两个线程可以在同一个端口中调用一个新的套接字（客户端套接字）吗？

Question

What should happen if a Thread instantiates a new client Socket while other Thread is handling another client Socket connected to the same port? This Threads are part of the same Java program.

Answer 1

Not sure if you asking about client or server socket. But I'll answer both. If you try to open a listening server socket on the same port - you'll get java.net.BindException thrown: Address already in use (Bind failed) or something similar. If you are connecting to a port - it works fine if the server accepts multiple clients.

And it does not matter if it is same thread, different thread or another process.

Answer 2

A Socket has two ports – a local port and a remote port . These are called source port and destination port in the TCP specification .

When you connect to a server, you specify the host name and the (remote) port. The server needs to know where to send the responses, which is why the client will also have a (local) port. The local port is assigned by the operating system from the pool of free ports.

Try this code sample:

import java.io.IOException;
import java.net.Socket;

public class SocketTest {
    public static void main(String[] args) {
        for (int i = 0; i < 10; i++) {
            final int j = i;
            new Thread(() -> {
                try {
                    Thread.sleep(100 * j);
                    try (Socket socket = new Socket("stackoverflow.com", 80)) {
                        System.out.printf("%d -> %d%n", socket.getLocalPort(), socket.getPort());
                    }
                } catch (InterruptedException | IOException ignored) {}
            }).start();
        }
    }
}

The output of one execution on my machine looks like this:

36699 -> 80
36700 -> 80
36701 -> 80
36703 -> 80
36708 -> 80
36709 -> 80
36710 -> 80
36712 -> 80
36713 -> 80
36714 -> 80

The local ports will generally vary each time.

In conclusion: The client doesn't care how many sockets are connected to the same remote address. There won't be any collision between the local ports, either.

Further reading:

• Lesson: All About Sockets (The Java™ Tutorials > Custom Networking)
• Java SE 8: Getting Started with Socket Programming