c++ - 如何从派生类调用重载的父 cout 友元类？

Question

设想如下设置。如何从派生类 cout 中调用基类 cout？我可以使用 getBrand() 方法，但我觉得我应该能够直接访问基类的 cout 友元函数。

我修改了一下，尝试了 this.Brand 和 Brand。运气不好。

class Brand {
public:
    Brand(std::string brand):brand_(brand) {};
    friend std::ostream & operator << (std::ostream & out, const Brand & b) {
        out << b.brand_ << ' ';
        return out;
    }   
    std::string getBrand()const { return brand_; }     
private:
    std::string brand_;
}

class Cheese : public Brand {
public:
    Cheese(std::string brand, std::string type):Brand(brand), type_(type) {};
    friend std::ostream & operator << (std::ostream & out, const Cheese & c) {
        out << /* THIS.BRAND?! BRAND?! getBrand() meh.. */ << ' ' << c.type_ << std::endl; // <-- HERE
        return out;
    }
private:
    std::string type_;
}

int main() {
    Cheese c("Cabot Clothbound", "Cheddar");
    std::cout << c << std::endl;
} 

期望的输出

Cabot Clothbound Cheddar

Answer 1

您可以调用重载的 operator <<来自派生类的基类。由于您将运算符声明为友元，因此您可以简单地将派生类转换为基类:

class Cheese : public Brand {
public:
    Cheese(std::string brand, std::string type):Brand(brand), type_(type) {};
    friend std::ostream & operator << (std::ostream & out, const Cheese & c) {

        //ADDED
        out << static_cast<const Brand&>(c) << c.type_ << std::endl;
        return out;
    }
private:
    std::string type_;
};

输出:

Cabot Clothbound Cheddar

See it Live