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Encrypt/Decrypt using Bouncy Castle in C#

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I am using the "BouncyCastle.Crypto.dll" for encrypt/decrypt a string in my app. I am using the following code from this blog:

  1. I have a class BCEngine, exactly the same as the one given in the link mentioned above.

    public class BCEngine
    {
       private readonly Encoding _encoding;
       private readonly IBlockCipher _blockCipher;
       private PaddedBufferedBlockCipher _cipher;
       private IBlockCipherPadding _padding;
    
       public BCEngine(IBlockCipher blockCipher, Encoding encoding)
       {
          _blockCipher = blockCipher;
          _encoding = encoding;
       }
    
       public void SetPadding(IBlockCipherPadding padding)
       {
           if (padding != null)
             _padding = padding;
       }
    
       public string Encrypt(string plain, string key)
       {
           byte[] result = BouncyCastleCrypto(true, _encoding.GetBytes(plain), key);
           return Convert.ToBase64String(result);
       }
    
       public string Decrypt(string cipher, string key)
       {
          byte[] result = BouncyCastleCrypto(false, Convert.FromBase64String(cipher), key);
          return _encoding.GetString(result);
       }
    
       /// <summary>
       ///
       /// </summary>
       /// <param name="forEncrypt"></param>
       /// <param name="input"></param>
       /// <param name="key"></param>
       /// <returns></returns>
       /// <exception cref="CryptoException"></exception>
       private byte[] BouncyCastleCrypto(bool forEncrypt, byte[] input, string key)
       {
           try
           {
               _cipher = _padding == null ? new PaddedBufferedBlockCipher(_blockCipher) : new PaddedBufferedBlockCipher(_blockCipher, _padding);
               byte[] keyByte = _encoding.GetBytes(key);
               _cipher.Init(forEncrypt, new KeyParameter(keyByte));
               return _cipher.DoFinal(input);
           }
           catch (Org.BouncyCastle.Crypto.CryptoException ex)
           {
               throw new CryptoException(ex.Message);
           }
       }
    }
    

I am using an asp.net form in which i have written code as given below:

    public partial class EncryptionForm : System.Web.UI.Page
    {
      Encoding _encoding;
      IBlockCipherPadding _padding;
      string key = "DFGFRT";
       string textToBeEncrypted = "Original text. Please encrypt me.";
       string txtEncryptedText = string.empty;
       string txtDecryptedText = string.empty;

      protected void Page_Load(object sender, EventArgs e)
      {
          _encoding = Encoding.ASCII; 
          Pkcs7Padding pkcs = new Pkcs7Padding();
          _padding = pkcs;   
      }

      protected void btnEncrypt_Click(object sender, EventArgs e)
      {
          txtEncryptedText = AESEncryption(textToBeEncrypted, key, true);
      }

      protected void btnDecrypt_Click(object sender, EventArgs e)
      {
          txtDecryptedText = AESDecryption(txtEncryptedText.Text, key, true);
      }

      public string AESEncryption(string plain, string key, bool fips)
      {
          BCEngine bcEngine = new BCEngine(new AesEngine(), _encoding);
          bcEngine.SetPadding(_padding);
          return bcEngine.Encrypt(plain, key);
      }

      public string AESDecryption(string cipher, string key, bool fips)
      {
          BCEngine bcEngine = new BCEngine(new AesEngine(), _encoding);
          bcEngine.SetPadding(_padding);
          return bcEngine.Decrypt(cipher, key);
      }
    }

Not sure, but due to some reason, I get an exception when I call the btnEncrypt_Click

"Key length not 128/192/256 bits."

Can anybody please guide? I am a complete newbie to this. Thanks in Advance.

like image 585
Hari Avatar asked May 06 '11 11:05

Hari


1 Answers

Your string key = "DFGFRT"; is not 128/192/256 bits.

DFGFRT is 6 characters, which is 6 (or 12?) bytes = 8*12 = 96 bits (at most).

To get a 128 bit key you need a 16 byte string, so I'd go on the safe side and use a 16 character string so it will be a 128 bit key if using single byte characters and 256 if using wide characters.

like image 113
Matt Avatar answered Sep 27 '22 01:09

Matt