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).