派生类的C++Typelist

Question

使用CRTP（奇怪的重复模板模式），您可以提供一个基类，其中包含派生自它的类的知识。创建一个数组来存储从基类派生的每个类的实例并不难（参见示例）

class Base{
    public:
        static std::vector<Base *> m_derivedInstances;
};

template <class Derived>
class CRTPBase : public Base {
    public:
        static bool m_temp;
        static bool addInstance()
        {
            m_derivedInstances.push_back(new Derived);
            return true;
        }
};
template <class Derived>
CRTPBase<Derived>::m_temp = CRTPBase<Derived>::addInstance();

我想知道是否可以创建一个Typelist（参见 http://www.research.ibm.com/designpatterns/pubs/ph-jun00.pdf）的所有类型的派生类。问题是，每当编译器看到一个派生自 Base 它需要在列表中追加一个新的类型，但是打字员是不可变的（可以创建一个新的列表，并将新的类型附加到它上面，但是就我所知，将元素添加到列表中是不可能的。最后我想有这样的东西:

struct DerivedClassHolder {
    typedef Loki::TL::MakeTypeList</*list all derived classes here*/>::Result DerivedTypes;
};

最终目标是能够迭代从 Base.

Answer 1

可以使用伪类型映射来完成。下面是一些使用boost::mpl的示例代码。"Implem"的显式定义可以用每个相应的implem头中的宏来完成。

#include <iostream>
#include <boost/mpl/vector.hpp>
#include <boost/mpl/eval_if.hpp>
#include <boost/mpl/identity.hpp>
#include <boost/mpl/for_each.hpp>
#include <boost/mpl/push_front.hpp>
#include <boost/mpl/empty_sequence.hpp>
#include <boost/type_traits/is_same.hpp>

using namespace boost::mpl;
using namespace boost;


// A type map. Implem #N of type Key is type (default: void)

template <typename Key, int N>
struct Implem
{
  public:
    typedef void type;
};


// Type vector building functions
// void, the default type, is used to stop the recursion

template <typename Key, int N = 1>
struct ImplemToList;

template <typename Key, typename Item, int N>
struct ImplemListItem
{
  public:
    typedef typename push_front<typename ImplemToList<Key, N + 1>::type, Item>::type type;
};

template <typename Key, int N>
struct ImplemToList
{
  public:
    typedef typename Implem<Key, N>::type item;
    typedef typename eval_if<is_same<item, void>,
                             identity<vector<> >,
                             ImplemListItem<Key, item, N> >::type type;
};


// Example code: an interface with two implems

class Interface
{
  public:
    virtual const char* name() const = 0;
};

class Implem1 : public Interface
{
  public:
    virtual const char* name() const { return "implem_1"; }
};

class Implem2 : public Interface
{
  public:
    virtual const char* name() const { return "implem_2"; }
};

template <>
struct Implem<Interface, 1>
{
  public:
    typedef Implem1 type;
};

template <>
struct Implem<Interface, 2>
{
  public:
    typedef Implem2 type;
};


void print(Interface const& i)
{
  std::cout << i.name() << std::endl;
}

int main()
{
  typedef ImplemToList<Interface>::type IList;
  for_each<IList>(&print);
}

Answer 2

您的typelist只能手工创建。你提到的问题，不变性，是不可逾越的。