嵌套有界通配符 [英] Nested Bounded Wildcard

问题描述

当我尝试编译以下代码时：

  LinkedList< List <？扩展Number>> numList = new LinkedList< List< Integer>>（）; 
  

我得到一个不兼容的类型错误：

 必需：LinkedList< java.util.list< extends java.lang.Number>> 
 Found：LinkedList< java.util.list< Integer>> 
  

如何获得 LinkedList 包含元素 List s元素扩展数字？

为了说清楚，我正在寻找以下列方式将列表添加到 numList 中：


numList.add（new LinkedList< Integer>（））;


解决方案

通配符捕获不会超过一个通用级别。因此，虽然这工作：

  LinkedList <？扩展Number> test = new LinkedList< Integer>（）; 
  

这不会：

 链表<名单，LT ;?扩展Number>> numList = new LinkedList< List< Integer>>（）; 
  

我能想到的最合理的解释是将泛型看作最外层的不变量。 A LinkedList< List< Integer>> 不是 LinkedList< List<即使 List< Integer> 是 List< ;?扩展Number>> 由于与 List< Dog> 不是 List< Animal> c $ c>即使 Dog 是 Animal 。在这里， Dog 是 Animal as List< Integer> 是列表<？扩展数字> 。

那么，Dog / Animal解决方案是？ ：

 列表< ;?延伸动物> animals = new List< Dog>（）; 
  

应用相同的推理，解决方法是另一个？扩展：

  LinkedList <？扩展List <？扩展Number>> numList = new LinkedList< List< Integer>>（）; 
  

但是，您将无法向此列表添加任何内容，因为第一个？延伸。引用类型变量 numList 不知道 List <>的哪个子类型？扩展Number> 它确实是;它可以是 ArrayList< Integer> ，所以Java不能提供类型安全性，以至于可以将这样的东西添加到这样的 LinkedList 。为了保持类型安全，编译器只允许添加 null 。您必须完全匹配泛型类型参数，而不使用通配符： LinkedList< List< Integer>> numList = new LinkedList< List< Integer>>（）; 。您可以为 LinkedList 添加 List< Integer> 。

When I try to compile the following code:

LinkedList<List<? extends Number>> numList = new LinkedList<List<Integer>>();

I get an incompatible type error:

Required: LinkedList <java.util.list<? extends java.lang.Number>>
Found: LinkedList <java.util.list<Integer>>

How can I achieve having a LinkedList which contains elements that are Lists with elements that extend Number?

To be clear, I'm looking to add lists to numList in the following fashion:

numList.add(new LinkedList<Integer>());

解决方案

Wildcard capture does not go more than one generic level deep. So while this works:

LinkedList<? extends Number> test = new LinkedList<Integer>();

This does not:

LinkedList<List<? extends Number>> numList = new LinkedList<List<Integer>>();

The most reasonable explanation I can think of is to think of generics as invariant at the outermost level. A LinkedList<List<Integer>> is not a LinkedList<List<? extends Number>>, even though a List<Integer> is a List<? extends Number>, for the same reason that a List<Dog> is not a List<Animal> even though a Dog is an Animal. Here, Dog is to Animal as List<Integer> is to List<? extends Number>.

Well, the Dog/Animal solution is ? extends:

List<? extends Animal> animals = new List<Dog>();

Applying the same reasoning, the workaround is another ? extends:

LinkedList<? extends List<? extends Number>> numList = new LinkedList<List<Integer>>();

However, you won't be able to add anything to this list because of the first ? extends. The reference type variable numList doesn't know which subtype of List<? extends Number> it really is; it could be ArrayList<Integer>, so Java cannot provide the type safety that such a thing can be added to such a LinkedList. To maintain type safety, the compiler will only allow adding null. You'll have to match the generic type parameters exactly, with no wildcards: LinkedList<List<Integer>> numList = new LinkedList<List<Integer>>();. You can add a List<Integer> to such a LinkedList.

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