泛型通配符实例化 [英] Generics wildcard instantiation

问题描述

我有一天正在审查别人的代码，并且遇到了一条引起关注的线路。为了简化，假设我有一个泛型类A和一个抽象类B.是否允许下面的实例化，如果是这样，为什么？

  Object obj = new A< ;?扩展B>（）; 
  

我个人从来没有见过像上面那样的实例，尽管声明如

  A <？扩展B> obj = null; 
  

肯定会成立。我一直使用泛型中的通配符来声明方法参数，所以我可能只是没有经验。

解决方案

其实新的A< ;?扩展B>（）不能编译。自从Java 5以来，它一直是非法的。

但我想你的原始示例是类似于 new A< X<扩展B>>（）。后者在最近版本的Java中是合法的。

这个想法是，在实例化对象时，类型参数的值可以是任何非通配符类型。 ？扩展B 是通配符类型，因此不允许使用。但 X <？扩展B> 不是通配符类型，尽管它具有通配符类型作为组件。所以你可以合法地称呼新的A< X<如果你这样想的话，这些规则是有意义的。最终它是更基本规则的副产品，像这样的通配符类型？扩展B 不能是字段或变量的声明类型。如果 A 定义为



  class A< T> {
 T值; 
} 
  

然后是假设的 new A< ;?扩展B>（）。value 将是一个声明为 类型的字段。扩展B 。既然这是非法的，那么实例化也是非法的。但新的A< X<扩展B>>（）没有这个问题。

I was reviewing someone else's code the other day and I came across a line that raised some concern. To simplify, say I have a generic Class A and an abstract Class B. Is the following instantiation allowed and if so, why?

Object obj = new A<? extends B>();

I personally have never seen an instantiation like the above, although a declaration such as

A<? extends B> obj = null;

would certainly hold. I've always used the wildcard in generics to declare method parameters, so I may just not have the experience.

解决方案

Actually new A<? extends B>() does not compile. It has been consistently illegal since Java 5.

But I guess your original example was something like new A<X<? extends B>>(). The latter is legal in recent versions of Java.

The idea is, when instantiating an object, the value for type parameters can be any non-wildcard type. ? extends B is a wildcard type, so it is disallowed. But X<? extends B> is not a wildcard type, though it has a wildcard type as a component. So you can say legally call new A<X<? extends B>>().

The rules makes sense if you think about it this way. Ultimately it is a byproduct of the more fundamental rule that a wildcard type like ? extends B cannot be the declared type of a field or variable. If A is defined as

class A<T> {
    T value;
}

then the hypothetical new A<? extends B>().value would be a field declared of type ? extends B. Since that is illegal, so is the instantiation. But new A<X<? extends B>>() does not have that problem.

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