&LT ;?延伸> Java语法 [英] <? extends > Java syntax

问题描述

此代码：

 列表< ;?扩展Reader> weirdList; 
 weirdList.add（new BufferedReader（null））; 
  

有一个编译错误：

列表中的方法add（capture＃1-of？extends Reader）不适用于
参数（BufferedReader）
blockquote>

为什么？ BufferedReader扩展了读者，为什么不是匹配？

解决方案

对于你给的变量：

 列表< ;?扩展Reader> weirdList; 
  

以下所有分配都是有效的：

  weirdList = new ArrayList< Reader>（）; 
 weirdList = new ArrayList< FileReader>（）; 
 weirdList = new ArrayList< BufferedReader>（）; 
 weirdList = new ArrayList< InputStreamReader>（）; 
  

希望这可以解释您的编译错误。如果 weirdList 保存的值类型 ArrayList< BufferedReader> ，但是不会使感觉类型 ArrayList< FileReader> 的值。由于类型 List <？>的变量扩展Reader> 可以保存任何一种类型的值（和更多！），Java会调用该错误。

Java中的泛型很难让你的头靠近。你可以想象 List<将Reader> 类型扩展为对方法中的赋值或参数类型非常有用，以便它们可以接受各种类型。对于常规使用，您可能更喜欢使用裸通用类型，如 List< Reader> 甚至 List< BufferedReader> code>。

This code:

List<? extends Reader> weirdList;
weirdList.add(new BufferedReader(null));

has a compile error of

The method add(capture#1-of ? extends Reader) in the type List is not applicable for the arguments (BufferedReader)

Why? BufferedReader extends reader, so why isn't that a "match"?

解决方案

For the variable you gave:

List<? extends Reader> weirdList;

All of the following assignments are valid:

weirdList = new ArrayList<Reader>();
weirdList = new ArrayList<FileReader>();
weirdList = new ArrayList<BufferedReader>();
weirdList = new ArrayList<InputStreamReader>();

Hopefully this explains your compile error. What you're trying makes sense if weirdList holds a value of type ArrayList<BufferedReader>, but doesn't make sense for a value of type ArrayList<FileReader>. Since a variable of type List<? extends Reader> can hold a value of either type (and more!), Java calls that an error.

Generics in Java are hard to get your head around. You can think of the List<? extends Reader> type as being mostly useful for assignment or parameter types in methods so that they can accept a wide variety of types. For "regular use", you're probably better off with a "bare" generic like List<Reader> or even List<BufferedReader>.

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