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;

}

推荐答案 2015-04-26

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;
}

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其他回答

第1个回答 2017-06-30

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本回答被网友采纳

第2个回答 2018-05-09

#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

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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？

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