通配符上的Java语言规范 [英] Java language specification on wildcards

问题描述

我正在通过此链接(第4章.类型，值和变量)，并且不理解以下几点:

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

通配符与已建立类型理论的关系是一个有趣的关系，我们在这里简要地提及.通配符是存在类型的一种受限形式. Given a generic type declaration G<T extends B>, G<?> is roughly analogous to Some X <: B. G<X>.

如果您提供了很好的例子来清楚地理解上述要点，我将不胜感激.

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

谢谢.

推荐答案

此语句的措词和格式有点不巧 ^* .实际上，答案中的链接涵盖了一般主题，但您可以尝试着重于特定情况Java和通配符在这里:

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:

通配符是存在类型的一种受限形式.给定一个通用类型声明G，G大致类似于Some X <:B. G.

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

这基本上表示G的类型参数是B的任何子类型.即使您没有明确说出来，这也是总是的情况.

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.

考虑以下片段，希望可以说明这一点:

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();
    }
}

通过调用g.get()获得的对象的类型为B，因为G<T extends B>的声明保证了任何类型参数(即使它是?通配符)总是至少"为B类型.

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.

(与此相反:如果仅声明是G<T>，则从g.get()获得的类型将仅是Object类型)

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

与类型理论符号的关系描述为"大致类似".您可能会想像这句话:如果声明为G<T extends B>，并且使用类型为G<?>，则这种大致(！)的含义是:存在类型为X extends B，此处的?表示该类型. (未知)，键入X.

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.

顺便说一句:请注意，这也指

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();

都有效.

^* 不吉利"的格式指的是该段的源代码实际读取的事实

^* 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> ...

更清楚地表明"Some"一词已经是在其中正式定义的 type 的一部分...

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

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