C ++中的命名空間類模板繼承

Question

在前面的問題中，我問了C ++中類模板繼承的問題。

我現在要添加一個額外的級別！

考慮下面的代碼。 （假設成員定義正確且正確）

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);
        };
    }
}

這個Ship類繼承自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);
}

我們剩下的是以下錯誤。

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

我得出的結論是，它必須與名稱空間的使用有關。 如果我想保留名稱空間，該如何解決？

這是方法定義的摘錄。 我是否必須dereference命名空間？ （我不認為這意味着什么哈哈）

/**
 * 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

朋友聲明（非模板非成員）與伙伴函數模板不匹配。 我建議您在類定義中提供實現：

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

這允許編譯器根據需要生成免費的非模板函數。 其他替代方法包括與您關心的模板或模板專業化做朋友。

當然，您可以完全忽略友誼，而只需在命名空間級別提供模板即可，因為它們是根據公共功能實現的……

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有關：

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