Question

Given a long with bytes WXYZ (where each letter is a byte), I would like some fast bit twiddling code that will create two longs with the same bytes as the original, but interleaved with the 0 byte.

For example, given the long with value ABCDEFGH (each letter being one byte), produce the two longs:

0A0B0C0D
0E0F0G0H

Something equivalent to, but faster than:

long result1 = expand((int)(input >>> 32));
long result2 = expand((int)input);

long expand(int inputInt) {
  long input = intputInt;
  return
    (input & 0x000000FF)       | 
    (input & 0x0000FF00) <<  8 | 
    (input & 0x00FF0000) << 16 | 
    (input & 0xFF000000) << 24;
}

Answer 1

The following is about 25% faster for me (Java 7, benchmarked using Google Caliper), YMMV may vary according to your compiler of course:

long a = (input | (input << 16));
long result = (a & 0xFF000000FFL) + ((a & 0xFF000000FF00L) <<8);

The idea is to use a bit of extra parallelism vs. the original approach.

The first line is a neat trick that produces garbage in bits 17-32, but you don’t care as you are going to mask it out anyway. 🙂