How can I convert the decimal representation of an

Question

Does anyone knows how to convert decimal notation of an IP address into binary form in Java? Please let me know...

Answer 1

Gathering your suggestions and some other sources, I found usefull to convert an InetAdress to an array of bit, as well as BitSet, which can help to compute and(), or(), xor() out of your binary representation.

Following sample shows how to convert ip to binary and binary to ip.

Enjoy!

public class IpConverter {
public static void main(String[] args) {
    String source = "192.168.1.1";
    InetAddress ip = null;
    try {
        ip = InetAddress.getByName(source);
    } catch (UnknownHostException e) {
        e.printStackTrace();
        return;
    }
    System.out.println( "source : " + ip);

    // To bit sequence ------------
    byte[] binaryIP = ip.getAddress();
    BitSet[] bitsets = new BitSet[binaryIP.length];
    int k = 0;

    System.out.print("to binary: ");
    for (byte b : binaryIP) {
        bitsets[k] = byteToBitSet(b);
        System.out.print( toString( bitsets[k] ) + ".");
        k++;
    }
    System.out.println();

    // Back to InetAdress ---------
    byte[] binaryIP2 = new byte[4];
    k = 0;
    for (BitSet b : bitsets) {
        binaryIP2[k] = bitSetToByte(b);
        k++;
    }

    InetAddress ip2 = null;
    try {
        ip2 = InetAddress.getByAddress(binaryIP2);
    } catch (UnknownHostException e) {
        e.printStackTrace();
        return;
    }

    System.out.println( "flipped back to : " + ip2);
}

public static BitSet byteToBitSet(byte b) {
    BitSet bits = new BitSet(8);
    for (int i = 0; i < 8; i++) {
        bits.set(i, ((b & (1 << i)) != 0) );
    }
    return bits;
}

public static byte bitSetToByte(BitSet bits) {
    int value = 0;
    for (int i = 0; i < 8; i++) {
        if (bits.get(i) == true) {
            value = value | (1 << i);
        }
    }
    return (byte) value;
}

public static byte bitsToByte(boolean[] bits) {
    int value = 0;
    for (int i = 0; i < 8; i++) {
        if (bits[i] == true) {
            value = value | (1 << i);
        }
    }
    return (byte) value;
}

public static boolean[] byteToBits(byte b) {
    boolean[] bits = new boolean[8];
    for (int i = 0; i < bits.length; i++) {
        bits[i] = ((b & (1 << i)) != 0);
    }
    return bits;
}

public static String toString(BitSet bits){
    String out = "";
    for (int i = 0; i < 8; i++) {
        out += bits.get(i)?"1":"0";         
    }
    return out;
}

}

Answer 2

You can use the java.net.InetAddress class. Two methods you should look at are getByName and getAddress. Here is a simple code example

import java.net.InetAddress;
import java.net.UnknownHostException;
/* ... */
String ip = "192.168.1.1";
InetAddress address = null;
try {
  address = InetAddress.getByName(ip);
} catch (UnknownHostException e) {
  //Your String wasn't a valid IP Address or host name
}
byte [] binaryIP = address.getAddress();

Answer 3

The open-source IPAddress Java library can do this for you. It can parse various IP address formats, including either IPv4 or IPv6, and has methods to produce various string formats, including one for binary. Disclaimer: I am the project manager of the IPAddress library.

This code will do it:

static void convert(String str) {
    IPAddressString string = new IPAddressString(str);
    IPAddress addr = string.getAddress();
    System.out.println(addr + " in binary is " + addr.toBinaryString());
}

Example:

convert("1.2.3.4");
convert("a:b:c:d:e:f:a:b");

The output is:

1.2.3.4 in binary is 00000001000000100000001100000100
a:b:c:d:e:f:a:b in binary is 00000000000010100000000000001011000000000000110000000000000011010000000000001110000000000000111100000000000010100000000000001011

Answer 4

An IP address written as a.b.c.d can be converted to a 32-bit integer value
using shift and bit-wise inclusive OR operators as,

(a << 24) | (b << 16) | (c << 8) | d

To be safe, each of a,b,c,d has valid range 0-255 -- you can check that in your conversion.
You can further validate the IP address using this regex example.