Exception in thread "main" java.net.BindException: Address already in use: bind

[greg mic]

Hello everyone!

I ran into a problem yet again, and for the life of me I can not figure out how to solve it.

What I'm trying to do now is set up a server-client communication using udp, but I get this message all the time: Exception in thread "main" java.net.BindException: Address already in use: bind

This is what my server code looks like:

public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
	
			
			byte [] message = new byte [1500];
			
			DatagramPacket p = new DatagramPacket(message, message.length);
			DatagramSocket s = new DatagramSocket(1234);
			
			
			
			s.receive(p);
			
			String text = new String(message, 0, message.length);
			
			System.out.println("Server recieved: " + text);
		}
	}

And this is what my client code looks like:

public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		
		String text  = "Hello";
		byte [] message = text.getBytes();
		InetAddress ip = InetAddress.getByName("localhost");
		
		DatagramPacket p = new DatagramPacket(message, message.length, ip, 1234);
		DatagramSocket s = new DatagramSocket(1234);
		s.send(p);

	}

}

So by running the server code first, and the client code after, I should get the message "Hello" on the console, but I just get an error message.

I've read that this error message means that the particular port is already in use, but I've tried a hundred other ports, and nothing seems to work. Few days ago, I had no problem with this, and the codes above are exactly how my professor's codes look like so I really don't know what goes wrong..

Thank you for your help in advance.

[Ron McLeod]

You have both the sender and receiver trying to bind to the same port.

Try setting the port in the sender to something different (for example:1235), or even better, don't specify a port at all, and have a random available port be allocated for you:
    DatagramSocket s = new DatagramSocket();

[Ron McLeod]

Also, if you are curious, you can see which port was allocated - it is visible at both sender and receiver:

public static void main(String[] args) throws IOException {
	String text = "Hello";
	byte[] message = text.getBytes();
	InetAddress ip = InetAddress.getByName("localhost");

	DatagramPacket p = new DatagramPacket(message, message.length, ip, 1234);
	DatagramSocket s = new DatagramSocket();
	System.out.println("My local port:" + s.getLocalPort());
	s.send(p);
}

public static void main(String[] args) throws IOException {
	byte[] message = new byte[1500];
	DatagramPacket p = new DatagramPacket(message, message.length);
	DatagramSocket s = new DatagramSocket(1234);

	s.receive(p);
	String text = new String(message, 0, message.length);
	System.out.println("Server recieved: " + text);
	System.out.println("Sender's port: " + p.getPort());
}

[greg mic]

Ron McLeod wrote:You have both the sender and receiver trying to bind to the same port.

Try setting the port in the sender to something different (for example:1235), or even better, don't specify a port at all, and have a random available port be allocated for you:
    DatagramSocket s = new DatagramSocket();

Thanks a lot Ron!!

Not specifying a port did solve the issue!