C++ 中类的前向声明

Question

我编写了以下代码，我将通过它来帮助我查看继承以及调度/双重调度在 C++ 中的工作原理，但它不会编译。我查看了类原型/前向声明，我已经完成了，但我仍然收到错误“B 是不完整的类型”、“SubB 是不完整的类型”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

当你有一个前向声明时，你可以使用只作为引用的类型：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;
    }