C++报错提示某类名不是类或命名空间名称

最新推荐文章于 2024-07-26 22:08:14 发布

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文章讨论了在C++中遇到的两个类互相调用时的错误，问题出在类的不完全类型上。通过理解不完全类型只能用于定义指针、引用以及函数声明，作者将函数实现移至类定义之后，解决了报错问题。同时，文章强调了正确实现深拷贝以避免重复释放的问题。

在学习C++的友元时，遇到一个问题:两个类互相调用时报错

如图：

#include <iostream>
#include <string>
using namespace std;
//成员函数作友元
class Home;//先声明Home类，之后再实现，方便写在前面的类GirlFriend调用
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
class GirlFriend
{
public:
	GirlFriend();
	GirlFriend(const GirlFriend& tmp);
	~GirlFriend();
	void visit();
public:
	Home* home = NULL;
};

GirlFriend::GirlFriend()
{
	home = new Home;	//用指针home接收一个在堆区创建的Home类对象
}
/*------------------------------------------------------------------------------*/
//为了避免浅拷贝带来的重复释放问题，需要自己编写拷贝构造函数实现深拷贝
GirlFriend::GirlFriend(const GirlFriend& tmp)
{
	home = new Home(*tmp.home);
}
/*------------------------------------------------------------------------------*/
GirlFriend::~GirlFriend()
{
	if (home != NULL)
	{
		delete home;
		home = NULL;
	}
}
/*------------------------------------------------------------------------------*/
void GirlFriend::visit()
{
	cout << "Your girl is visiting:" << home->m_LivingRoom << endl;
	//因为visit()函数作为了Home类的友元，所以能够访问其私有成员m_LivingRoom
	cout << "Your girl is visiting:" << home->m_BedRoom << endl;
}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
class Home
{
	//把GirlFriend类中的成员函数visit()作为Home类的友元，以便访问Home类中的私有属性
	friend void GirlFriend::visit();
public:
	Home();

public:
	string m_LivingRoom;
private:
	string m_BedRoom;
};
/*------------------------------------------------------------------------------*/
Home::Home()
{
	m_BedRoom = "BedRoom";
	m_LivingRoom = "LivingRoom";
}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
void test10()
{
	GirlFriend myGirl;
	myGirl.visit();
}
/*------------------------------------------------------------------------------*/
int main()
{
	test10();

	system("pause");
	return 0;
}

报错如下

再参考https://blog.youkuaiyun.com/strive_max/article/details/127947510的方法以后，发现原来是在类的声明和定义之间，类是一个不完全类型，而不完全类型只能用于“定义指向该类型的指针及引用，或者用于声明(而不是定义)使用该类型作为形参类型或返回类型的函数”

同时在参考https://blog.youkuaiyun.com/chuomei5332/article/details/109675366和https://blog.youkuaiyun.com/qq_43915356/article/details/107577450两位大佬的说明后，将代码修改成以下状态（将除开不完全类型能用的操作之外的函数全部移到类的定义之后）

#include <iostream>
#include <string>
using namespace std;
//成员函数作友元
class Home;//先声明Home类，之后再实现，方便写在前面的类GirlFriend调用
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
class GirlFriend
{
public:
	GirlFriend();
	GirlFriend(const GirlFriend& tmp);
	~GirlFriend();
	void visit();
public:
	Home* home = NULL;
};
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
class Home
{
	//把GirlFriend类中的成员函数visit()作为Home类的友元，以便访问Home类中的私有属性
	friend void GirlFriend::visit();
public:
	Home();

public:
	string m_LivingRoom;
private:
	string m_BedRoom;
};
/*------------------------------------------------------------------------------*/
Home::Home()
{
	m_BedRoom = "BedRoom";
	m_LivingRoom = "LivingRoom";
}
/*++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
/*----------------------------不完全类型不允许的操作----------------------------*/
GirlFriend::GirlFriend()
{
	home = new Home;	//用指针home接收一个在堆区创建的Home类对象
}
/*------------------------------------------------------------------------------*/
//为了避免浅拷贝带来的重复释放问题，需要自己编写拷贝构造函数实现深拷贝
GirlFriend::GirlFriend(const GirlFriend& tmp)
{
	home = new Home(*tmp.home);
}
/*------------------------------------------------------------------------------*/
GirlFriend::~GirlFriend()
{
	if (home != NULL)
	{
		delete home;
		home = NULL;
	}
}
/*------------------------------------------------------------------------------*/
void GirlFriend::visit()
{
	cout << "Your girl is visiting:" << home->m_LivingRoom << endl;
	//因为visit()函数作为了Home类的友元，所以能够访问其私有成员m_LivingRoom
	cout << "Your girl is visiting:" << home->m_BedRoom << endl;
}
/*----------------------------不完全类型不允许的操作----------------------------*/
void test10()
{
	GirlFriend myGirl;
	myGirl.visit();
}
/*------------------------------------------------------------------------------*/
int main()
{
	test10();

	system("pause");
	return 0;
}

果然，再次运行，成功，没有报错