Why TransformBlock(byte[1]) twice not the same as TransformBlock(byte[2])?

Question

In the C# implementation of the salsa20, if I call the method TransformBlock two times on a block of size 1, it is different than calling it a single time on a block of size 2, this is problematic when you use this class to encrypt objects sent via BinaryFormatter.

Is this by design?

To remedy this I wrapped salsa20 in another class (decorator design pattern) that generates and caches blocks of 64 bytes at a time, basically like this (in simplified pseudo-code):

private Queue<byte> queue;
private ICryptoTransform salsa20CryptoTransform;
public int TransformBlock(byte[] input, byte[] output){
    while(input.Length > queue.Count){
        byte[] temp1 = new byte[64];
        byte[] temp2 = new byte[64];
        salsa20CryptoTransform.TransformBlock(temp1, temp2);
        foreach(byte b in  temp2){
            queue.Enqueue(b);
        }
    }
    for(int i = 0;i<input.Length;i++){
        output[i] = intput[i] ^ queue.Dequeue();
    }
}

Is there anything I should be concerned about here in terms of security?

Answer 1

Blocks are blocks; a lot of compression and encryption processes have special demarcation of blocks, where they essentially reset a few things - making a block a unit (potentially also the minimal unit that can be unscrambled - an important consideration for streaming APIs). The important question, though, is not "is the encrypted data the same" - frankly, it doesn't need to be. In many cases, it would be entirely legitimate if it encrypted it differently every time you called it (via some internal randomisation). The only important question is:

If I decrypt this appropriately, with the correct keys: do I get back the original data?