Sending the same but modifed object over ObjectOutputStream

Question

I have the following code that shows either a bug or a misunderstanding on my part.

I sent the same list, but modified over an ObjectOutputStream. Once as [0] and other as [1]. But when I read it, I get [0] twice. I think this is caused by the fact that I am sending over the same object and ObjectOutputStream must be caching them somehow.

Is this work as it should, or should I file a bug?

import java.io.*;
import java.net.*;
import java.util.*;

public class OOS {

    public static void main(String[] args) throws Exception {
        Thread t1 = new Thread(new Runnable() {
            public void run() {
                try {
                    ServerSocket ss = new ServerSocket(12344);
                    Socket s= ss.accept();

                    ObjectOutputStream oos = new ObjectOutputStream(s.getOutputStream());
                    List same = new ArrayList();
                    same.add(0);
                    oos.writeObject(same);
                    same.clear();
                    same.add(1);
                    oos.writeObject(same);

                } catch(Exception e) {
                    e.printStackTrace();
                }
            }
        });
        t1.start();

        Socket s = new Socket("localhost", 12344);
        ObjectInputStream ois = new ObjectInputStream(s.getInputStream());

        // outputs [0] as expected
        System.out.println(ois.readObject());

        // outputs [0], but expected [1]
        System.out.println(ois.readObject());
        System.exit(0);
    }
}

Answer 1

The stream has a reference graph, so an object which is sent twice will not give two objects on the other end, you will only get one. And sending the same object twice separately will give you the same instance twice (each with the same data - which is what you're seeing).

See the reset() method if you want to reset the graph.

Answer 2

What you probably want is:

ObjectOutputStream oos = new ObjectOutputStream(s.getOutputStream());
List same = new ArrayList();
same.add(0);
oos.writeObject(same);
oos.flush();  // flush the stream here
same.clear();
same.add(1);
oos.writeObject(same);

Otherwise the same object will be flushed twice when the stream is closed or its buffer runs out.

Just FYI, when you deserialize the objects into, let's say o1 and o2, o1 != o2.

Answer 3

Max is correct, but you can also use:

public void writeUnshared(Object obj);

See comment below for caveat