Real vs Apparent classes in Java

Question

class A { public static void main(String[] args) 
{ A a = new A(); 
  B b = new B(); 
  A ab = new B(); 
  System.out.format("%d %d %d %d %d %d", a.x, b.x, ab.x, a.y, b.y, ab.y); } 
  int x = 2; 
  int y = 3;
  A(int x) { this.x = x; } 
  A() { this(7); } } 

class B extends A { 
  int y = 4;
  B() { super(6); 
  }

Hey all, I was just looking through some examples from my course and came across this problem that stumped me.
I realize that that this code should print out "7 6 6 3 4 3"

But why is ab.y equal to 3? Isn't the "real" type of object ab of class B? Which then would lead me to believe that ab.y be 4?

Answer 1

Because you are accessing the fields directly, and not via getter methods.

You cannot override fields, only methods.

Class B has a field y in addition to the one in parent class A. The two fields do not interfere with each-other, which one gets picked up is determined at compile-time (by the type known to the compiler).

If you say

   A ab = new B(); 

then

  ab.y

will be compiled to look at the field y declared in class A. This is not dispatched at runtime to the real instance's class. It would be the same with a static method.

If you do

    B ab = new B();

then you get the field in class B.