Question

Is there a way one could avoid type erasure and get access to a type parameter?

public class Foo<T extends Enum<?> & Bar> {
    public Foo() {
        // access the template class here?
        // i.e. :
        baz(T.class); // obviously doesn't work
    }

    private void baz(Class<T> qux) { 
        // do stuff like
        T[] constants = qux.getEnumConstants();
        ...
    } 
}

I need to know about T, and do things with it. Is it possible, and if so, how can it be done without passing in the class in the constructor or anywhere besides the parameter?

EDIT: The main purpose of this question is to find out if there is any practical way around type erasure.

Answer 1

AFACT, there is no practical way around type erasure because you can’t ask for something which the runtime doesn’t have access to. Assuming of course you agree that sub-classing generic classes for each enum which implements Bar interface is a practical work around.

enum Test implements Bar {
    ONE, TWO
}

class Foo<T> extends FooAbstract<Test> {
    public Foo() {
        ParameterizedType genericSuperclass =
                (ParameterizedType) getClass().getGenericSuperclass();
        baz((Class<T>) genericSuperclass.getActualTypeArguments()[0]);
    }

    private void baz(Class<T> qux) {
        T[] constants = qux.getEnumConstants();
        System.out.println(Arrays.toString(constants)); // print [ONE, TWO]
    }
}

interface Bar {
}

class FooAbstract<T extends Enum<?> & Bar> {
}