How can I learn actual type argument of an generic class? [duplicate]

Question

I have a parameterized class :

class ParameterizedClass<T extends AbstractSomething> {
}

calling:

new ParameterizedClass<Something>();

So how can I get actual type Something of T by using Java Generics?

Answer 1

It can be done, but type erasure can make it very hard. As the other answers discuss, you either have to make a subclass of ParameterizedClass or add a field of type T to ParameterizedClass, and the reflection code you need to do it is convoluted.

What I recommend in these circumstances, is to work around the issue like this:

class ParameterizedClass<T> {
    private Class<T> type;

    /** Factory method */
    public static <T> ParameterizedClass<T> of(Class<T> type) {
        return new ParameterizedClass<T>(type);
    }

    /** Private constructor; use the factory method instead */
    private ParameterizedClass(Class<T> type) {
        this.type = type;
    }

    // Do something useful with type
}

Due to Java's type inference for static methods, you can construct your class without too much extra boilerplate:

ParameterizedClass<Something> foo = ParameterizedClass.of(Something.class);

That way, your ParameterizedClass is fully generic and type-safe, and you still have access to the class object.

EDIT: Addressed the comment