Synchronization code output in wrong order in Java

Question

I am currently learning Java, and I have reached the topic of Synchronization.

For some reasons, the code below (based on the code from The Complete Reference JAVA - Herbert Schildt 7th Edition, Pg no. 239-240) does not give the required output.

Code:

package package1;
class Call{
    synchronized public void call(String msg){
        System.out.print("[" + msg);
        try{
            Thread.sleep(100);
        }catch (InterruptedException e){
            System.out.println("Interrupted Exception Caught");
        }
        System.out.println("]");
    }
}

class CallMe implements Runnable{
    String msg;
    Call target;
    Thread t;

    public CallMe(Call targ, String message){
        target = targ;
        msg = message;
        t = new Thread(this);
        t.start();
    }

    public void run(){
        target.call(msg);
    }
}

public class Synchronization {
    public static void main(String[] args) {
        Call target = new Call();
        CallMe obj1 = new CallMe(target, "Hello");
        CallMe obj2 = new CallMe(target, "Synchronized");
        CallMe obj3 = new CallMe(target, "World");

        try{
            obj1.t.join();
            obj2.t.join();
            obj3.t.join();
        }catch (InterruptedException e){
            System.out.println("Interrupted Exception Caught");
        }
    }
}

The Required Output:

[Hello]
[Synchronized]
[World]

Actual Output (I am using Eclipse on Macbook Pro late 2013):

[Hello]
[World]
[Synchronized]

I have read that all these topic's output differ from computer to computer.

Could anyone please explain me why does not this work?

Yes, there is nothing in this code that would guarantee a specific order. — Oliver Charlesworth
– Oliver Charlesworth, Commented Mar 23, 2014 at 12:47

Peter Keller · Accepted Answer · 2014-03-23 12:47:24Z

3

Because the different CallMe instances are invoked in parallel and you have no guarantee in which order they are processed. The only thing you know is, that Call.call(String) is not invoked in parallel as this method is synchronized.

answered Mar 23, 2014 at 12:47

Peter Keller

7,6442 gold badges28 silver badges31 bronze badges

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

2 Comments

BigFatProgrammer Over a year ago

So this there no way to work around this? Because the book got the "required output" ..

Oliver Charlesworth Over a year ago

@BigFatProgrammer: The naive solution is to not use threads. If you want sequential behaviour, then don't use parallelisation.

N8sBug · Accepted Answer · 2014-03-23 13:06:03Z

0

To piggy-back a little on Peter's answer. All the synchronization does is prevent things from going crazy.

Trying this code without the synchronized call and you can see that the call class can hit .out as it pleases without regard for the other objects. I get an output like this, with "synchronized" removed:

[Hello[World[Synchronized]
]
]

With synchronized in you get something like what you see,.. but it doesn't guarantee order. Sometimes you get:

[Hello]
[World]
[Synchronized]

Sometimes you get:

[Hello]
[Synchronized]
[World]

Run it over and over again and you will see what I mean.
To really throw it off try using:

Thread.sleep(1);

answered Mar 23, 2014 at 13:06

N8sBug

3523 silver badges9 bronze badges

Collectives™ on Stack Overflow

Synchronization code output in wrong order in Java

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related