Java language specification on wildcards

问题

I am going through this link (Chapter 4. Types, Values, and Variables) and did not understand below point:

The relationship of wildcards to established type theory is an interesting one, which we briefly allude to here. 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>.

I appreciate if you provide good example to understand above point clearly.

Thanks in advance.

回答1:

The wording and formatting of this statement are a bit unlucky^*. The link in the answer by Maouven actually covers the general topic pretty well, but one can try to focus on the particular case of Java and Wildcards here:

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

This basically says that the type parameter of the G is any subtype of B. And this is always the case, even when you don't say it explicitly.

Consider the following snippet, which hopefully illustrates the point:

class B { } 
class G<T extends B> 
{
    T get() { return null; }
}

public class Example
{
    public static void main(String[] args)
    {
        G<?> g = null;

        // This works, even though "G<?>" seemingly does not say 
        // anything about the type parameter:
        B b = g.get();
    }
}

The object that you obtain by calling g.get() is of type B, because the declaration of G<T extends B> guarantees that any type parameter (even if it is the ? wildcard) always be "at least" of type B.

(In contrast to that: If the declaration only was G<T>, then the type obtained from g.get() would only be of type Object)

---

The relationship is described as "roughly analogous" to the type theoretic notation. You can probably imagine this as saying: If the declaration is G<T extends B>, and you use the type G<?>, then this roughly (!) means: There exists a type X extends B, and the ? here stands for this (unknown) type X.

---

An aside: Note that this also refers to Insersection Types. If you declared the class as class G<T extends B & Runnable>, then the statements

B b = g.get();
Runnable x = g.get();

would both be valid.

---

^* The "unlucky" formatting referred to the fact that the source code of this paragraph actually reads

    ... is roughly analogous to <span class="type">Some <span class="type">X</span> ...

making clearer that the word "Some" already is part of the type that is being defined there formally...

回答2:

Wildcards are a restricted form of existential types, in the way they incorporate the principle of existential types in Java. You can refer to the link here which provide explanatory examples: What is an existential type?

来源：https://stackoverflow.com/questions/35577532/java-language-specification-on-wildcards