Java中的Rfc2898DeriveBytes [英] Rfc2898DeriveBytes in java

问题描述

我正在尝试在Java中实现以下代码:

I am trying to implement the following code in java:

var keyGenerator = new Rfc2898DeriveBytes(password, salt, 1000);
byte[] key = keyGenerator.GetBytes(32);
byte[] iv = keyGenerator.GetBytes(16);
using (AesManaged aes = new AesManaged())
{
    using (ICryptoTransform encryptor = aes.CreateEncryptor(key, iv)) 
    {
        byte[] result = encryptor.TransformFinalBlock(content, 0, content.Length);
    }
}

使用以下一项:

SecretKeyFactory factory = SecretKeyFactory.getInstance("PBKDF2WithHmacSHA1");
PBEKeySpec keyspec = new PBEKeySpec(password, salt, 1000, 256);
Key key = factory.generateSecret(keyspec);
SecretKeySpec secret = new SecretKeySpec(key.getEncoded(), "AES");
byte[] iv = "how_to_generate_in_java_as_in_c".getBytes();
AlgorithmParameterSpec ivSpec = new IvParameterSpec(iv);
Cipher cipher = Cipher.getInstance("AES/CBC/PKCS5Padding");
cipher.init(Cipher.ENCRYPT_MODE, secret, ivSpec);
byte[] result = cipher.doFinal("asdfasdf".getBytes("UTF-8"));

我浏览了许多关于SO的示例和问题，但没有找到正确生成iv []的方法(与C中的值相同).似乎没有办法做到这一点，因为java允许仅以随机(而不是C语言中可用的伪随机)创建此值.这是正确的吗?有人可以帮忙解决这个问题吗?

I looked through a lot of examples and questions of SO and did not find a way to generate correctly iv[] (in the same value as in C). And it seems there is not way to do it, as java allow to create this value only as random (not pseudo random as available in C). Is this right? Could somebody please help with this issue?

推荐答案

从地穴专家那里得到了提示，并找到了正确的解决方案:

Got a prompt from crypt specialist and found the right solution:

SecretKeyFactory factory = SecretKeyFactory.getInstance("PBKDF2WithHmacSHA1");
PBEKeySpec pbeKeySpec = new PBEKeySpec(password, salt, 1000, 384);
Key secretKey = factory.generateSecret(pbeKeySpec);
byte[] key = new byte[32];
byte[] iv = new byte[16];
System.arraycopy(secretKey.getEncoded(), 0, key, 0, 32);
System.arraycopy(secretKey.getEncoded(), 32, iv, 0, 16);

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