Question

link

Multiple threads can publish and receive each other’s message:

1. whenever a thread publishes a message, all the other threads can receive and print out that message;

2. if multiple message get published, the messages should be queued or whatever recorded and other threads can print out the message one by one;

3. no published message should be missed by any other threads.

Solution

Suggested by level 13:

给每个thread分一个blocking queue就完了。这题延伸开来是个很好的设计题, pubsub, messaging framework etc.

Using a blocking queue to store messages for each server. It’s pretty much like consumer-producer. But here, the server takes on both roles. Read my code below, or [Java OOP] PubSub (Publish–subscribe) pattern to learn about PubSub.

Code

Below is my code, it may not be the correct solution. If you would like to discuss, please leave me a comment!

public class MessagingServer implements Runnable {

    String serverId;
    List<MessagingServer> servers;
    BlockingQueue<String> messages;
    boolean isFinished;

    public MessagingServer(String id, List<MessagingServer> list, int qSize) {
        this.serverId = id;
        this.servers = list;
        messages = new ArrayBlockingQueue<String>(qSize);
        isFinished = false;
    }

    public void run() {
        // this would be the consumer
        // everything that is added to BlockingQueue<String> messages is printed
        while (!isFinished) {
            String msg;
            try {
                msg = messages.take();
                System.out.println(serverId + " says: " + msg);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }

    public void sendMessage(String msg) {
        // this is the producer

        // insert this msg in the blockingQ of all other servers
        for (MessagingServer server : servers) {
            server.messages.add(msg + " (received from " + this.serverId + ")");
        }
    }

    public void terminate() {
        this.isFinished = true;
    }
}