Question

class Test1<T> {
  Test1(Class<T> type) {}
}

class Test2 extends Test1<Class> {
  Test2() {
    super(Class.class);
  }
}

This works, but, I am warned about use of a raw generic type in “Test1<Class>“. I understand that, but, changing it to “Class<?>” (which should be equivalent?) gives a compiler error in the call to super() — Class.class is apparently of type “Class<Class>” instead of “Class<Class<?>>“.

Variants on these all seem to result in a compile error.

Can anyone see how to resolve this, such that I am always using a type param with Class, and hence don’t get an IDE warning? It seems better that way, even if it works as-is.
I am open to changes in the structure of these two classes two as long as it preserves the basic intent: Test2 is a special type of Test1, specialized for “Class” objects.

Answer 1

Your warnings stem from the fact that Class is itself a paramterized class: public final class Class<T>. The simplest way to remove the warnings is to make the T the type parameter instead of Class<T>:

class Test1<T> {
    Test1(T type) { Class<?> classType = type.getClass();}
}

class Test2 extends Test1<String> {
    Test2() {
        super("");
    }
}