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copied!<p>Use a lookup table. There are only 256 possible values after XORing, so it's not exactly going to take a long time. Unlike izb's solution though, I wouldn't suggest manually putting all the values in though - compute the lookup table <em>once</em> at startup using one of the looping answers.</p> <p>For example:</p> <pre><code>public static class ByteArrayHelpers { private static readonly int[] LookupTable = Enumerable.Range(0, 256).Select(CountBits).ToArray(); private static int CountBits(int value) { int count = 0; for (int i=0; i < 8; i++) { count += (value >> i) & 1; } return count; } public static int CountBitsAfterXor(byte[] array) { int xor = 0; foreach (byte b in array) { xor ^= b; } return LookupTable[xor]; } } </code></pre> <p>(You <em>could</em> make it an extension method if you really wanted...)</p> <p>Note the use of <code>byte[]</code> in the <code>CountBitsAfterXor</code> method - you <em>could</em> make it an <code>IEnumerable<byte></code> for more generality, but iterating over an array (which is known to be an array at compile-time) will be faster. Probably only microscopically faster, but hey, you asked for the <em>fastest</em> way :)</p> <p>I would almost certainly <em>actually</em> express it as</p> <pre><code>public static int CountBitsAfterXor(IEnumerable<byte> data) </code></pre> <p>in real life, but see which works better for you.</p> <p>Also note the type of the <code>xor</code> variable as an <code>int</code>. In fact, there's no XOR operator defined for <code>byte</code> values, and if you made <code>xor</code> a <code>byte</code> it would still compile due to the nature of compound assignment operators, but it would be performing a cast on each iteration - at least in the IL. It's quite possible that the JIT would take care of this, but there's no need to even ask it to :)</p>

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