如何覆盖更丰富类型的接口方法中定义的参数？

Question

我有这些:

public class TennisPlayer
{

}

public class RogerFederer : TennisPlayer
{

}

public class RafaelNadal : TennisPlayer
{

}

然后我有一些方法的类,如下所示:

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)'

我想要实现的是,我想要有子类,Manager但是每个这样的类都将绑定到一个玩家,因此这些子类(Manager类)中的方法将特定于该玩家.换句话说Roger应该有他自己的ScheduleFriendlies,而Rafael应该有他自己的.

我该怎么做？替代设计会做,但请记住:

我能改变的是:遗产,路线等

但是没有足够的奢侈:取消孩子经理课程并合并成一个(换句话说,我需要RogerFederer和RogerFedererManager作为单独的课程).

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?
    }
}

Answer 1

你很接近,但你需要类泛型,而不仅仅是方法:

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)
     {}
}