Question

My code is:

class Foo {
  public int a=3;
  public void addFive() {
    a+=5;
    System.out.print("f ");
  }
}

class Bar extends Foo {
  public int a=8;
  public void addFive() {
    this.a += 5;
    System.out.print("b ");
  }
}

public class TestClass {
  public static void main(String[]args) {
    Foo f = new Bar();
    f.addFive();
    System.out.println(f.a);
  }
}

Output:

b 3

Please explain to me, why is the output for this question “b 3” and not “b 13” since the method has been overridden?

Answer 1

You cannot override variables in Java, hence you actually have two a variables – one in Foo and one in Bar. On the other hand addFive() method is polymorphic, thus it modifies Bar.a (Bar.addFive() is called, despite static type of f being Foo).

But in the end you access f.a and this reference is resolved during compilation using known type of f, which is Foo. And therefore Foo.a was never touched.

BTW non-final variables in Java should never be public.