Java Inheritance Puzzle

Imagine we have one interface and these classes:

interface SuperType {
    void test();
}


class DefaultSuperType implements SuperType{
    @Override
    void test(){
        System.out.println("0")
    }
}

class ChildOfSuperType extends DefaultSuperType{
    @Override
    void test(){
        System.out.println("1")
    }
}

And then we execute this piece of code

SuperType superType = new ChildOfSuperType();
DefaultSuperType defaultSuperType = (DefaultSuperType) superType;
defaultSuperType.test()

Do you know what will be printed?

Comments

[Sam Rose]

I think it's 1, and it's because the object is always of type ChildOfSuperType no matter what type you store it in, so when the method is dispatched it will always resolve to the version on ChildOfSuperType.

/me goes to check.

Spoiler here for anyone wondering what the answer is: repl.it/@samwho/KnownEnormousWeara...

[Jason R Tibbetts]

1. It doesn't matter whether superType is cast as DefaultSuperType, because it's still a ChildOfSuperType, and the function dispatch table will still point to ChildOfSuperType.test().

[Meghan (she/her)]

The others were right in saying "1", but you also have the annotation wrong. It's just @Override

[Ildarov]

Thank you. I corrected it.

[Ananthu]

Despite the variable defaultSuperType being of type DefaultSuperType, it refers to an object of type ChildOfSuperType. Therefore, when test() is called on defaultSuperType, it invokes the overridden version of test() in ChildOfSuperType, resulting in 1

[Navneet Karnani]

"1". Because, in Java, the actual object's implementation is used, and not the typecast.