问题描述

例如

BaseClass MyBase()
{
    public int Add(BaseClass next)
    {
        if (this is InheritedA && next is InheritedA)
            return 1;
        else if (this is InheritedA && next is InheritedB)
            return 2;
        else if (this is InheritedB && next is InheritedA)
            return 3;
        else if (this is InheritedB && next is InheritedB)
            return 4;      
     }
}

其中InheritedA和InheritedB是它的继承类.其实继承的类更多，Add根据其操作数的顺序和类型返回不同的结果.

where InheritedA, and InheritedB are its inherited classes. In fact, there are more Inherited classes, and the Add returns different results based on the order and types of its operand.

我正在考虑使用多态性和重载来重写它，但是它变得相当复杂，我必须引入一个辅助方法来解析任一端的类型.

I am thinking of rewriting it using Polymorphism and overloading, however, it becomes rather complicated, I have to introduce an helper method to resolve the type of either end.

例如

InheritedA myA()
{
    public override int Add(BaseClass next)
    {
        return next.AddTo(this);
    }
}

现在我必须将 AddTo 放入 BaseClass，并在继承的类中覆盖它.

Now I have to put AddTo into BaseClass, and override it in inherited class as well.

InheritedA myA()
{
    public override int AddTo(InheritedA next) { return 1; }
    public override int AddTo(InheritedB next) { return 3; }
}

BaseClass myBase()
{
    public abstract int Add(BaseClass next);
    public abstract int AddTo(InheritedA next);
    public abstract int AddTo(InheritedB next);
}

有没有更好的方法?

推荐答案

您正在实现的模式称为双虚拟调度.

The pattern you are implementing is called double virtual dispatch.

单个虚拟调度根据接收者的运行时类型和编译时类型选择调用哪个方法的论点.这是一个传统的虚拟调度:

A single virtual dispatch chooses which method to call on the basis of the run-time type of the receiver and the compile time type of the arguments. This is a traditional virtual dispatch:

abstract class Animal {}
class Tiger : Animal {}
class Giraffe : Animal {} 
class B
{
    public virtual void M(Tiger x) {}
    public virtual void M(Animal x) {}
}
class D : B
{
    public override void M(Tiger x) {}
    public override void M(Animal x) {}
}
...
B b = whatever;
Animal a = new Tiger();
b.M(a);

调用哪个方法?B.M(Tiger) 和 D.M(Tiger) 没有被选中；我们根据参数的编译时间类型拒绝它们，即Animal.但是我们根据 whatever 是否为 new B() 来选择在运行时调用 BM(Animal) 还是 DM(Animal) 或 new D().

dynamic b = whatever;
dynamic a = new Tiger();
b.M(a);