当类实现Serializable接口时，为什么会得到java.io.NotSerializableException？

Question

As I try to serialize an object with a call , stashCon.stash() I get java.io.NotSerializableException even when the class StashCon implements Serializable interface.

What could be the reason for this ?

public boolean connect(String username,String password) {
    try {
        Openfire.connection.connect();
        Openfire.connection.login(username,password);
        stashCon = new StashCon(Openfire.connection);
        stashCon.stash(); // CALL THAT ATTEMPTS TO SERIALIZE THE OBJECT
    }catch(Exception exc){
        exc.printStackTrace();
        return false;
    }
    return true;
}

Following method is of the class StashCon

public void stash() {
    try {
        FileOutputStream outputStream = new FileOutputStream(new File(Constants.BLAB_CONNECTION_FILE));
        ObjectOutputStream objectOutputStream = new ObjectOutputStream(outputStream);
        objectOutputStream.writeObject(this); // LINE 33
        objectOutputStream.close();
        outputStream.close();
    }catch(Exception exc) {
        exc.printStackTrace();
    } 
}

Exception

java.io.NotSerializableException: org.jivesoftware.smack.XMPPConnection
at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1156)
at java.io.ObjectOutputStream.defaultWriteFields(ObjectOutputStream.java:1509)
at java.io.ObjectOutputStream.writeSerialData(ObjectOutputStream.java:1474)
at java.io.ObjectOutputStream.writeOrdinaryObject(ObjectOutputStream.java:1392)
at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1150)
at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:326)
at blab.StashCon.stash(StashCon.java:33)
at blab.Openfire.connect(Openfire.java:27)
at blab.ext.gui.SignIn$4.run(SignIn.java:214)
at java.lang.Thread.run(Thread.java:619)

Answer 1

Your object contains member variables which are themselves not serializable (an instance of org.jivesoftware.smack.XMPPConnection ).

If you really want to serialize your object, you'll have to do something about that member variable. One option would be to declare that variable as transient so that it is not serialized.

On deserialization however, you'll have to handle that member (like reastablishing the connection). For this you could define the method readObject which is called during deserialization. In there you can (and probably should) initialize all transient member variables to set your object to a good state.

Here is also a good question discussing serialization.

Answer 2

XMPPConnection is coupled to physical resources (TCP sockets) on the machine in which it resides, therefore it cannot be made serializable. It is also has an identitiy associated with the connection to the server which cannot be duplicated, since a full JID can only be connected to the same server once, a second connection would force the other one to get disconnected.

The connection should not be put into the session, you have to manage it outside of that scope.