CODEWITHSUNDEEP
java
Matteo Moroni
Matteo Moroni

Reputation: 41

Exercise on method call in class hierarchy

I'm trying to solve a problem about method calls in Java that seems really strange, I hope someone can help about it. Here it is the code of the three classes, A is the top class with C inheriting from A and D inheriting from C.

class A {
    int f(double x) {
        return 1;
    }

    int f(boolean x) {
        return 2;
    }
}

class C extends A {
    int f(double x) {
        return 3;
    }

    int f(int x) {
        return 4;
    }
} 

class D extends C {
    int f(double x) {
        return 5;
    }   

    int f(int x) {
        return 6;
    }

} 

public class test {
    public static void main(String[] args) {
        C b = new D();
        A d = new D();

        System.out.println(b.f(33));
        System.out.println(d.f(33));
    }
}

When I call method of object b with an int I get 6, which means that the method f(int x) from class D is called. But when I call the same method on object d I get 5, which means that the method f(double x) from class D is called. Why does it behave like that? Why an integer is considered an integer in one case and a double in the other? It is something related to how I declared the two objects but I can't get it.

Upvotes: 0

Views: 138

Answers (1)

Elan Hamburger
Elan Hamburger

Reputation: 2177

The difference is in the type of the objects. b's reference is of type C and d's reference is of type A. In cases of inheritance, the JVM binds the methods at run-time, meaning that you can only call methods of the reference type, but the overridden method is the one that will run.

When you call the function on d, the JVM automatically casts to a double because A has no corresponding method for an int argument. Then, the method trickles down to D because the object is an instance of D, but it still trickles down as a double.

Upvotes: 1

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