In Neal Gafter's "super type token" pattern (http://gafter.blogspot.com/2006/12/super-type-tokens.html), an anonymous object was used to pass in the parameterized type :
class ReferenceType<T>{}
/* anonymous subclass of "ReferenceType" */
ReferenceType<List<Integer>> referenceType = new ReferenceType<List<Integer>>(){
};
Type superClass = b.getClass().getGenericSuperclass();
System.out.println("super type : " + superClass);
Type genericType = ((ParameterizedType)superClass).getActualTypeArguments()[0];
System.out.println("actual parameterized type : " + genericType);
Then result is :
super type : com.superluli.test.ReferenceType<java.util.List<java.lang.Integer>>
actual parameterized type : java.util.List<java.lang.Integer>
My question is, what the magic does the anonymous object "referenceType" do to make it work? If I define a explicit subclass of "ReferenceType" and use it instead of the anonymous style, it doesn't as expected.
class ReferenceType<T>{}
class ReferenceTypeSub<T> extends ReferenceType<T>{}
/* explicitly(or, named) defined subclass of "ReferenceType" */
ReferenceType<List<Integer>> b = new ReferenceTypeSub<List<Integer>>();
Type superClass = b.getClass().getGenericSuperclass();
System.out.println("super type : " + superClass);
Type genericType = ((ParameterizedType)superClass).getActualTypeArguments()[0];
System.out.println("actual parameterized type : " + genericType);
The result is :
super type : com.superluli.test.ReferenceType<T>
actual parameterized type : T
>" was saved in the anonymous class's .class file, while in the 2nd example "ReferenceType" was saved.
– superluli May 01 '14 at 06:48