息。示例消息是“这是我的秘密字符串，lorem ipsum”，密码是“password”这是用C代码加密的。下面的C代码可以很好地解密这条消息

private static readonly int SALT_SIZE = 256;
public static void Decrytor(){
// Encrypted Message
           var cipherText = "i+EKwmlAF0VYh4GwDd+bGf3+yreYsPJW2Oq/w9FXjsp7RI3VqRiqtnqiAD4n6U0JJSTe2ct4B7lgrG+dHxeGcXYEYIERXvU0xnUdH+z3mRwmgYOqCU9HRUKy/z3GKISTm8qH030KTYm3YMBjnKpU8gaRcoDPP/nCiB3o5fPdyspgJgT/qt5BuvwYq7n0qg6ez/Wi4447gq/qHwG3wuuYLSBUCfmIkgGaO1KXqv3SsR8EAhrmMBmPDJfjc3sydNqs5B8J9/JvZFEZULTb8rLQZKQvgHhH9/53Bzs3zmoq0RFbgSueUbyeWb9rLAzYieTz8Yj0srG4GtwPrTPoItc6/hvx5stZ6pX8tgyk9Y3baT0JFMtGgxve7yduy8idTCQdAwRc5NOo4+CBk7P/sIw6+Q==";
            var key = "password";
            // Extract the salt from our cipherText
            var allTheBytes = Convert.FromBase64String(cipherText);
            var saltBytes = allTheBytes.Take(SALT_SIZE).ToArray();
            var cipherTextBytes = allTheBytes.Skip(SALT_SIZE).Take(allTheBytes.Length - SALT_SIZE).ToArray();

            var keyDerivationFunction = new Rfc2898DeriveBytes(key, saltBytes);
            // Derive the previous IV from the Key and Salt
            var keyBytes = keyDerivationFunction.GetBytes(32);
            var ivBytes = keyDerivationFunction.GetBytes(16);

            // Create a decrytor to perform the stream transform.
            // Create the streams used for decryption.
            // The default Cipher Mode is CBC and the Padding is PKCS7 which are both good
            var aesManaged = new AesManaged();
            var decryptor = aesManaged.CreateDecryptor(keyBytes, ivBytes);
            var memoryStream = new MemoryStream(cipherTextBytes);
            var cryptoStream = new CryptoStream(memoryStream, decryptor, CryptoStreamMode.Read);
            var streamReader = new StreamReader(cryptoStream);

            // Return the decrypted bytes from the decrypting stream.
            Console.WriteLine("\n{0}\n", streamReader.ReadToEnd());
        }

输出为：“这是我的秘密字符串，lorem ipsum”

但是当我试图通过Python2.7等效的实现来解密消息时，它没有正确地解密前几个字符

import base64
from Crypto.Cipher import AES
from Crypto.Protocol import KDF

def p_decrypt( self, text ):
    text_dec = base64.b64decode(text)
    salt = text_dec[:256]
    enc_txt = text_dec[256:]
    key_bytes = KDF.PBKDF2(self.key, salt, dkLen=32)
    iv = KDF.PBKDF2(self.key, salt)
    cipher = AES.new(key_bytes, AES.MODE_CBC, iv)
    return cipher.decrypt(enc_txt)

输出为：“�增��"J�����“t弦，lorem ipsum”

预期输出：“这是我的秘密字符串，lorem ipsum”

我试图找到这个问题，当我使用C#RFC2898DeriveBytes方法生成的keyBytes和IV时，该方法也适用于精细的python代码，但python代码不能使用PBKDF2生成的keyBytes和IV正确解密整个消息。

C#RFC2898DeriveBytes和python PBKDF2都使用HMACSHA1哈希算法生成密钥字节，但C#RFC2898DeriveBytes方法生成的密钥字节和IV不同，而python PBKDF2为IV调用返回生成的密钥字节的前16个字节。

请给我一些有用的指导方针。

谢谢，默尔

Answer 1

# coding:utf8
# python3
# pip3 install pycryptodome

import base64
from Crypto.Cipher import AES
from Crypto.Protocol import KDF
from Crypto.Util.Padding import unpad

cipherText = 'i+EKwmlAF0VYh4GwDd+bGf3+yreYsPJW2Oq/w9FXjsp7RI3VqRiqtnqiAD4n6U0JJSTe2ct4B7lgrG+dHxeGcXYEYIERXvU0xnUdH+z3mRwmgYOqCU9HRUKy/z3GKISTm8qH030KTYm3YMBjnKpU8gaRcoDPP/nCiB3o5fPdyspgJgT/qt5BuvwYq7n0qg6ez/Wi4447gq/qHwG3wuuYLSBUCfmIkgGaO1KXqv3SsR8EAhrmMBmPDJfjc3sydNqs5B8J9/JvZFEZULTb8rLQZKQvgHhH9/53Bzs3zmoq0RFbgSueUbyeWb9rLAzYieTz8Yj0srG4GtwPrTPoItc6/hvx5stZ6pX8tgyk9Y3baT0JFMtGgxve7yduy8idTCQdAwRc5NOo4+CBk7P/sIw6+Q=='
the_pass = 'password'

text_dec = base64.b64decode(cipherText)
salt = text_dec[:256]
enc_txt = text_dec[256:]
key_iv_bytes = KDF.PBKDF2(the_pass, salt, dkLen=48)

key = key_iv_bytes[:32]
iv = key_iv_bytes[32:]
cipher = AES.new(key, AES.MODE_CBC, iv)
plaintext = cipher.decrypt(enc_txt)
plaintext = unpad(plaintext, 16)

print(plaintext)
# b'This is my secret string, lorem ipsum'

Answer 2

Rfc2898DeriveBytes是一个流式响应对象，因此连接两个连续调用与执行一个将两个长度相加的调用相同。

var pbkdf2WithTwoCalls = new Rfc2898DeriveBytes(...)
var pbkdf2WithOneCall = new Rfc2898DeriveBytes(sameParametersAsAbove);

byte[] twoCallA = pbkdf2WithTwoCalls.GetBytes(32);
byte[] twoCallB = pbkdf2WithTwoCalls.GetBytes(16);

byte[] oneCall = pbkdf2WithOneCall.GetBytes(32 + 16);

if (!oneCall.SequenceEquals(twoCallA.Concat(twoCallB))
    throw new TheUniverseMakesNoSenseException();

因此，Python中的解决方案是对PBKDF2进行一个48字节的调用，然后将其拆分为32字节的AES密钥和16字节的IV。

您的解密响应表明密钥正确，但IV不正确。