What happens when an object is assigned to another object

public class DrumKitTestDrive {
/**
 * @param args
 */
public static void main(String[] args) {
    // TODO Auto-generated method stub
    Echo e1 = new Echo();
    Echo e2 = new Echo();
//      **e2 = e1;**

    int x=0;
    while( x < 4 ){
        e1.hello();
        e1.count = e1.count + 1;
        if(x==3){
            e2.count = e2.count + 1;
        }
        if(x>0){
            e2.count = e2.count + e1.count;
        }
        x = x + 1;
    }
    System.out.print(e2.count);
    }
}

class Echo {
    int  count = 0;

    void hello(){
        System.out.println("Hellooooo....");
    }
}

The output of this code will be:

Hellooooo....
Hellooooo....
Hellooooo....
Hellooooo....
10

But if I remove the comments from // e2= e1; when I run the code the system will print 24 instead of 10. I don't understand why is it so?

What I understand is the system will just copy the values of e1 instance into e2 instance. And if the system will do so, the result shall remain the same because both objects are of the same class.

Solution 1:

What I understand is the system will just copy the values of e1 instance into e2 instance.

No, when you do e2 = e1 you're copying object references - you're not making a copy of the object - and so the variables e1 and e2 will both point to the same object.

And so when you do your increments, they're all incrementing the same count field.

It's only without the assignment e2 = e1 that the increments are happening on two different instances.

Solution 2:

This is one of the basic principles of Java. Each variable, is just a handle to some object. Executing

e1=e2

doesn't copy the values of the objects, it points both handles to the second object that was created. The first Echo object that you instantiated is now not referenced at all, and will be garbage collected the next time gc runs.