Question

C++ 一个类的成员函数作为另一个类的友元函数，为什么错了？

//问题见注释

#include <iostream>

using namespace std;

class Teacher; //这里编译不成功，提示incomplete type 'Teacher' named in nested name specifier

class Student

{

friend void printStudent(Student grade);

friend void Teacher::printStudentInTeacher(Student &s);

public:

private:

int grade;

};

class Teacher

{

public:

void printStudentInTeacher(Student &s);

};

void printStudent(Student s)

{

cout<<"outer function print:";

cout << s.grade << endl;

}

void Teacher::printStudentInTeacher(Student &s)

{

cout <<"Teacher print:";

cout << s.grade << endl;

}

Answer 1

顺序应该是这样：

//class Teacher;  //这里编译不成功，提示incomplete type 'Teacher' named in nested name specifier

class Student;//add

class Teacher
{
public:
	void printStudentInTeacher(Student &s);
};


class Student
{
	friend void printStudent(Student grade);
	friend void Teacher::printStudentInTeacher(Student &s);
public:
private:
	int grade;
};


void printStudent(Student s)
{
	cout << "outer function print:";
	cout << s.grade << endl;
}

void Teacher::printStudentInTeacher(Student &s)
{
	cout << "Teacher print:";
	cout << s.grade << endl;
}

Answer 2

worker(){}默认构造函数，其作用是把类的私有成员变量初始化为string类的默认值。
也就是空字符串。

~worker(){}是析构函数，其作用是在超出对象的作用域后，将对象所占的内存给释放掉。
避免物理内存被用尽

Answer 3

#include <iostream>
using namespace std;
class Student;
class Teacher
{
public:
void printStudentInTeacher(Student &s);
};
class Student
{
//public:
friend void printStudent(Student grade);
friend void Teacher::printStudentInTeacher(Student &s);
private:
int grade;
};
void printStudent(Student s)
{
cout<<"outer function print:";
cout << s.grade << endl;
}
void Teacher::printStudentInTeacher(Student &s)
{
cout <<"Teacher print:";
cout << s.grade << endl;
}
顺序应该是这样子的