C++中的命名空间类模板继承

Question

在之前的一个问题中，我问了class template inheritance in C++。

我现在有了一个额外的关卡要添加！

考虑下面的代码。(假设成员定义存在且准确)

namespace Game
{
    namespace Object
    {
        template<typename T>
        class Packable
        {
        public:

            /**
             * 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, T &t);
            friend sf::Packet& operator <<(sf::Packet& packet, T *t);

            /**
             * Unpacks a <class T> from a Packet (Packet >> T)
             * Required for chaining packet unpacking
             *************************************************/
            virtual sf::Packet& operator >>(sf::Packet& packet) = 0; // Work-horse, must be defined by child-class
            friend sf::Packet& operator >>(sf::Packet& packet, T &t);
            friend sf::Packet& operator >>(sf::Packet& packet, T *t);

            /**
             * Unpacks a <class T> from a Packet (T <<= Packet)
             * Returns the <class T> for convienence
             *************************************************/
            //friend T& operator <<=(T t, sf::Packet& packet); // Returning reference to cut down on copying (they're already passing us our own copy)
            friend T& operator <<=(T &t, sf::Packet& packet);
            friend T* operator <<=(T *t, sf::Packet& packet);
        };
    }
}

这个Ship类继承自Game::Object::Packable

class Ship : public Game::Object::Base<Ship>, public Game::Object::Packable<Ship>
{
    public:
        Ship( void );
        //...

        // For packing and unpackinng packets
        sf::Packet& operator <<(sf::Packet& packet);
        sf::Packet& operator >>(sf::Packet& packet);
}

我们剩下的是下面的错误。

(null): "Game::Object::operator<<(sf::Packet&, Ship*)", referenced from:

我得出的结论是，它一定与名称空间的使用有关。如果我想保留名称空间，解决方案是什么？

下面是方法定义的摘录。我必须对名称空间执行dereference操作吗？(我甚至不认为这意味着什么，哈哈)

/**
 * Packs a <class T> into a Packet (Packet << T)
 * Required for chaining packet packing
 *************************************************/
template<class T>
sf::Packet& operator <<(sf::Packet& packet, T *t)
{
    // Call the actual one, but basically do nothing... this needs to be overrided
    return packet << *t;
}

template<class T>
sf::Packet& operator <<(sf::Packet& packet, T &t)
{
    // Call the pointer one, but basically do nothing... this needs to be overrided
    return packet << &t;
}

// ... other definitions etc.

Answer 1

友元声明(非模板非成员)与友好函数模板不匹配。我建议您在类定义中提供实现：

    template<typename T>
    class Packable
        friend sf::Packet& operator <<(sf::Packet& packet, T &t) {
           return packet << t;
        }
    //...

这允许一个免费的非模板函数，它将由编译器按需生成。其他替代方案包括与模板或您所关心的模板专门化成为好友。

当然，您可以完全忽略友情，只提供名称空间级别的模板，因为它们是在公共函数…中实现的

相关信息：

• https://stackoverflow.com/a/23718008/36565
• https://stackoverflow.com/a/4661372/36565