Runtime Polymorphism giving wrong output

Question

To my understanding, the following code should print a as per my knowledge of run time polymorphism.

However, when I run the following code it is printing b:

Per JLS 8.4.8.1, B1.m1 does not override A1.m1, and so when A1.m1 is invoked, B1.m1 should not be selected

package a;

public interface I1 {
    public Object m1();
}

public class A1 {
    Object m1() {
        return "a";
    }
}

public class C1 extends b.B1 implements I1 {
    public static void main(String[] args) {
        a.A1 a = new a.C1();
        System.out.println(a.m1());
    }
}

package b;

public class B1 extends a.A1 {
    public String m1() {
        return "b";
    }
}

Can some one help me understand this behavior.

Answer 1

The expected output is indeed b.

When you declare your object a as being of the type A1, that class defines only the interface of the methods. It defines that m1 returns a String, but the implementation of that method is defined by the Class used to build the object, which is Test1. And Test1 extends B1, which overrides the method m1, so that is the implementation of m1 used for your object.

The output of that call m1() must be indeed the B1's.

EDIT: This answer was written for the first version of the question. OP changed a lot of the code, but the root of the explanation is still the same.

Answer 2

After adding the packages, the question is much more difficult. I've tried this, and I changed your main program to

public class C1 extends b.B1 implements I1 {
    public static void main(String[] args) {
        a.A1 a = new a.C1();
        System.out.println(a.m1());
        a.A1 b = new b.B1();
        System.out.println(b.m1());
    }
}

(I actually used different package names to avoid conflicts with the variable names. So my code looks a bit different from the above.)

I've confirmed that this prints "b" and "a". That is, if I create a new B1, its m1 method does not override the one in A1. Thus if I print b.m1(), since b is of type A1, polymorphism doesn't kick in, and the method declared in A1 is called. So what's going on with C1?

C1 inherits the m1 method from B1. But even though the m1 method in B1 doesn't override the one in A1, the m1 method in C1, which it inherits from B1, actually does override the one in A1. I think it has to do with this clause in 8.4.8.1:

mA is declared with package access in the same package as C, and either C declares mC or mA is a member of the direct superclass of C.

Here C is your C1 class. mC is the m1 that's inherited from B1. In this case, "C declares mC" is false, because C1 doesn't declare m1, it inherits it. However, I believe "mA is a member of the direct superclass of C" is true. As I understand it, B1 has all the members that A1 has. B1 declares its own m1, and since it's not overriding, it's a new m1 that causes the m1 it inherits from A1 to be hidden. But even though it's hidden, it's still a member. Thus the condition that mA is a member of the direct superclass of C (which is B1) is satisfied, and thus all the conditions of 8.4.8.1 are satisfied and thus the inherited m1 in C1 overrides the one in A1.

Answer 3

The following line A1 a = new Test1(); simply means build a Test1 instance and store it in a A1 box.

So the instance will be a Test1, but you will only have access to the method/variable of A1, but every overriden method in Test1 will be accessed.

This is polymorpish.

By reading the JLS about 8.4.8.1. Overriding (by Instance Methods) about the accesor

An instance method mC declared in or inherited by class C, overrides from C another method mA declared in class A, iff all of the following are true:

• A is a superclass of C.
• The signature of mC is a subsignature (§8.4.2) of the signature of mA.
• mA is public.

You can find more information about the access modifiers in 8.4.8.3. Requirements in Overriding and Hiding

The access modifier (§6.6) of an overriding or hiding method must provide at least as much access as the overridden or hidden method, as follows:

• If the overridden or hidden method is public, then the overriding or hiding method must be public; otherwise, a compile-time error occurs.
• If the overridden or hidden method is protected, then the overriding or hiding method must be protected or public; otherwise, a compile-time error occurs.
• If the overridden or hidden method has package access, then the overriding or hiding method must not be private; otherwise, a compile-time error occurs.

EDIT :

Now, with your package added.

Having C1 to implement m1 (because of the interface), you are hiding the method of A1 with the implementation you find in B1, this method is indeed a valid definition for the interface contract.

You can see you are not overriding the method (you can't call super.m1 or even add @Override on the B1.m1. But the call a.m1() is valid as it is define in the class itself.

Answer 4

You are Overriding. Include the @Override annotation and you can see that. As long as your class extending can override the parent class method, you can increase the access, but not decrease the access.

If you tried to make B#m1 private, then somebody could just cast to an A and use the method.

Conversely, if you make A#m1 private, then B cannot override it and you can end up with an object having two methods with the same signature.

static class A{

    private String get(){
        return "a";
    }

}

static class B extends A{

    public String get(){
        return "b";
    }
}
public static void main (String[] args) throws java.lang.Exception
{
    A b = new B();
    System.out.println(b.get());
    System.out.println(((B)b).get());
    // your code goes here
}

This will output:

• a
• b

Answer 5

You have an interface I1, which is implemented by A1 Class B1 extends A1 Class C1 extends B1 (and therefore implicitly extends A1).

So an instance of C1 is also of type B1, A1 & I1, however it remains an instance of C1 regardless of whether you assign it to one of the other types.

If you have:

 I1 instance = new C1();
    String value = instance.m1();

The first m1 method going up the inheritance tree from the real type (C1) will be called, which will be in B1 and return "b".

Answer 6

All comments and answers are mostly correct. They explain things in term of language mechanisms. I think, instead, that to realise the real meaning of inheritance and polymorphism and how to use them, you have to take a more conceptual approach.

First of all inheritance is a relationship between two things and the relationship is of the type “is a”. In other words when you write the statement class C1 extends B1 you mean C1 is a B1. Of course this won’t work with A1, B1 and C1. Let me change them in something more real. For example:

A1 = Animal

B1 = Feline

C1 = Cat and C2 = Lion (polymorphism)

At this point you will have class Cat extends Feline, and you can conceptually read it as: Cat is a Feline. I suggest to challenge your inheritance formal correctness using the “is a” test. If it doesn't work, it is better to reconsider or rethink the inheritance. Your resulting code will be like the following:

public interface IAnimal {
    public Object saySome();
}

public class Animal {
    Object saySome() {
        return "I am an animal";
    }
}

public class Feline extends Animal {
    public String saySome() {
        return "I am a feline";
    }

public class Cat extends Feline implements IAnimal {
    Object saySome() {
        return "meow";
    }
}

public class Lion extends Feline implements IAnimal {
    Object saySome() {
        return "roar";
    }
}

class Aplication {
    public static void main(String[] args) {
        Animal anAnimal = new Cat();
        Animal anotherAnimal = new Lion();
        System.out.println(anAnimal.saySome());
        System.out.println(anotherAnimal.saySome());
    }
}

And clearly the output will be

meow

roar

I hope this will help.