Java and “& 0xFF” example

By mkyong | Updated: November 25, 2013

Viewed: 961 (+53 pv/w)

Tags: bit shifting bitwise ip address java

Before you understand what is & 0xFF, make sure you know following stuffs :

1. Bitwise AND operator, link.
2. Converts hex to/from binary, and decimal to/from binary.

In short, & 0xFF is used to make sure you always get the last 8 bits. Let’s see an example to convert an IP address to/from decimal number.

1. Convert IP Address To Decimal

It’s a common practice to convert IpAddress to decimal and store it into database for better calculation and comparison.

Testing IP address = 192.168.1.2
Decimal Number     = 3232235778

To convert 192.168.1.2 to decimal (base 10) the formula is:

192 x (256)^3 + 168 x (256)^2 + 1 x (256)^1 + 2 (256)^0 = ?
3221225472 + 11010048 + 256 + 2 = 3232235778

P.S A standard IP is “base 256”, source.

2. Convert Decimal to IP Address, with & 0xFF

To convert decimal back to an IP address, we use bit shifting operator and “mask” it with & 0xff.

  long ipAddress = 3232235778L;
  String binary = Long.toBinaryString(ipAddress);
  System.out.println(binary);
  System.out.println((ipAddress>>24) & 0xFF);
  System.out.println((ipAddress>>16) & 0xFF);
  System.out.println((ipAddress>>8) & 0xFF);
  System.out.println((ipAddress) & 0xFF);

Output

11000000101010000000000100000010
192
168
1
2

The question is why (ipAddress>>24) & 0xFF will return 192? Let dive into the binary bit shifting theory below :

2.1 (ipAddress>>24) & 0xFF = 192

Decimal   = 3232235778
Binary    = 11000000 10101000 00000001 00000010
IpAddress = 192      168      1        2

(ipAddress>>24)
            -------------------------->24
Binary    = 00000000 00000000 00000000 11000000

(ipAddress>>24) & 0xFF
Binary    = 00000000 00000000 00000000 11000000
& 0xFF    = 00000000 00000000 00000000 11111111
Result    = 00000000 00000000 00000000 11000000 = 192 (2^7 + 2^6)

In this case, the 0xFF is optional.

2.2 (ipAddress>>16) & 0xFF = 168

Decimal = 3232235778
Binary  = 11000000 10101000 00000001 00000010

(ipAddress>>16)
          ----------------->16
Binary  = 00000000 00000000 11000000 10101000 

(ipAddress>>16) & 0xFF
Binary  = 00000000 00000000 11000000 10101000 = 49320 (2^14 + 2^15 + 2^7 + 2^5 + 2^3)
& 0xFF  = 00000000 00000000 00000000 11111111
Result  = 00000000 00000000 00000000 10101000 = 168

Before & 0xFF, you get 49320, after & 0xFF, you get the correct 168. Now, you should understand the usage of & 0xFF, it just makes sure you always get the last 8 bits.

2.3 (ipAddress>>8) & 0xFF = 1

Decimal = 3232235778
Binary  = 11000000 10101000 00000001 00000010

(ipAddress>>8)
          -------->8
Binary  = 00000000 11000000 10101000 00000001

(ipAddress>>8) & 0xFF
Binary  = 00000000 11000000 10101000 00000001 = 12625921
& 0xFF  = 00000000 00000000 00000000 11111111
Result  = 00000000 00000000 00000000 00000001 = 1

2.4 (ipAddress) & 0xFF = 2

Decimal = 3232235778
Binary  = 11000000 10101000 00000001 00000010

(ipAddress)
Binary  = 11000000 10101000 00000001 00000010

(ipAddress) & 0xFF
Binary  = 11000000 10101000 00000001 00000010 = 3232235778
& 0xFF  = 00000000 00000000 00000000 11111111
Result  = 00000000 00000000 00000000 00000010 = 2

3. Java Source Code

Full Java example to demonstrate above scenario.

package com.mkyong.core;

public class BitwiseExample {

  public static void main(String[] args) {

	BitwiseExample obj = new BitwiseExample();
	long ipAddressInLong = obj.ipToLong("192.168.1.2");
	System.out.println(ipAddressInLong);

	String binary = Long.toBinaryString(ipAddressInLong);
	printPrettyBinary(binary);

	String ipAddressInString = obj.longToIp(ipAddressInLong);
	System.out.println(ipAddressInString);

  }

  public long ipToLong(String ipAddress) {

	String[] addrArray = ipAddress.split("\\.");

	long num = 0;
	for (int i = 0; i < addrArray.length; i++) {

		int power = 3 - i;

		// 1. (192 % 256) * 256 pow 3
		// 2. (168 % 256) * 256 pow 2
		// 3. (2 % 256) * 256 pow 1
		// 4. (1 % 256) * 256 pow 0
		num += ((Integer.parseInt(addrArray[i]) % 256 * Math.pow(256, power)));

	}

	return num;
  }

  public String longToIp(long i) {

	return ((i >> 24) & 0xFF) + "." + ((i >> 16) & 0xFF) + "." + ((i >> 8) & 0xFF) + "." + (i & 0xFF);

  }

  //print pretty binary code, padding left zero 
  private static void printPrettyBinary(String binary) {

	String s1 = String.format("%32s", binary).replace(' ', '0');
	System.out.format("%8s %8s %8s %8s %n", 
                s1.substring(0, 8), s1.substring(8, 16), 
                s1.substring(16, 24), s1.substring(24, 32));
  }

}

Output

3232235778
11000000 10101000 00000001 00000010 
192.168.1.2

Photo Credit : http://chortle.ccsu.edu/AssemblyTutorial/Chapter-12/lshiftLeft.gif

References

1. Java – Convert IP to Decimal example
2. What does “& 0xFF” do?
3. Stackoverflow : Absolute Beginner’s Guide to Bit Shifting?
4. Stackoverflow : Calculate whether an ip is in a specified range in java
5. Wikipedia : Two’s complement
6. Wikipedia : Bitwise operation
7. Convert base 10 to ip
8. Microsoft : IP Addressing