Question

Consider this:

class A  {
    int x =5;
}

class B extends A{
        int x =6;
    }
public class CovariantTest {

    public A getObject() {
        return new A();
    }

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        CovariantTest c1 = new SubCovariantTest();
        System.out.println(c1.getObject().x);
    }

}

class SubCovariantTest extends CovariantTest {
    public B getObject(){
        return new B();
    }
}

As far as I know, the JVM chooses a method based on the true type of its object. Here the true type is SubCovariantTest, which has defined an overriding method getObject.

The program prints 5, instead of 6. Why?

Answer 1

The method is indeed chosen by the runtime type of the object. What is not chosen by the runtime type is the integer field x. Two copies of x exist for the B object, one for A.x and one for B.x. You are statically choosing the field from A class, as the compile-time type of the object returned by getObject is A. This fact can be verified by adding a method to A and B:

class A  {
    public String print() {
        return "A";
    }
}

class B extends A {
    public String print() {
        return "B";
    }
}

and changing the test expression to:

System.out.println(c1.getObject().print());