DatagramSocket.getLocalAddress().getHostAddress() returns 127.0.0.1

Question

Hello, I am new here, I need to bind an UDP socket to a clients machine through an applet and notify the IP:Port I am bound to.
When running the applet locally I perfectly get 10.100.100.22, but when I run it remotely (Other machine opens the web page on my machine) I get 127.0.0.1.
the code part that creates the DatagramSocket and notifies the address and port is the followin:
//Create Socket
this.mSocket=new DatagramSocket(new InetSocketAddress(InetAddress.getLocalHost(), this.mPort));
//Get Listening port
this.mPort=this.mSocket.getLocalPort();
//Get Listening IP
this.mAddress=this.mSocket.getLocalAddress().getHostAddress();
//Notify
this.notifyListening();
inside notify
I read the contents of the member variable mAddress.
Could Anybody point me out in the right direction.
Thanks, Frank

Answer 1

Sylvia thanks for replying after all.
May I ask why the workarround. I seen a similar one, and one with a tcp socket connecting to the host given by the getCodeBase().getHost() accessor of applet.
But anyway... I don't expect you to answer this really but isn't it time for a fix up of this behavior.
let me show you a very ugly code that I ripped out from a sample class that I downloaded yesterday.
I works every time exept when the web programmers debug the asp.net application locally, in that situation it returns 127.0.0.1 but no one has a problem then because it works fine in the implementation.
I tried creating the socket the same way but I get the same result as .
socket = new DatagramSocket();
port = socket.getLocalPort();
for(int i = 0; i < listeners.size(); i++)
((UDPListener)listeners.get(i)).portOpened(port, socket.getLocalAddress().getHostAddress());
seem unreal. I know.
Thanks, Frank.