Question

Java tutorial says that <?> and <T> are interchangeable. Why then can I compile line 1 below, while I cannot compile line [2]?

abstract class A<K extends Number>{
  abstract public A<?> f(A<?> k); //[1]
  abstract public <S> A<S> f(A<S> k); //[2]
} 

Answer 1

After many hours of reading and searching, I have eventually found the answer to my own question.
First of all, I have to say about <?>, there is some information here.

So what is the supertype of all kinds of collections? It’s written
Collection<?> (pronounced “collection of unknown”), that is, a
collection whose element type matches anything.(Java tutorials)

Ok, so our A<?> is super type of all kinds of As.
And parametr A<?> can accept any of As (polymorphically) as a argument therefore line 1 compiles.

Java specification tells us:

Wildcards are a restricted form of existential types. Given a generic
type declaration G<T extends B>, G<?> is roughly analogous
to Some X<:B. G<X>

B stands for bounds; X<:B indicates that the subtype relation holds between
types X and B.

Therefore A<?> is indeed auto-restricted;

When we declare type argument <S>, it is, in some way, as if we declared class S{}, type S don’t have any relations to Number(our bound) and cast to one would fail, so we should declare that “S extends Number” to have the same effect as for <?>.