通配符上的Java语言规范

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:

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:

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，这也指插入类型.如果您将该类声明为class G<T extends B & Runnable>，则该语句

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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