Going from 127.0.0.1 to 2130706433, and back again

String to int:

int pack(byte[] bytes) {
  int val = 0;
  for (int i = 0; i < bytes.length; i++) {
    val <<= 8;
    val |= bytes[i] & 0xff;
  }
  return val;
}

pack(InetAddress.getByName(dottedString).getAddress());

Int to string:

byte[] unpack(int bytes) {
  return new byte[] {
    (byte)((bytes >>> 24) & 0xff),
    (byte)((bytes >>> 16) & 0xff),
    (byte)((bytes >>>  8) & 0xff),
    (byte)((bytes       ) & 0xff)
  };
}


InetAddress.getByAddress(unpack(packedBytes)).getHostAddress()

I've modified my original answer. In Sun's implementation of InetAddress, the hashCode method produces the integer representation of the IPv4 address, but as the commenters correctly pointed out, this is not guaranteed by the JavaDoc. Therefore, I decided to use the ByteBuffer class to calculate the value of the IPv4 address instead.

import java.net.InetAddress;
import java.nio.ByteBuffer;

// ...

try {
    // Convert from integer to an IPv4 address
    InetAddress foo = InetAddress.getByName("2130706433");
    String address = foo.getHostAddress();
    System.out.println(address);

    // Convert from an IPv4 address to an integer
    InetAddress bar = InetAddress.getByName("127.0.0.1");
    int value = ByteBuffer.wrap(bar.getAddress()).getInt();
    System.out.println(value);

} catch (Exception e) {
    e.printStackTrace();
}

The output will be:

127.0.0.1
2130706433

I've not tried it wrt. performance, but the simplest way is probably to use the NIO ByteBuffer.

e.g.

 byteBuffer.put(integer).array();

would return you a byte array representing the integer. You may need to modify the byte order.