Java中的协方差和重载

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 Bf(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.

Answer 3

Since you are using the object type of A, you can call only f(AA). Because B overrides it.

Answer 4

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

Think otherwise :-

try commenting below in class B (the overriding code)

boolean f(A a) { return false; } // override A.f(A)

and add syso in Class A in this method --> boolean f(A a){....}

Then you'll see, ab.f(A a) or ab.f(B b) will invoke f(A a) method of class A only.

As ab is of type A.

Also please note - You'll not be able to call any method of class B as well from object ab.

Hope this further adds more clarification to above brilliant answers.

Lastly, you may seek this now --> Why do we assign a parent reference to the child object in Java?