Acquiring local address for datagram socket created by DatagramChannel.open

Question

I've found that when I open an anonymous DatagramChannel (i.e. unbound, unconnected), the socket that it creates has no local address info I.e. no local port. Hopefully the example below will explain the problem...
import java.nio.channels.DatagramChannel;
import java.io.IOException;
import java.net.DatagramSocket;
import java.net.DatagramPacket;
import java.net.InetAddress;
public class QuickTest {
    public static void main( String[] args )    {
        try {
            DatagramChannel dc = DatagramChannel.open();
            DatagramSocket ds = dc.socket();
            byte[] ba = new byte[]{0,1,2,3};
            DatagramPacket dp = new DatagramPacket(ba,4,InetAddress.getLocalHost(),8000);
            ds.send(dp);
            System.out.println( "DatagramSocket from DatagramChannel");
            System.out.println();
            System.out.println( "          getInetAddress(): " + ds.getInetAddress());
            System.out.println( "         getLocalAddress(): " + ds.getLocalAddress ());
            System.out.println( "            getLocalPort(): " + ds.getLocalPort());
            System.out.println( "   getLocalSocketAddress(): " + ds.getLocalSocketAddress());
            System.out.println( "                 getPort(): " + ds.getPort());
            System.out.println( "  getRemoteSocketAddress(): " + ds.getRemoteSocketAddress());
            ds = new DatagramSocket();
            System.out.println();
            System.out.println();
            System.out.println();
            System.out.println( "DatagramSocket from new DatagramSocket()");
            System.out.println();
            System.out.println( "          getInetAddress(): " + ds.getInetAddress());
            System.out.println( "         getLocalAddress(): " + ds.getLocalAddress ());
            System.out.println( "            getLocalPort(): " + ds.getLocalPort());
            System.out.println( "   getLocalSocketAddress(): "    + ds.getLocalSocketAddress());
            System.out.println( "                 getPort(): " + ds.getPort());
            System.out.println( "  getRemoteSocketAddress(): " + ds.getRemoteSocketAddress());
        } catch (IOException e) {
            e.printStackTrace();  //To change body of catch statement use File | Settings | File Templates.
}Running this program yields the following information....
DatagramSocket from DatagramChannel
getInetAddress(): null
getLocalAddress(): 0.0.0.0/0.0.0.0
getLocalPort(): 0
getLocalSocketAddress(): 0.0.0.0/0.0.0.0:0
getPort(): -1
getRemoteSocketAddress(): null
DatagramSocket from new DatagramSocket()
getInetAddress(): null
getLocalAddress(): 0.0.0.0/0.0.0.0
getLocalPort(): 1194
getLocalSocketAddress(): 0.0.0.0/0.0.0.0:1194
getPort(): -1
getRemoteSocketAddress(): null
I understand that the sockets address will be included in any packets sent, but why can I determine the port for one and not the other? I would like to continue using an anonymous channel, but I also would like to have the port for use in my test harness and for logging info...
Any info and possible work around for this behavior would be much appreciated.
tx in advance.

Answer 1

This looks like a bug in the socket returned by DatagramSocketChannel and you should report it.
In the meantime you could either just use DatagramSocket, or if you must have NIO you could try binding this socket to null or new InetSocketAddress(null,0) and see whether getLocalPort() now returns the right data.