Why is ? used in generics, and is it equivalent to T/E?

Question

Java

I met such a design . I never thought about why I need it . Is the code above equivalent to this: . If so, why use . Function<? super T, ? extends U>
?
Function<E super T, E extends U>
?

Answer 1

Quoting approximately Eckel:
<?> - unlimited mask, means "anything", in most cases the compiler does not see the difference between T and <?>.
Without ? you can do more with parameters, but with ? - accept a wider range of parameters.

Answer 2

? extends U - some type that is a descendant of U, ? super T - some type that is an ancestor of T. Your scheme (as you have written) is usually applied: the first part for the receiver, the second for the source when processing something.