运算符重载：从成员函数调用友元函数

Question

我有一个学生班。我想重载+ 运算符，这样我就可以在类中添加一个双变量。这是Student类：

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;
}; 

及实施：

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 是通过朋友函数完成的，Student + double 是通过成员函数完成的。当我编译我得到这个：

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]

但是，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;
}

Answer 1

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

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

Example

使用:: 可确保从您当前所在的全局命名空间发生重载解析。

另一种方式（我认为可读性较差）是将友元函数添加到重载解析集

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

或者，正如 Jarod 建议的那样，更具可读性，

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

---

编辑：至于“为什么”：[class.friend]/p7 和[basic.lookup.argdep]/p3 说明成员函数用重载解析期间同名“hides”友元函数的名称。