title: Open tcp InetSocketAddress to application path (not just host + port)
body: <p>I'm attempting to connect to a TCP socket in Android.</p>
  
  <p>I know the socket service works because I can connect and interact with it in a browser (in JavaScript) as follows:</p>
  
  <pre><code>var ws = window.WebSocket || window.MozWebSocket;  
  window.ws = new wsImpl('ws://foo.bar.com:8282/MySocketService', 'my-protocol');  
  ...
  </code></pre>
  
  <hr>
  
  <p>So, in my Android app:</p>
  
  <p>This connects successfully, but I never receive messages from it:  </p>
  
  <pre><code>SocketAddress sa = new InetSocketAddress("foo.bar.com", 8282);
  </code></pre>
  
  <p>This fails to connect:  </p>
  
  <pre><code>SocketAddress sa = new InetSocketAddress("foo.bar.com/MySocketService", 8282);
  </code></pre>
  
  <p>and I receive an error like:  </p>
  
  <blockquote>
    <p><em>java.net.UnknownHostException: Host is unresolved: foo.bar.com/MySocketService:8282</em></p>
  </blockquote>
  
  <p>Is there any way to indicate the application path for a TCP service?</p>