Issue with Rule Round-up Game

[Sidharth Pallai]

I didn't get the logic behind the output of this code i found in Rule Round-up Game, Question #159. The output is correct,but i've a doubt.

class A
{
int x=5;
void method(){ System.out.println(x); }
}
class B extends A
{
int x=10
public static void main(String a[])
{
new A().method();
}
}

Though class B inherits method(),why it shouldn't access its own member variable x, rather than priting 10, why it prints 9.Can anyone explain.

[Bill Cruise]

Because the object you're calling the method on is type A.

[Ilja Preuss]

Originally posted by Bill Cruise:
Because the object you're calling the method on is type A.

Correct.

But even

new B().method();

would print 5. And *that* is because *fields* are *not* polymorphic, only methods are.

[Sidharth Pallai]

Thank Preuss,
But i didn't get idea of how fields being non-polymorphic.

[Ernest Friedman-Hill]

The variable A.x and the variable B.x are different variables. Even though they have the same name, they're not related to each other, and a B object contains one of each. The code in method() knows that, when it is compiled, "x" means "A.x", and therefore, no matter how that method is called, it's the x in class A that will be used.

Compare your code to this version of class B:

class B extends A
{
public B() { x = 10; }
    public static void main(String a[])
    {
    new B().method();
    }
}

This version prints "10", for two reasons. First, because we're using an instance of class B, not A. But as Ilja said, that wouldn't be enough with the original code, which still prints "5" even if you create an instance of B. In this version, instead of creating a new, separate variable, we just assign a new value to the existing one; and then method() sees that new version.