Decoding Hex: What does this line do (len & 0x01) != 0

Question

I was going through a piece of code in the Apache commons library and was wondering what these conditions do exactly.

public static byte[] decodeHex(final char[         


        

Answer 1

This checks if the last digit in the binary writing of len is a 1.

  xxxxxxxy
& 00000001

gives 1 if y is 1, 0 if y is 0, ignoring the other digits.

If y is 1, the length of the char array is odd, which shouldn't happen in this hex writing, hence the exception.

Another solution would have been

if (len%2 != 0) {

which would have been clearer in my opinion. I doubt the slight performance increase just before a loop really matters.

Answer 2

It's a 1337 (high performance) way of coding:

if (len % 2 == 1)

i.e. is len odd. It works because the binary representation of every odd integer has its least significant (ie last) bit set. Performaning a bitwise AND with 1 masks all other bits, leaving a result of either 1 if it's odd or 0 if even.

It's a carryover from C, where you can code simply:

if (len & 1)

Answer 3

This line checks if len is an odd number or not. If len isn't odd, len & 1 will be equal to 0. (1 and 0x01 are the same value, 0x01 is just the hexadecimal notation)