11

How does Java ServerSocket get a new socket bound to the same local port after a...

 3 years ago
source link: https://www.codesd.com/item/how-does-java-serversocket-get-a-new-socket-bound-to-the-same-local-port-after-accepting-the-client.html
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
neoserver,ios ssh client

How does Java ServerSocket get a new socket bound to the same local port after accepting the client?

advertisements

I am confused about Socket and ServerSocket port usages. Oracle's java tutorial about sockets says the following :

What Is a Socket?

Normally, a server runs on a specific computer and has a socket that is bound to a specific port number. The server just waits, listening to the socket for a client to make a connection request. On the client-side: The client knows the hostname of the machine on which the server is running and the port number on which the server is listening. To make a connection request, the client tries to rendezvous with the server on the server's machine and port. The client also needs to identify itself to the server so it binds to a local port number that it will use during this connection. This is usually assigned by the system.

If everything goes well, the server accepts the connection. Upon acceptance, the server gets a new socket bound to the same local port and also has its remote endpoint set to the address and port of the client. It needs a new socket so that it can continue to listen to the original socket for connection requests while tending to the needs of the connected client.

On the client side, if the connection is accepted, a socket is successfully created and the client can use the socket to communicate with the server. The client and server can now communicate by writing to or reading from their sockets.

I tried the following code to test. But it throws an exception.

try {
    ServerSocket serverSocket = new ServerSocket(8080);

    Socket socket = serverSocket.accept();

    // This prints 8080
    System.out.println("Local port of accepted socket : " + socket.getLocalPort());  

    // But this throws java.net.BindException: Address already in use: JVM_Bind
    Socket myClientSocket = new Socket("www.google.com", 80, null, 8080);

} catch (Exception e) {
    e.printStackTrace();
}

My question is obvious. While the socket returned from serverSocket.accept() can use the same local port (8080) why the socket that I created cannot use it?

To keep it simple, the listening TCP socket is already explicitly bound to the port, so you can't explicitly bind a second TCP socket to the same port (unless both sockets are also explicitly bound to different IP addresses, not including INADDR_ANY).

The accepted sockets don't go through an explicit 'bind' process. They acquire their local IP address and port automagically, as does an outbound socket if you connect it without binding.


report

Recommend

About Joyk


Aggregate valuable and interesting links.
Joyk means Joy of geeK