Explain how variable hiding is working in this Java code

Question

Consider below code

class A
{
    int x = 5;
    void foo()
    {
        System.out.println(this.x);
    }
}
class B extends A
{
    int x = 6;
    // some extra stuff
}
class C
{
    public static void main(String args[])
    {
         B b = new B();
         System.out.println(b.x);
         System.out.println(((A)b).x);
         b.foo();
    }
 }  

Output of the program is

6
5
5

I understand the first two but can't get my head around the last one. How does b.foo() print 5. B class will inherit the foo method. But shouldn't it print what b.x would print? What exactly is happening here?

Answer 1

Yes, the B class inherits the foo method. But the variable x in B hides the x in A; it doesn't replace it.

This is an issue of scope. The foo method in A sees only the variables that are in scope. The only variable in scope is the instance variable x in A.

The foo method is inherited, but not overridden, in B. If you were to explicitly override foo with the same exact code:

class B extends A
{
    int x = 6;

    @Override
    void foo()
    {
        System.out.println(this.x);
    }
}

Then the variable that would be in scope when referred to by this.x would be B's x, and 6 would be printed. While the text of the method is the same, the reference is different because of scope.

Incidentally, if you really wanted to refer to A's x in the B class, you can use super.x.