Field shadowing or overriding?

Question

I have two classes (A and B) and B extends A.

public class A {
   protected int i = 1;
}

public class B extends A{
   protected int i = 2;
}

In this case the program writes 1.

A a = new B();
System.out.println(a.i); //1

But if I assign value i in the constructor it writes 2.

public class B extends A{
   public B(){
      i=2;
   }
}

A a = new B();
System.out.println(a.i); //2

Why?

Your first example, `B.i` is hiding `A.i`. There is no overriding fields, only hiding them. — Sotirios Delimanolis, Feb 26 '14 at 21:17
Fields cannot be overridden. Only non-static and non-private methods can be overridden. — Jesper, Feb 26 '14 at 21:17
But if B.i is hiding A.i, why not writes 2 because of the dynamic binding? — user2693979, Feb 26 '14 at 21:18
There is no dynamic binding with fields. They are resolved based on the static type of the reference. — Sotirios Delimanolis, Feb 26 '14 at 21:19
But if in the second case there isn't dynamic binding, why does the program write B.i? — user2693979, Feb 26 '14 at 21:21
So in both case the program writes A.i, but in the second case I changed A.i to 2? — user2693979, Feb 26 '14 at 21:23

score 2 · Answer 1 · edited May 23 '17 at 12:28

From the official doc:

Within a class, a field that has the same name as a field in the superclass hides the superclass's field, even if their types are different. Within the subclass, the field in the superclass cannot be referenced by its simple name. Instead, the field must be accessed through super, which is covered in the next section. Generally speaking, we don't recommend hiding fields as it makes code difficult to read.

Fields aren't polymorphic, and you keep a reference on a A object. By executing i = 2, you change the value of the field, hence the result modified.

Related question: Hiding Fields in Java Inheritance

Field shadowing or overriding?

1 Answers1