Question

I got this from a berkley cs data structures webcast:

class A {
  void f() {System.out.println("A.f");}
  void g() {f();}
 // static void g(A y) {y.f();}
}

class B extends A {
  void f(){
    System.out.println("B.f");
  }
}

class C {
  static void main (String[] args){
    B aB = new B();
    h (aB);
  }

  static void h (A x) {x.g();}
  //static void h (A x) {A.g(x);}  what if this were h 

}

Can you tell me what prints out and why? The instructor said B.f but I do not understand why. I thought it was A.f. Thank you (no, I am not in the class, just trying to learn.)

edit: sorry for the mistakes I was copying it from a video lecture.

Answer 1

The reason “B.f” prints is because the implementation of a method is determined by an object’s run-time type, not its compile-time type. It’s like the virtual keyword in C++.

Once you construct a B, you know that calling its f method will print “B.f”, even if the B is being referred to as an A.

Also, your A.f method is missing a close brace.