问题
I get the following error when I try to create a IV initialization vector for TripleDES encryptor.
Please see the code example:
TripleDESCryptoServiceProvider tripDES = new TripleDESCryptoServiceProvider();
byte[] key = Encoding.ASCII.GetBytes("SomeKey132123ABC");
byte[] v4 = key;
byte[] connectionString = Encoding.ASCII.GetBytes("SomeConnectionStringValue");
byte[] encryptedConnectionString = Encoding.ASCII.GetBytes("");
// Read the key and convert it to byte stream
tripDES.Key = key;
tripDES.IV = v4;
This is the exception that I get from the VS.
Specified initialization vector (IV) does not match the block size for this algorithm.
Where am I going wrong?
Thank you
回答1:
I've upvoted every answer (well the ones that are here before mine!) here as they're all correct.
However there's a bigger mistake you're making (one which I also made v.early on) - DO NOT USE A STRING TO SEED THE IV OR KEY!!!
A compile-time string literal is a unicode string and, despite the fact that you will not be getting either a random or wide-enough spread of byte values (because even a random string contains lots of repeating bytes due to the narrow byte range of printable characters), it's very easy to get a character which actually requires 2 bytes instead of 1 - try using 8 of some of the more exotic characters on the keyboard and you'll see what I mean - when converted to bytes you can end up with more than 8 bytes.
Okay - so you're using ASCII Encoding - but that doesn't solve the non-random problem.
Instead you should use RNGCryptoServiceProvider to initialise your IV and Key and, if you need to capture a constant value for this for future use, then you should still use that class - but capture the result as a hex string or Base-64 encoded value (I prefer hex, though).
To achieve this simply, I've written a macro that I use in VS (bound to the keyboard shortcut CTRL+SHIFT+G, CTRL+SHIFT+H) which uses the .Net PRNG to produce a hex string:
Public Sub GenerateHexKey()
Dim result As String = InputBox("How many bits?", "Key Generator", 128)
Dim len As Int32 = 128
If String.IsNullOrEmpty(result) Then Return
If System.Int32.TryParse(result, len) = False Then
Return
End If
Dim oldCursor As Cursor = Cursor.Current
Cursor.Current = Cursors.WaitCursor
Dim buff((len / 8) - 1) As Byte
Dim rng As New System.Security.Cryptography.RNGCryptoServiceProvider()
rng.GetBytes(buff)
Dim sb As New StringBuilder(CType((len / 8) * 2, Integer))
For Each b In buff
sb.AppendFormat("{0:X2}", b)
Next
Dim selection As EnvDTE.TextSelection = DTE.ActiveDocument.Selection
Dim editPoint As EnvDTE.EditPoint
selection.Insert(sb.ToString())
Cursor.Current = oldCursor
End Sub
Now all you need to do is to turn your hex string literal into a byte array - I do this with a helpful extension method:
public static byte[] FromHexString(this string str)
{
//null check a good idea
int NumberChars = str.Length;
byte[] bytes = new byte[NumberChars / 2];
for (int i = 0; i < NumberChars; i += 2)
bytes[i / 2] = Convert.ToByte(str.Substring(i, 2), 16);
return bytes;
}
There are probably better ways of doing that bit - but it works for me.
回答2:
MSDN explicitly states that:
...The size of the IV property must be the same as the BlockSize property.
For Triple DES it is 64 bits.
回答3:
The size of the initialization vector must match the block size - 64 bit in case of TripleDES. Your initialization vector is much longer than eight bytes.
Further you should really use a key derivation function like PBKDF2 to create strong keys and initialization vectors from password phrases.
回答4:
Key should be 24 bytes and IV should be 8 bytes.
tripDES.Key = Encoding.ASCII.GetBytes("123456789012345678901234");
tripDES.IV = Encoding.ASCII.GetBytes("12345678");
回答5:
The IV must be the same length (in bits) as tripDES.BlockSize
. This will be 8 bytes (64 bits) for TripleDES.
回答6:
I do it like this:
var derivedForIv = new Rfc2898DeriveBytes(passwordBytes, _saltBytes, 3);
_encryptionAlgorithm.IV = derivedForIv.GetBytes(_encryptionAlgorithm.LegalBlockSizes[0].MaxSize / 8);
The IV gets bytes from the derive bytes 'smusher' using the block size as described by the algorithm itself via the LegalBlockSizes property.
来源:https://stackoverflow.com/questions/3179304/c-sharp-cant-generate-initialization-vector-iv