Trying to understand the details of converting an int to a byte[] and back

Question

I found the following code to convert ints into bytes and wanted to understand it better:
public static final byte[] intToByteArray(int value) {
return new byte[]{
(byte)(value >>> 24), (byte)(value >> 16 & 0xff), (byte)(value >> 8 & 0xff), (byte)(value & 0xff) };
}I understand that an int requires 4 bytes and that each byte allows for a power of 8. But, sadly, I can't figure out how to directly translate the code above? Can someone recommend a site that explains the following:
1. >>> and >>
2. Oxff.
3. Masks vs. casts.
thanks so much.

Answer 1

By the way, people often introduce this redundancy (as in your example above):
(byte)(i >> 8 & 0xFF)When this suffices:
(byte)(i >> 8)Since by [JLS 5.1.3|http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#25363] :
A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T.Casting to a byte is merely keeping only the lower order 8 bits, and (anything & 0xFF) is simply clearing all but the lower order 8 bits. So it's redundant. Or I could just show you the more simple proof: try absolutely every value of int as an input and see if they ever differ:
   public static void main(String[] args) {
      for ( int i = Integer.MIN_VALUE;; i++ ) {
         if ( i % 100000000 == 0 ) {
            //report progress
            System.out.println("i=" + i);
         if ( (byte)(i >> 8 & 0xff) != (byte)(i >> 8) ) {
            System.out.println("ANOMOLY: " + i);
         if ( i == Integer.MAX_VALUE ) {
            break;
   }Needless to say, they don't ever differ. Where masking does matter is when you're widening (say from a byte to an int):
   public static void main(String[] args) {
      byte b = -1;
      int i = b;
      int j = b & 0xFF;
      System.out.println("i: " + i + " j: " + j);
   }That's because bytes are signed, and when you widen a signed negative integer to a wider integer, you get 1s in the higher bits, not 0s due to sign-extension. See [http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#5.1.2]
Edited by: endasil on 3-Dec-2009 4:53 PM