Question

In the following snippet consider replacing line 8 with commented equivalent

1. private static String ipToText(byte[] ip) {
2.  StringBuffer result = new StringBuffer();
3.
4.  for (int i = 0; i < ip.length; i++) {
5.      if (i > 0)
6.          result.append(".");
7.
8.      result.append(ip[i]); // compare with result.append(0xff & ip[i]);
9.  }
10.
11.     return result.toString();
12. }

.equals() test confirms that adding 0xff does not change anything. Is there a reason for this mask to be applied?

Answer 1

byte in Java is a number between −128 and 127 (signed, like every integer in Java (except for char if you want to count it)). By anding with 0xff you’re forcing it to be a positive int between 0 and 255.

It works because Java will perform a widening conversion to int, using sign extension, so instead of a negative byte you will have a negative int. Masking with 0xff will leave only the lower 8 bits, thus making the number positive again (and what you initially intended).

You probably didn’t notice the difference because you tested with a byte[] with only values smaller than 128.

Small example:

public class A {
    public static void main(String[] args) {
        int[] ip = new int[] {192, 168, 101, 23};
        byte[] ipb = new byte[4];
        for (int i =0; i < 4; i++) {
            ipb[i] = (byte)ip[i];
        }

        for (int i =0; i < 4; i++) {
            System.out.println("Byte: " + ipb[i] + ", And: " + (0xff & ipb[i]));
        }
    }
}

This prints

Byte: -64, And: 192
Byte: -88, And: 168
Byte: 101, And: 101
Byte: 23, And: 23

showing the difference between what’s in the byte, what went into the byte when it still was an int and what the result of the & operation is.