如何覆盖更丰富类型的接口方法中定义的参数？ [英] How to override parameter defined in interface method with richer type?

问题描述

我有这些：

public class TennisPlayer
{

}

public class RogerFederer : TennisPlayer
{

}

public class RafaelNadal : TennisPlayer
{

}

然后我有一些方法，比如：

And then I have some classes with methods, like these:

public abstract class Manager
{
    protected abstract void ScheduleFriendlies(TennisPlayer player);
}

public class RafaelNadalManager : Manager
{
    public void ScheduleFriendlies(RafaelNadal rn)
    {
        //throw new NotClayException();
    }
}

public class RogerFedererManager : Manager
{
    public void ScheduleFriendlies(RogerFederer rf)
    {
        //throw new NotGrassException();
    }
}

//'RafaelNadalManager' does not implement inherited abstract member 'Manager.ScheduleFriendlies(TennisPlayer)'

我想要达到的目标是，我希望拥有经理的子课程这样的类将绑定到一个玩家，因此这些子类中的方法（ Manager 类）将特定于该玩家。换句话说， Roger 应该有自己的 ScheduleFriendlies ，而 Rafael 应该有他自己的。

The thing I'm trying to achieve is that I would want to have child classes of Manager but every such class will be tied to one player and hence the methods in those child classes (Manager classes) will be specific to that player. In other words Roger should have his own ScheduleFriendlies while Rafael should have his own.

我该如何解决这个问题？替代设计会做，但请记住：

我能改变什么：遗产，路线等

但是没有奢侈品：取消儿童经理课程并合并成一个（换句话说，我需要 RogerFederer 和 RogerFedererManager 作为单独的类）。

But do not have the luxury to: Take off child manager classes and merge into one (in other words, I need RogerFederer and RogerFedererManager as separate classes).

我试过这个：

public abstract class Manager
{
    protected abstract bool ScheduleFriendlies<T>(T player) where T : TennisPlayer;
}

public class RafaelNadalManager : Manager
{
    protected override bool ScheduleFriendlies<T>(T player)
    {
        //but how do I enforce the caller that only RafaelNadal object can be 
        //passed to this but not RogerFederer?
    }
}

推荐答案

你很接近，但你需要类泛型，而不仅仅是方法：

You were close, but you need the class generic, not just the method:

public abstract class Manager<T> where T : TennisPlayer
{
    protected abstract bool ScheduleFriendlies(T player);
}

然后你可以使用：

public class RafaelNadalManager : Manager<RafaelNadal>
{
     protected override bool ScheduleFriendlies(RafaelNadal player)
     {}
}

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