Can't send and receive on the same port

Question

Hello,
I have a socket client which send a message to serverSocket then it should wait for a response.
The problem I have is that the client send the message but when it tries to open a connection as socketserver on the same port it can't , a have this ERROR CannotGetConnectToTheHost
this is the code of my client :
try {
echoSocket = new Socket("localhost", 3001);
out = new PrintWriter(echoSocket.getOutputStream(), true);
out.println(hdr + testContent );
echoSocket.getInputStream()));
} catch (UnknownHostException e) {
          System.err.println("Don't know about host: localhost.");
          System.exit(1);
} catch (IOException e) {
     System.err.println(
          "Couldn't get I/O for "
     + "the connection to: USE host : "
          + e.toString());
          System.exit(1);
     out.close();
     echoSocket.close();
int compteur = 0;
try {
ServerSocket oreille = new ServerSocket(4001);
Socket socket = (Socket)null;
System.out.println("Ready to accept connections");
while (true) {
     socket = oreille.accept();
     System.out.println("\n Nouveau message r�ceptionn� :\n");
     InputStream entree = socket.getInputStream();
     int c;
     String msgAdh = "";
     while ((c = entree.read()) != -1) {
     msgAdh += (char) c; //lecture byte par byte
     compteur++;
     System.out.println("Ready to accept Another Message ");
     entree.close();
     socket.close();
     oreille.close();
} catch (IOException e) {
System.out.println("CannotGetConnectToTheHost");
thinks in advance

Answer 1

When printing out an exception, print the exception object too, e.g.System.out.println("..whatever...:" + e); It almost always contains useful information. For extra credit, print out the stack trace too.
If I understand correctly what the program does:
    while (true) {
        socket = oreille.accept();
        oreille.close();
    }That can't be right: after the first loop, you close the socket, and then try to use it again on the second loop. If you want to keep using a socket, keep it open.