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CRTP 如何使派生的 class 有一个基础容器 class

[英]CRTP how to make a derived class have a container of base class

I would like to mimic the following behavior using CRTP:我想使用 CRTP 模仿以下行为:

#include <vector>
#include <memory>

class GameNumber {
public:
    ~GameNumber(){}
};

class GameNumber_real : public GameNumber {
public:
    GameNumber_real (float m) {n = m;}
    ~GameNumber_real(){}
private:
    float n;
};

class GameNumber_sets : public GameNumber {
public:
    GameNumber_sets (float n, float m) {
        left.push_back (std::make_shared<GameNumber_real>
            (GameNumber_real(n)));
        right.push_back (std::make_shared<GameNumber_real>
            (GameNumber_real(m)));
    }
    ~GameNumber_sets(){}
private:
    std::vector <std::shared_ptr<GameNumber>> left;
    std::vector <std::shared_ptr<GameNumber>> right;
};

My attempt was:我的尝试是:

template <class T>
class GameNumber {}

class GameNumber_real : public GameNumber<GameNumber_real> {
public:
    GameNumber_real (float m) {n = m;}
private:
    float n;
};

class GameNumber_sets : public GameNumber<GameNumber_sets> {
public:
    GameNumber_sets (float n, float m) {
        left.push_back (std::make_shared<GameNumber_real>
            (GameNumber_real(n)));
        right.push_back (std::make_shared<GameNumber_real>
            (GameNumber_real(m)));
    }
private:
    std::vector <std::shared_ptr<GameNumber>> left;
    std::vector <std::shared_ptr<GameNumber>> right;
};

The compiler error is:编译器错误是:

include/game_number.hpp:44:33: error: no matching function for call to
‘std::vector<std::shared_ptr<GameNumber<GameNumber_sets>>>
::push_back(std::shared_ptr<GameNumber_real>)’  
             (GameNumber_real(n)));

From the message I understand that inside GameNumber_sets, GameNumber resolves to GameNumber_GameNumber_sets_.从消息中我了解到,在 GameNumber_sets 中,GameNumber 解析为 GameNumber_GameNumber_sets_。 I would be really glad if you could tell me how achieve the behavior I expected.如果您能告诉我如何实现我所期望的行为,我将非常高兴。

You can explicitly specify template argument like GameNumber<GameNumber_real> and it will solve your problem.您可以明确指定像GameNumber<GameNumber_real>这样的模板参数,它将解决您的问题。

Also you can pass constructor arguments directly to std::make_shared .您也可以将构造函数 arguments 直接传递给std::make_shared

Improved code:改进的代码:

template <class T>
class GameNumber {};

class GameNumber_real : public GameNumber<GameNumber_real> {
public:
    GameNumber_real (float m) {n = m;}
private:
    float n;
};

class GameNumber_sets : public GameNumber<GameNumber_sets> {
public:
    GameNumber_sets (float n, float m) {
        left.push_back (std::make_shared<GameNumber_real>(n));
        right.push_back (std::make_shared<GameNumber_real>(m));
    }
private:
    std::vector <std::shared_ptr<GameNumber<GameNumber_real>>> left;
    std::vector <std::shared_ptr<GameNumber<GameNumber_real>>> right;
};

However I don't understand why do you need CRTP in here, there is no use of it in your example.但是我不明白为什么在这里需要 CRTP,在您的示例中没有使用它。

Edit:编辑:

As far as I know you can't use CRTP as interface.据我所知,您不能使用 CRTP 作为接口。 However there is std::any ( https://en.cppreference.com/w/cpp/utility/any ) which may help if c++17 is ok for you.但是,如果 c++17 适合您,则有std::anyhttps://en.cppreference.com/w/cpp/utility/any )可能会有所帮助。

Code would look like:代码如下所示:

template <class T>
class GameNumber {};

class GameNumber_real : public GameNumber<GameNumber_real> {
public:
    GameNumber_real (float m) {n = m;}
private:
    float n;
};

class GameNumber_sets : public GameNumber<GameNumber_sets> {
public:
    GameNumber_sets (float n, float m) {
        left.push_back (std::make_shared<std::any>
            (GameNumber_real(n)));
        right.push_back (std::make_shared<std::any>
            (GameNumber_real(m)));
    }
private:
    std::vector <std::shared_ptr<std::any>> left;
    std::vector <std::shared_ptr<std::any>> right;
};

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