C++ 重载运算符，取消引用参考

Question

我在整个项目/类中继承了这个小实用程序类（模板化）。

这个想法是它允许轻松地将各种成员打包进出类实例（对于网络等，并不完全重要）

我得到的如下

    template<typename T>
    struct Packable
        /**
         * Packs a <class T> into a Packet (Packet << T)
         * Required for chaining packet packing
         *************************************************/
        virtual sf::Packet& operator <<(sf::Packet& packet) = 0; // Work-horse, must be defined by child-class

        friend sf::Packet& operator <<(sf::Packet& packet, const T *t)
        {
            // Call the actual one, but basically do nothing...
            return packet << *t;
        }

        friend sf::Packet& operator <<(sf::Packet& packet, const T &t)
        {
            // Call the actual one, but basically do nothing...
            return packet << &t;
        }

        friend sf::Packet& operator <<(sf::Packet& packet, T *t)
        {
            // Call the actual one, but basically do nothing...
            return packet << *t;
        }

        friend sf::Packet& operator <<(sf::Packet& packet, T &t)
        {
            // Call the actual one, but basically do nothing...
            return packet << &t;
        }
    };

我想要做的，简而言之，就是让它只需要在子类中指定/充实一种方法（用“虚拟”词表示）。

我想要的是提供采用各种形式的类的其他方法，并根据需要取消引用它们以使用编译类时将存在的虚拟方法。

问题是我似乎创建了一些无限循环。

        friend sf::Packet& operator <<(sf::Packet& packet, T &t)
        {
            // Call the actual one, but basically do nothing...
            return packet << &t;
        }

只是一遍又一遍地调用自己。如何取消引用对其对象的引用？

Answer 1

operator << 有 4 个重载，它们相互依赖。

1. const T *，取决于 2
2. const T&，取决于 1
3. T *，取决于 4
4. T&，取决于 3

所以它引起了无休止的回避。您应该至少将其中一个实现为independent functionality，然后其余的都依赖于此。

如下图即可

template<typename T>
struct Packable
    /**
     * Packs a <class T> into a Packet (Packet << T)
     * Required for chaining packet packing
     *************************************************/
    virtual sf::Packet& operator <<(sf::Packet& packet) = 0; // Work-horse, must be defined by child-class

    friend sf::Packet& operator <<(sf::Packet& packet, const T *t)
    {
        // Call the actual one, but basically do nothing...
        return packet << *t;
    }

    friend sf::Packet& operator <<(sf::Packet& packet, const T &t)
    {
        // Call the actual one, but basically do nothing...
        //return packet << &t;

        // Serialize the contents into packet stream
        t.serialize(packet);
        return packet;
    }

    friend sf::Packet& operator <<(sf::Packet& packet, T *t)
    {
        // Call the actual one, but basically do nothing...
        return packet << const_cast<const T*>(t);
    }

    friend sf::Packet& operator <<(sf::Packet& packet, T &t)
    {
        // Call the actual one, but basically do nothing...
        return packet << &t;
    }
};

这个函数serialize应该在每个T类型的类中实现，否则你会看到编译时错误。