与泛型类实现可比较 [英] Implementing Comparable with a generic class

问题描述

我想定义一个实现泛型Comparable接口的类。在我的课程中，我还定义了一个泛型类型元素 T 。为了实现接口，我将比较委托给 T 。这是我的代码：

  public class Item< T extends Comparable< T>>实现可比< Item> {
 
 private int s; 
私人T t; 
 
 public T getT（）{
 return t; 
} 
 
 @Override 
 public int compareTo（Item o）{
 return getT（）。compareTo（o.getT（））; 
 
 
  

当我尝试编译它时，我得到以下错误信息：

  Item.java:11：error：方法compareTo在接口Comparable< T＃2>不能应用于给定的类型; 
 return getT（）。compareTo（o.getT（））; 
 $ 
 required：T＃1 
 found：Comparable 
 reason：实际参数Comparable不能通过方法调用转换转换为T＃1 
其中T＃1， T＃2是类型变量：
 T＃1扩展了Comparable< T＃1>在类Item中声明
 T＃2 extends接口中声明的对象Comparable 
 1错误
  

<有人可以告诉我为什么以及如何修复它吗？
基本上编译器不能证明泛型类型 o 与这个的类型是相同的 T 。

举例来说，我可以这样做：

 新项目< String>（）。compareTo（new Item< Integer>（））; 
  

这显然是错误的，并且是编译器阻止的场景。我认为你只需要nix原始类型：

  public class Item< T extends Comparable< T>> 
实现了Comparable< Item< T>> {
 
 
 
 @Override 
 public int compareTo（Item< T>）{
 return getT（）。compareTo（o。 GETT（））; 
} 
} 
  

I want to define a class that implements the generic Comparable interface. While in my class I also defined a generic type element T. In order to implement the interface, I delegate the comparison to T. Here is my code:

public class Item<T extends Comparable<T>> implements Comparable<Item> {

    private int s;
    private T t;

    public T getT() {
        return t;
    }

    @Override
    public int compareTo(Item o) {
        return getT().compareTo(o.getT());
    }
}

When I try to compile it, I get the following error information:

Item.java:11: error: method compareTo in interface Comparable<T#2> cannot be applied to given types;
        return getT().compareTo(o.getT());
                     ^
  required: T#1
  found: Comparable
  reason: actual argument Comparable cannot be converted to T#1 by method invocation conversion
  where T#1,T#2 are type-variables:
    T#1 extends Comparable<T#1> declared in class Item
    T#2 extends Object declared in interface Comparable
1 error

Can anybody tell me why and how to fix it?

解决方案

Basically the compiler can't prove that the generic type of o is the same T as the type of this.

For example I could do the following:

new Item<String>().compareTo(new Item<Integer>());

That is clearly wrong and is the scenario the compiler is preventing. I think you just need to nix the raw types:

public class Item<T extends Comparable<T>>
implements Comparable<Item<T>> {

    ...

    @Override
    public int compareTo(Item<T> o) {
        return getT().compareTo(o.getT());
    }
}

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