将Generic类的子类分配给该类的超类 [英] Assign a subclass of a Generic class to a super class of this class

问题描述

我有几个提供的接口

  public interface Finder< S extends Stack< T>，T extends Item> {
 public S find（S s，int a）; 
} 
 
 public interface Stack< T extends Item> {
 Stack< T> getCopy（）; 
} 
  

以及实现第一个类的类：

  public class SimpleFinder< S extends Stack< T>，T extends Item>实现Finder< S，T> {
 
 public S find（S s，int a）{
 S stack = ....; 
 ... 
 stack = s.getCopy（）; \\错误：不兼容的类型
 \\ required：S 
 \\ found：Stack< T> 
 ..... 
返回堆栈; 
} 
 
 
} 
  

如果我不行改变任何接口什么是最好的行为，同时保持实现尽可能通用？

编辑
其他一些代码我不能破坏实例化 SimpleFinder< ClassA，ClassB> ，所以我应该在实现中有两个泛型类型。

Stack< T> 不是 S扩展Stack< T> 。 Java是强类型的，不会让你做这样的事情。

您可以转换为 Stack< T> ，在这种情况下，您仍然会收到关于未经检查转换的警告。这意味着这种转换是不安全的。

  public class SimpleFinder< S extends Stack< T> ;, T扩展项目>实现Finder< S，T> {
 
 @Override 
 public S find（S s，int a）{
 Stack< T> stack = s.getCopy（）; 
返回（S）堆栈; 
} 
 
} 
  

或简单地使用 Stack< T> 代替 S延伸Stack< T> ，这是我的建议：

  public class SimpleFinder< T extends Item>实现Finder< Stack> T>，T> {
 
 @Override 
 public Stack< T> find（Stack< T> s，int a）{
 Stack< T> stack = s.getCopy（）; 
返回堆栈; 
} 
 
} 
  

I have couple of supplied interfaces

public interface Finder<S extends Stack<T>,T extends Item> {
    public S find(S s, int a);
}

public interface Stack<T extends Item> {
    Stack<T> getCopy();
}

and a class that implements the first:

public class SimpleFinder<S extends Stack<T>,T extends Item> implements Finder<S,T>{

    public S find(S s, int a){
         S stack = ....;
         ...
         stack = s.getCopy(); \\Error: incompatible types
                              \\       required: S
                              \\       found:    Stack<T>
        .....
        return stack;
    }


}

If I cannot change any interface what would be the best course of action while keeping the implementation as generic as possible?

EDIT Some other code which I cannot break instantiates SimpleFinder<ClassA,ClassB> so I should have two generic types in the implementation as well.

解决方案

The problem is that obviously Stack<T> is not S extends Stack<T>. Java is strongly typed and won't let you do such things.

You can either cast to Stack<T>, in which case you will still get a warning about unchecked conversion. This means this conversion is unsafe.

public class SimpleFinder<S extends Stack<T>, T extends Item> implements Finder<S, T> {

    @Override
    public S find(S s, int a) {
        Stack<T> stack = s.getCopy();
        return (S) stack;
    }

}

or simply use Stack<T> instead of S extends Stack<T>, which is my recommendation:

public class SimpleFinder<T extends Item> implements Finder<Stack<T>, T> {

    @Override
    public Stack<T> find(Stack<T> s, int a) {
        Stack<T> stack = s.getCopy();
        return stack;
    }

}

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