Getting a generic parameter from a supertype class

Let's say I have a parent interface / class similar to

interface Parent<T> {}


And a series of implementation interfaces that capture the generic type.

interface Child extends Parent<Type> {}


Is it possible to use reflection to get an instance Class

representing T

if I have an object Class

for Child

. Something like that:

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


I am currently passing tType

as a parameter, but I would like to simplify everything if I can.


source to share

3 answers

Yes, despite what others have said, this information is available if you have access to the subclass object Class

. You must use getGenericSuperclass

together with getActualTypeArguments


ParameterizedType superClass = (ParameterizedType)childClass.getGenericSuperclass();


In your example, an argument of type "actual" should return Class

for Type




If you need to do something non-trivial with generic types at runtime, consider Guava TypeToken

. He can answer your question (and more!) By addressing some of the nuanced issues raised by commentators:

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"




I do not think so. Read about erasure dash : generic types are only used for compile-time checking and then discarded. They are not stored in compiled class files, so they are not available at runtime.



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