Covariance and Overloading in Java

Question

class A {    boolean f(A a) { return true; } } class B extends A {    boolean f(A a) { return false; } // override A.f(A)    boolean f(B b) { return true; }  // overload A.f }  void f() {      A a = new A();    A ab = new B();    B b = new B();    ab.f(a); ab.f(ab); ab.f(b); //(1) false, false, *false*    b.f(a); b.f(ab); b.f(b);    //(2) false, false, true } 

Can you please explain the first line last false output, why it is not true?

Answer 1

can you please explain the first line last false output, why it is not true?

The compile-time type of ab is A, so the compiler - which is what performs overload resolution - determines that the only valid method signature is f(A a), so it calls that.

At execution time, that method signature is executed as B.f(A a) because B overrides it.

Always remember that the signature is chosen at compile time (overloading) but the implementation is chosen at execution time (overriding).

Answer 2

Well..because you are calling on type of A. So you can call only version of f(A a). that's returning false in B.