Find generic method with actual types from getDeclaredMethods

Question

I have generic classes that look like:

interface X<Input, Output>
{
  Output process(Input input);
}

class Y implements X<Integer, Float> 
{   
  Float process(Integer input); 
}

I use getDeclaredMethods to find process on Y just with its name (not the arguments, on purpose). When I look at the return Method[], process shows up twice, with Input=Object, Output=Object, and then with the actual instantiation types: Integer and Float.

Please note: I can see 1 function with Object,Object in the Method[] AND 1 function with the actual types I use to instantiate, like Integer,Float. So, the second function, which I'm interested in, is accessible from Method[].

What is the best way to get only the method with the actual types?

Do you have actual subclasses of `X` with hardcoded values for `Input` and `Output`? — Louis Wasserman, Jan 03 '13 at 18:53
Possible duplicate of http://stackoverflow.com/questions/3403909/get-generic-type-of-class-at-runtime — sorencito, Jan 03 '13 at 19:37
Note also, that I slightly modified the question, from class to interface, which is actually my real case. Sorry for the confusion. — Frank, Jan 03 '13 at 20:17
Curious why you are looking for this. You *can* call the generic method (with types erased) using reflection. — Miserable Variable, Jan 03 '13 at 20:35

score 5 · Accepted Answer · answered Jan 03 '13 at 20:19

and then with the actual instantiation types.

That's not possible. You should see a single method whose parameter types are Object. You can also look at the method's generic parameter and return types, and you will find that they are both type variables (Input and Output are type variables).

Subclasses of this class that inherit from it with specific type arguments for the type parameters, will have two methods: a method with the more specific parameter and return types, and a bridge method, with the original class's method's parameter and return types, in order to override it. If you are asking how to ignore the bridge method, simply check if it's a bridge method (.isBridge())

Check it out. Note my change of the original question to interface/class. — Frank, Jan 03 '13 at 20:23

score 0 · Answer 2 · answered Jan 03 '13 at 18:33

0

Generic types are known only at compilation time. Object is the actual type of generics at runtime, so you can't get them while running the program.

answered Jan 03 '13 at 18:33

user1871166

258
2
9

Are you saying that Object is the dynamic type of generic parameters? – Frank Jan 03 '13 at 18:35
Due to Java's type erasure concept, that's the way generics work at runtime. – sorencito Jan 03 '13 at 19:34
@Frank: parameter type is a compile-time (static) thing. "dynamic type of generic parameters" does not make sense. – newacct Jan 03 '13 at 20:15

Find generic method with actual types from getDeclaredMethods

2 Answers2