What's the fastest way to do a right bit rotation/circular shift on a byte array

Question

If I have the array:

{01101111,11110000,00001111} // {111, 240, 15}

The result for a 1 bit shift is:

{10110111,11111000,00000111} // {183, 248, 7}

The array size is not fixed, and the shifting will be from 1 to 7 inclusive. Currently I have the following code (which works fine):

private static void shiftBitsRight(byte[] bytes, final int rightShifts) {
   assert rightShifts >= 1 && rightShifts <= 7;

   final int leftShifts = 8 - rightShifts;

   byte previousByte = bytes[0]; // keep the byte before modification
   bytes[0] = (byte) (((bytes[0] & 0xff) >> rightShifts) | ((bytes[bytes.length - 1] & 0xff) << leftShifts));
   for (int i = 1; i < bytes.length; i++) {
      byte tmp = bytes[i];
      bytes[i] = (byte) (((bytes[i] & 0xff) >> rightShifts) | ((previousByte & 0xff) << leftShifts));
      previousByte = tmp;
   }
}

Is there a faster way to achieve this than my current approach?

Answer 1

The only way to find out is with thorough benchmarking, and the fastest implementations will vary from platform to platfrm. Use a tool like Caliper if you really need to optimize this.