Java - 递归泛型类型定义 [英] Java - recursive generic type definitions

问题描述

我尝试使用< E扩展可比< E>> 来创建界面 ISortableStack ，但是我不能前进。以下是做什么的？

 < E扩展了Comparable< E>> 
  

我试过这个，但它没有帮助。

解决方案

< E延伸Comparable< E>> 意味着 E 必须是类型，知道如何与本身进行比较，因此，递归类型定义。

  public class Comparables {
 
 static class User实现Comparable< User> {
 @Override 
 public int compareTo（用户用户）{
 return 0; 
} 
} 
 
 / ** 
 *该类不能与Collections.sort一起使用，因为
 * UncomparableUser与自身无法比较。但是，请注意
 *，您只是为了实现
 * Comparable< String>而没有编译器错误。 
 * / 
 static class UncomparableUser implements Comparable< String> {
 @Override 
 public int compareTo（String user）{
 return 0; 
 
 
 
 public static void main（String [] args）{
 List< User> users = Arrays.asList（new User（））; 
 
 //使用这将导致编译器错误
 // List< UncomparableUser> users = Arrays.asList（new UncomparableUser（））; 
 
 Collections.sort（users）; 
} 
} 
  

I tried to create an interface ISortableStack using <E extends comparable <E>> but I can't move forward. What does the following do?

<E extends Comparable<E>> 

I've tried this, but it doesn't help.

解决方案

<E extends Comparable<E>> means that E must be a type that knows how to compare to itself, hence, the recursive type definition.

public class Comparables {

  static class User implements Comparable<User> {
    @Override
    public int compareTo(User user) {
      return 0;
    }
  }

  /**
   * This class cannot be used with Collections.sort because an
   * UncomparableUser is not comparable with itself. However, notice
   * that you get no compiler error just for implementing
   * Comparable<String>.
   */
  static class UncomparableUser implements Comparable<String> {
    @Override
    public int compareTo(String user) {
      return 0;
    }
  }

  public static void main(String[] args) {
    List<User> users = Arrays.asList(new User());

    // Using this would cause a compiler error
    // List<UncomparableUser> users = Arrays.asList(new UncomparableUser());

    Collections.sort(users);
  }
}

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