Specified initialization vector (IV) does not match the block size for this algorithm (Microsoft Example)

Question

I copied the code from here and i changed something that's not crucial and I now get this error:

Specified initialization vector (IV) does not match the block size for this algorithm

byte[] publicKey = {214,46,220,83,160,73,40,39,201,155,19,202,3,11,191,178,56,
            74,90,36,248,103,18,144,170,163,145,87,54,61,34,220,222,
            207,137,149,173,14,92,120,206,222,158,28,40,24,30,16,175,
            108,128,35,230,118,40,121,113,125,216,130,11,24,90,48,194,
            240,105,44,76,34,57,249,228,125,80,38,9,136,29,117,207,139,
            168,181,85,137,126,10,126,242,120,247,121,8,100,12,201,171,
            38,226,193,180,190,117,177,87,143,242,213,11,44,180,113,93,
            106,99,179,68,175,211,164,116,64,148,226,254,172,147};

byte[] exponent = { 1, 0, 1 };

RSACryptoServiceProvider rsa = new RSACryptoServiceProvider();
MemoryStream ms = new MemoryStream();
RijndaelManaged rm = new RijndaelManaged();

RSAParameters rsaKeyInfo = new RSAParameters
{
    Modulus = publicKey,
    Exponent = exponent
};

rsa.ImportParameters(rsaKeyInfo);

byte[] encryptedSymmetricKey = rsa.Encrypt(rm.Key, false);
byte[] encryptedSymmetricIV = rsa.Encrypt(rm.IV, false);

CryptoStream cryptoStream = new CryptoStream(ms, rm.CreateEncryptor(encryptedSymmetricKey, encryptedSymmetricIV), CryptoStreamMode.Write);

(I only added the CryptoStream)

Why in the microsoft article they encrypt key and iv if then throws exception of matching the size? for what purpose the encrypt this key and iv, how can i use this key and iv encrypted ?

The Key and IV are `rm.Key` and `rm.IV`. Using the encrypted Key and encrypted IV in `rm.CreateEncryptor()` doesn't make sense. Also, encrypting the IV doesn't make sense either as it provides no security benefit but is expensive. — President James K. Polk, Jul 25 '19 at 21:02
//Encrypt the symmetric key and IV. encryptedSymmetricKey = rsa.Encrypt(rm.Key, false); encryptedSymmetricIV = rsa.Encrypt(rm.IV, false); — Susta250, Jul 26 '19 at 15:24
Yes it is, the last 2 lines of code from Asymmetric Encryption — Susta250, Jul 26 '19 at 15:24
Those lines in your code don't throw an exception, do they? The exception in thrown in the final line with the `CryptoStream`. — President James K. Polk, Jul 26 '19 at 15:47
No they don´t, but why in the Microsoft article encrypt key and IV??? — Susta250, Jul 26 '19 at 16:16

Specified initialization vector (IV) does not match the block size for this algorithm (Microsoft Example)

0 Answers0