Java procedures of extending abstract class

Question

I am not clear about the procedure of extending a class. Given the following piece of code, why the output is 32?

class Rts {
    public static void main(String[] args) {
        System.out.println(zorg(new RtsC()));
    }

    static int zorg(RtsA x) {
        return x.f()*10 + x.a;
    }
}

abstract class RtsA {
    int a = 2;
    int f() { return a; }
}

class RtsB extends RtsA {
    int a = 3;
    int f() { return a; }
}

class RtsC extends RtsB {
    int a = 4;
}

Answer 1

First off, fields aren't overridden, so all this is equivalent to

public class Rts {
    public static void main(String[] args) {
        System.out.println(zorg(new RtsC()));
    }
    static int zorg(RtsA x) {
        return x.f()*10 + x.a;
    } 
}

abstract class RtsA {
  int a = 2;
  int f() { return a; }
}
class RtsB extends RtsA {
  int b = 3;
  int f() { return b; }
}
class RtsC extends RtsB {
  int c = 4;
}

The implementation of f() for an object of type RtsC comes from RtsB, since that is the lowest-level class that overrides f(), so its implementation is used, and that returns b, which is 3. That's multiplied by 10, and then added to the a from RtsA, since zorg only knows that x is of type RtsA, so that field is used. That's 3 * 10 + 2 = 32.

(Note that the fact that RtsA is abstract didn't come into this at all; that mostly only matters when you have abstract methods to worry about.)