Your code has one big problem: It is mixing the character encodings! In Java you are calling key.getBytes(), without arguments. This method returns the UTF-8 or CP1252/ISO 8859-1 encoded data depending on your operating system and the default charset in Java. On C# side you are using Encoding.Unicode.GetBytes(key) – “Unicode” in .Net is a synonym … Read more
The encrypted data starts with a 8 byte “magic” indicating that there is a salt (the ASCII encoding of “Salted__”). Then the next 8 bytes is the salt. Now the bad news: Node.js does not seem to use the salt for the EVP_BytesToKey method: int key_len = EVP_BytesToKey(cipher, EVP_md5(), NULL, (unsigned char*) key_buf, key_buf_len, 1, … Read more
I’m not terribly familiar with this stuff, but it seems like trying MCRYPT_RIJNDAEL_256 in place of MCRYPT_RIJNDAEL_128 would be an obvious next step… Edit: You’re right — this isn’t what you need. MCRYPT_RIJNDAEL_128 is in fact the right choice. According to the link you provided, your key and IV are twice as long as they … Read more
You will have to implement OpenSSL’s EVP_BytesToKey, because that is what CryptoJS uses to derive the key and IV from the provided password, but pyCrypto only supports the key+IV type encryption. CryptoJS also generates a random salt which also must be send to the server. If the ciphertext object is converted to a string, then … Read more
I would use your suggested code, but with a slight simplification: KeyGenerator keyGen = KeyGenerator.getInstance(“AES”); keyGen.init(256); // for example SecretKey secretKey = keyGen.generateKey(); Let the provider select how it plans to obtain randomness – don’t define something that may not be as good as what the provider has already selected. This code example assumes (as … Read more
As you can see from the other answers, having a unique IV per encrypted file is crucial, but why is that? First – let’s review why a unique IV per encrypted file is important. (Wikipedia on IV). The IV adds randomness to your start of your encryption process. When using a chained block encryption mode … Read more
AES has a fixed block size of 16-bytes regardless key size. Assuming you use PKCS 5/7 padding, use this formula, cipherLen = (clearLen/16 + 1) * 16; Please note that if the clear-text is multiple of block size, a whole new block is needed for padding. Say you clear-text is 16 bytes. The cipher-text will … Read more
MCRYPT_RIJNDAEL_256 isn’t AES-256. It’s the Rijndael cipher with a block size of 256 (hence the error). AES is a subset of the Rijndael cipher using block size of 128 bits and key sizes of 128, 192 and 256 bits. This is also reflected in the IV size. To create an AES-256 encrypted ciphertext you can … Read more
There is no “quick check” mechanism that is secure, by definition. The whole point of using PBKDF2 or related techniques is to make password checking slow, to foil password cracking programs. If you added a quick check system, password crackers would be able to guess passwords in bulk very quickly.
A quick Google Search brings up the following functions: I’ve not tested whether the output they produce is correct, however they do appear to compile correctly. Public Function AES_Encrypt(ByVal input As String, ByVal pass As String) As String Dim AES As New System.Security.Cryptography.RijndaelManaged Dim Hash_AES As New System.Security.Cryptography.MD5CryptoServiceProvider Dim encrypted As String = “” Try … Read more