c++ - 使用友元函数重载算术运算符

Question

我找到了如何重载算术运算符的示例 using friend functions ，并且重载运算符函数在类内部定义，并带有注释:

/* This function is not considered a member of the class, even though the definition is 
    inside the class */

这是示例:

#include <iostream>

class Cents
{
private:
    int m_cents {};

public:
    Cents(int cents) : m_cents{ cents } { }

    // add Cents + Cents using a friend function
        // This function is not considered a member of the class, even though the definition is inside the class
    friend Cents operator+(const Cents& c1, const Cents& c2)
    {
        // use the Cents constructor and operator+(int, int)
        // we can access m_cents directly because this is a friend function
        return Cents{c1.m_cents + c2.m_cents};
    }

    int getCents() const { return m_cents; }
};

int main()
{
    Cents cents1{ 6 };
    Cents cents2{ 8 };
    Cents centsSum{ cents1 + cents2 };
    std::cout << "I have " << centsSum.getCents() << " cents.\n";

    return 0;
}

这个函数真的不是该类的成员吗？并且都是在类内部定义的友元函数，但不是该类的成员，或者它仅适用于使用friend关键字的重载函数。

Answer 1

Is this function really not a member of that class?

是的。

and are all friend functions defined inside the class but not a member of that class

是的。

简单来说，你不是你的 friend 。你的 friend 可能可以访问你的所有东西(访问类成员)，并且住在你的房子里(在类体内定义)，但他们仍然不是你(类的成员)。它们与类是分离的，唯一的耦合是它们可以访问类成员。