C++发送派生指针作为参数

Question

我正在尝试通过基类的另一个函数发送指向基类函数的派生指针，但由于某种原因，它抱怨：错误：在第 8 行上无效使用不完整的类型"结构派生"。

#include <iostream>
using namespace std;
class Derived;
class Base
{
public:
    void getsomething(Derived *derived){derived->saysomething();} //This is line 8
    void recieveit(Derived *derived){getsomething(&*derived);}
};
class Derived : public Base
{
public:
    void giveself(){recieveit(this);};
    void saysomething(){cout << "something" << endl;}
};

int main()
{
    Base *b = new Base;
    Derived *d = new Derived;
    d->giveself();
    return 0;
}

你知道我该怎么解决这个问题吗？

Answer 1

当编译器需要有关类成员的信息时，不能使用前向声明。

前向声明仅用于告诉编译器具有该名称的类确实存在，并且稍后将声明和定义。

所以喜欢以下：

class Derived ;
class Base
{
public:
    void getsomething(Derived *derived); 
    void recieveit(Derived *derived);
};
class Derived : public Base
{
public:
    void giveself(){recieveit(this);};
    void saysomething(){cout << "something" << endl;}
};
void Base::getsomething(Derived *derived){derived->saysomething();} 
void Base::recieveit(Derived *derived){getsomething(&*derived);}

Answer 2

唯一的方法是将函数定义从类声明中取出并将它们放在Derived声明之后。在你尝试使用它们的时候，可怜的编译器甚至还不知道Derived上存在什么方法。