socket programming serialized objects

Question

i am having a very weird situation in my code which i dont understand i am sending an object lets say O through a socket then i am changing the value of a variable in the object and sending it again but the second time when i print it on the client side, i am getting the same values as in the 1st object.

client code:

 while(true){
             try{
             order=(Order)ois.readObject();

            System.out.println(order);

             }

server code:

public void sendOrder(Order o){
    try {
        out.writeObject(o);

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

main method:

Server_Socket ss=new Server_Socket();
    ss.sendOrder(o);
    o.add(r2);
    ss.sendOrder(o);

The value is definitely changing on the serverside before i send it, but i dont understand why on the client side its not showing that r2 added in the object.

user207421 · Accepted Answer · 2012-09-26 01:46:29Z

5

The objects are being cached by the ObjectOutputStream. To prevent this, call ObjectOutputStream.reset() after each write. If you are sending simple objects that don't contain other objects, use writeUnshared() instead of writeObject().

answered Sep 26, 2012 at 1:46

user207421

312k45 gold badges324 silver badges493 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

yahh Over a year ago

would using reset everytime make that much of a difference on the timing

user207421 Over a year ago

@yahh Call it whenever you want to send a fresh copy rather than every time if you like, but the overhead is pretty minimal.

Alex Lynch Over a year ago

as a note on this, java caches for the objects for a reason. it's java's way to telling you that it's not super good practice to be writing mutable objects over a socket. sadly, i couldn't tell you exactly why.

user207421 Over a year ago

@AlexLynch No, it isn't "Java's way of telling you it's not super good practice to be writing mutable objects" at all. It is Java's way of implementing the Object Serialization Specification, which provides that object graphs are transmitted correctly, including cyclic graphs. It is impossible to implement that without some kind of 'handle' system that remembers what has already been transmitted and plants handles to those objects instead of retransmitting them.

Alex Lynch Over a year ago

@EJP can you point me to a resource that explains the process and reasoning more in-depth? i am interested, but don't fully understand based on your comment.

|

allingeek · Accepted Answer · 2012-09-26 01:44:32Z

0

The objects are being cached by the IOStreams. To fix this, create a deep clone on the server prior to sending the object back. When the client pulls the object from the stream, it will have a different instance id and it will actually deserialize and instantiate the object on the client side.

Fun stuff.

answered Sep 26, 2012 at 1:44

allingeek

1,3988 silver badges16 bronze badges

1 Comment

user207421 Over a year ago

Much too complex and error-prone. A far simpler solution exists.

Collectives™ on Stack Overflow

socket programming serialized objects

2 Answers 2

6 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related