Firewall burrowing/chat program help

Question

I'm trying to develop a person-server-person chat program with java with the ability to burrow through firewalls. I'm using the example program from 1996 at: http://developer.java.sun.com/developer/technicalArticles/InnerWorkings/Burrowing/ as a guide.
I can get the applet up and running, but it won't connect to the server. The server says "Ready...", but neither connects to each other. They are both supposedly using port 6006.
When looking through the code of SimpleChatServer.java, I can't find where it designates an IP address, and can't figure out just how the applet is supposed to connect to the server.
Thanks everyone,
Mark

Answer 1

(Most if this code is from the example in the URL I first gave)
Here's the important code (trying 8080 this time):
Server:
int port = 8080;
     // Get the port number
     if (argv.length == 1)
     port = Integer.parseInt(argv[0]);
     // Create a server socket on the given port
     ServerSocket ss = null;
     try {
     ss = new ServerSocket(port);
     } catch (IOException e) {
     System.out.println("Could not listen on port: " +
port + ", " + e);
     System.exit(1);
     // Allocate a vector for all open connections to clients.
     Vector connections = new Vector();
     System.out.println("port: "+port+"Serversocket: "+ss);
     System.out.println("Ready...");
Output from DOS prompt is:
port: 8080Serversocket: ServerSocket[addr=0.0.0.0/0.0.0.0,port=0,localport=8080]
Ready...
From the Client/Applet side, run from IE 5:
String str = getParameter("port");
     int port = (str != null) ? Integer.parseInt(str) : 8080;
     connection = new ServerConnection(getCodeBase().getHost(),port);
System.out.println("connection: "+connection);
Output from java console:
connection: ServerConnection@de9ac4
Error getting a new connection to read from
I have no idea what ServerConnection@de9ac4 means (I know where it came from, just not what it means).
Thanks,
Mark