Question

I was trying following example:

class BaseClass {
    public void methodA(Class<?> cl) {
        System.out.println("Base.methodA()");
    }
}

class SubClass extends BaseClass {
    public void methodA(Class cl) {
        System.out.println("Sub.methodA()");
    }
}

public class OverrideEx {

    public static void main(String[] args) {
        BaseClass b = new BaseClass();
        BaseClass s = new SubClass();

        b.methodA(Class.class);
        s.methodA(Class.class);
    }
}

Output:
Base.methodA()
Sub.methodA()

But if I change over riding method argument other way around as follows:

class BaseClass {
    public void methodA(Class cl) {
        System.out.println("Base.methodA()");
    }
}

class SubClass extends BaseClass {
    public void methodA(Class<?> cl) {
        System.out.println("Sub.methodA()");
    }
}

public class OverrideEx {

    public static void main(String[] args) {
        BaseClass b = new BaseClass();
        BaseClass s = new SubClass();

        b.methodA(Class.class);
        s.methodA(Class.class);
    }
}

I get compilation error. It says “Name clash: The method methodA(Class) of type SubClass has the same erasure as methodA(Class) of type BaseClass but does not override it”.

Why is that?

Answer 1

The parameterized type Class<?> is a subtype of the raw type Class (§4.10.2).

Thus for any x the call s.methodA(x) in the first example would also be valid if rewritten:

((BaseClass)s).methodA(x)

This is not true for the second example.

Imagine x being of type Class which is not a subtype of Class<?>. That call
would be illegal because the argument’s type is not a subtype of the formal parameter’s type.

This means that methodA of SubClass does not override methodA of BaseClass.