Java (1.6) generics wildcard
generics, java
Solution
The way you wrote it, there is no difference.
At compile-time, both have the same type; namely, a `List<>` of some unknown type that extends `A`. This is why you can't add anything to either list – you have no idea whether it's actually a `List<B>` or a `List<C>` or of some other type.
At runtime, they also have the same type; namely, `List`. Due to type erasure, the type parameter doesn't exist at runtime.
Because you don't save any more strongly-typed references to it, no-one can tell that one of the lists is actually a `List<B>`.
The point of these wildcards is for functions. You can make a function that takes a `List<? extends A>`, then pass it a `List<B>` or a `List<C>`, and the calling code can keep using the original list with its original type.
Problem
Consider flowing situation: ``` class A { } class B extends A { } List <? extends A> x = new ArrayList<A>(); List <? extends A> xx = new ArrayList<B>(); ``` both 'x' and 'xx' are legal declarations in Java 6 and Java 7 (I know that in Java 7 you can also substitute the parameterized type of the constructor with an empty set of type parameters (<>). However, I wonder, what is the difference between 'x' and 'xx' in Java 6?