C++ 中 类 的前向声明

Question

我已经编写了下面的代码，我打算 运行 通过它来帮助我回顾继承以及 dispatching/double 调度在 C++ 中的工作方式，但它无法编译。我查看了 class prototyping/forward 声明，我已经这样做了，但我仍然收到错误 "B is an incomplete type"、"SubB is an incomplete type" 等。有什么问题？

#include <iostream>

class B;
class SubB;

class A { 
    public:
        void talkTo(B b){
            std::cout << "A talking to instance of B" << std::endl;
        }
        void talkTo(SubB sb){
            std::cout << "A talking to instance of SubB" << std::endl;
        }
};
class SubA : A {
    public:
        void talkTo(B b){
            std::cout << "SubA talking to instance of B" << std::endl;
        }
        void talkTo(SubB sb){
            std::cout << "SubA talking to instance of SubB" << std::endl;
        }
};
class B { 
    public:
        void talkTo(A a){
            std::cout << "B talking to instance of A" << std::endl;
        }
        void talkTo(SubA sa){
            std::cout << "B talking to instance of SubA" << std::endl;
        }
};
class SubB : B {
    public:
        void talkTo(A a){
            std::cout << "SubB talking to instance of A" << std::endl;
        }
        void talkTo(SubA sa){
            std::cout << "SubB talking to instance of SubA" << std::endl;
        }
};

编辑

将参数更改为引用使这项工作（来自 R Sahu 的帮助）但为什么现在不工作？

class A { 
    public:
        void talkTo(B &b){
            //std::cout << "A talking to instance of B" << std::endl;
            b.talkTo(this);
        }
        void talkTo(SubB &sb){
            //std::cout << "A talking to instance of SubB" << std::endl;
            sb.talkTo(this);
        }
};
class B { 
    public:
        void talkTo(A &a){
            std::cout << "B talking to instance of A" << std::endl;
        }
        void talkTo(SubA &sa){
            std::cout << "B talking to instance of SubA" << std::endl;
        }
};
class SubB : B {
    public:
        void talkTo(A &a){
            std::cout << "SubB talking to instance of A" << std::endl;
        }
        void talkTo(SubA &sa){
            std::cout << "SubB talking to instance of SubA" << std::endl;
        }
};

A a;
SubA subA;
B b;
SubB subB;

a.talkTo(b);
a.talkTo(subB);

Answer 1

当你有前向声明时，你只能使用引用类型：指针和引用是最明显的引用。

而不是

    void talkTo(B b){
        std::cout << "A talking to instance of B" << std::endl;
    }
    void talkTo(SubB sb){
        std::cout << "A talking to instance of SubB" << std::endl;
    }

使用

    void talkTo(B const& b){
        std::cout << "A talking to instance of B" << std::endl;
    }
    void talkTo(SubB const& sb){
        std::cout << "A talking to instance of SubB" << std::endl;
    }