Getting generic parameter from supertype class

Question

Say I have a parent interface/class like so

interface Parent<T> {}

And a number of implementing interfaces that fix the generic type.

interface Child extends Parent<Type> {}

Can I use reflection to get the instance of Class representing T if I have the Class object for Child. Something like this:

<T, I extends Parent<T>> I create(Class<I> type) {
    Class<T> tType = ...
    ...
}

Currently I'm having tType be passed in as a parameter, but I'd like to simplify things if I can.

Answer 1

Yes, despite what the others have said, this info is available if you have access to the subclass' Class object. You need to use getGenericSuperclass along with getActualTypeArguments.

ParameterizedType superClass = (ParameterizedType)childClass.getGenericSuperclass();
System.out.println(superClass.getActualTypeArguments()[0]);

In your example, the "actual" type argument should return the Class for Type.

Answer 2

If you need to do anything non-trivial with generic types at runtime, consider Guava's TypeToken. It can answer your question (and many more!) while addressing some of the nuanced concerns raised by commenters:

private interface Parent<T> {}
private interface Intermediate<U, V> extends Parent<V> {}
private interface Child<Z> extends Comparable<Double>, Intermediate<Z, Iterable<String>> {}

public void exploreGuavaTypeTokens() {
    final TypeToken<? super Child> token = TypeToken.of(Child.class).getSupertype(Parent.class);
    final TypeToken<?> resolved = token.resolveType(Parent.class.getTypeParameters()[0]);
    System.out.println(resolved); // "java.lang.Iterable<java.lang.String>"
    final Class<?> raw = resolved.getRawType();
    System.out.println(raw); // "interface java.lang.Iterable"
}