How to encode 8 4bit values into an Int32 and then get those back?

Question

How can I put 8 4bit (0..15 or 0..F if you like hex ;) ) values into an Int32 (int) in C# and get those back? I need these for a game that I am writing or more specifically, editor for it (which will feature vertex manipulation like in Notch's failed 0x10c) and since there's already some legacy code and few levels built, I really can't break the format so I can use Int64 and encode 8 8bit values using BitConverter class ;).

I'm not very good at bitwise ops and those who tried to explain these to me so far have failed so yeah. Need help.

Answer 1

The main idea is to use left << shift and | to encode and & and right >> shift to decode. Suppose the values are organized as array:

 byte[] data = new byte[] { 1, 2, 3, 4, 5, 6, 7, 8 };

 int result = data
   .Select((value, i) => value << (4 * i))
   .Aggregate((x, y) => x | y); // or just .Sum() - Jeppe Stig Nielsen's idea

Edit: for loop implementation:

 int result = 0;

 for (int i = 0; i < data.Length; ++i)
   result |= data[i] << (4 * i);

To retrieve the index-th value (index is in [0..7] range)

 byte item = (byte) ((result >> (index * 4)) & 0xF);