C++11转换可变模板来推断类类型

Question

我想从类属性列表中生成类。对于每个属性，通过模板推导实现。

为了解决这个问题，我将尝试创建一个类来处理具有两个属性的虚拟实体:疗愈者和商人。

#include <iostream>

// Base class for entities
class entity
{
    public:
    virtual void do_walk() { std::cout << "I like walking" << std::endl; };
};

// A class property for healer characters
class healer
{
    public:
    // interface
    virtual void do_heal() = 0;
};

// A class property for healer npcs
class merchant
{
    public:
    // interface
    virtual void do_sell() = 0;
};

// implementation of the healer property
class healer_implementation : public healer
{
    public:
    virtual void do_heal() { std::cout << "I heal people" << std::endl; }
};



// implementation of the healer property
class merchant_implementation : public merchant
{
    public:
    virtual void do_sell() { std::cout << "I sell things" << std::endl; }
};

// To deduce the implementation of a property, we'll use the template property which will be specialized for each property and reference an implementation
template<typename T>
class property
{};

template<>
class property<merchant>
{
public: typedef merchant_implementation implementation;
};

template<>
class property<healer>
{
public: typedef healer_implementation implementation;
};


// This class is a class helper to deduce the right class type from a list of property for an entity
template<typename PROPERTY0, typename PROPERTY1=void>
class factory
{
public:
    typedef typename property<PROPERTY0>::implementation base0;
    typedef typename property<PROPERTY1>::implementation base1;


    class custom : public base0, public base1, public entity {};

};

int main()
{
    entity* bob = new factory<healer, merchant>::custom();

    // bob will try to sell things
    merchant* m = dynamic_cast<merchant*>(bob);
    if (m)
        m->do_sell();

    // bob will try to heal people
    healer* h = dynamic_cast<healer*>(bob);
    if (h)
        h->do_heal();

    // and as an entity, bob can walk
    bob->do_walk();

return 1;
}

如果我执行这段代码，就会得到预期的结果：

I sell things
I heal people
I like walking

现在，如果我想创建一个只有一个属性的实体，我必须为工厂添加一个专门化：

template<typename PROPERTY0>
class factory<PROPERTY0, void>
{
public:
    typedef typename property<PROPERTY0>::implementation base0;

    class custom : public base0, public entity {};

};

我可以这样测试：

entity* joe = new factory<merchant>::custom();
h = dynamic_cast<healer*>(joe);
if (!h)
   std::cout << "Joe is not an healer." << std::endl;


m = dynamic_cast<merchant*>(joe);
if (m)
    m->do_sell();

然后我得到：

Joe is not an healer
I sell things

我的问题是:是否有一种方法可以用各种模板对类工厂进行编码以处理任意数量的属性，或者，如果我需要一个具有最多N个属性的实体，那么必须创建工厂的N个专门化吗？

Answer 1

如果您可以不使用base0、base1等类型的代码，那么您确实可以通过多种方式来实现这一点：

template <class... Property>
class factory
{
public:
    typedef typename property<PROPERTY0>::implementation base0;
    typedef typename property<PROPERTY1>::implementation base1;


    class custom : public property<Property>::implementation..., public entity {};

};

[实例]