运算符重载:从成员函数调用友元函数

Operator overloading: calling friend function from member function

我有一个学生class。我想重载 + 运算符,以便我可以向 class 添加一个双精度变量。这里是 Student class:

class Student {
private:
    std::string firstName;
    double grade;
public:
    Student(const std::string &firstName, double grade);

    double getGrade() const;

    friend Student operator+(double grade, const Student &student);

    Student operator+(double grade) const;
}; 

和实现:

Student::Student(const std::string &firstName, double grade) {
    this->firstName = firstName;
    this->grade = grade;
}

double Student::getGrade() const {
    return grade;
}

Student operator+(double grade, const Student &student) {
    return Student(student.firstName, student.grade + grade);
}

Student Student::operator+(double grade) const {
    return operator+(grade, *this);
}

double + Student 通过友元函数完成,Student + double 通过成员函数完成。当我编译时,我得到这个:

error: no matching function for call to ‘Student::operator+(double&, const Student&) const’
     return operator+(grade, *this);
                                  ^
note: candidate is:
note: Student Student::operator+(double) const
 Student Student::operator+(double grade) const {
         ^
note:   candidate expects 1 argument, 2 provided

为什么我不能从成员函数调用友元函数?

[更新]

然而,当我重载 << 运算符时,我可以从成员函数中调用它而无需预先挂起 ::.

friend std::ostream &operator<<(std::ostream &os, const Student &student);

和实施:

std::ostream &operator<<(std::ostream &os, const Student &student) {
    os << student.grade;
    return os;
}

您正在尝试调用成员函数,而不是友元函数(参见 C++11 7.3.1.2/3)。你应该写

Student Student::operator+(double grade) const {
    return ::operator+(grade, *this);
}

Example

使用 :: 确保从您当前所在的全局命名空间发生重载解析。

另一种方法(在我看来可读性较差)是将友元函数添加到重载决议集中

Student Student::operator+(double grade) const {
    using ::operator+;
    return operator+(grade, *this);
}

或者,按照 Jarod 的建议,更具可读性,

Student Student::operator+(double grade) const {
    return grade + *this;
}

编辑:对于 "why":[class.friend]/p7[basic.lookup.argdep]/p3 解释重载解析时同名成员函数“hides”的友元函数名称。