用部分实例化的祖先类代替完全实例化的类

Substitute fully instantiated classes with partially instantiated ancestor classes

本文关键字：实例化 祖先 用部      更新时间：2023-10-16

我希望accepted_dense_vector<??>::value返回'true'时，我给模板参数的形式:

当C为uvector, dynamic_array, T为std::is_arithmetic时，

C<T>。

当T为std::is_arithmetic时

std::array<T,S>。

当accepted_dense_vector<C>::value为'true'时，container_reference<C>

所有工作正常，但在派生的，完全实例化类A, C, D中，我想删除解决parent定义。

我该怎么做?

#include <iostream>
#include <array>
using namespace std;
// Definition of used types.
template<typename T> struct dynamic_array {};
template<typename T> struct uvector {};
struct A : public uvector<double> { typedef uvector<double> parent; };
struct C : public A { typedef A parent; };
struct D : public std::array<double,5> { typedef std::array<double,5> parent; };
template<typename T> struct B : public uvector<T> { typedef uvector<T> parent; };
template<typename T> struct container_reference {};
// Catches 'false', A, C, D  ======== HERE IT IS !!! ========
template<typename C> struct accepted_dense_vector
{
    template<typename C1 = C>
    static typename std::enable_if<accepted_dense_vector<typename C1::parent>::value, std::true_type>::type helper(const C&);
    static std::false_type helper(...);
    constexpr static bool value = decltype(helper(std::declval<C>()))::value;
};
// Catches C<T>
template<template<typename> class C, typename T>
struct accepted_dense_vector<C<T>>
{
    constexpr static bool value = std::is_arithmetic<T>::value &&
        (std::is_base_of<uvector<T>, C<T>>::value || std::is_base_of<dynamic_array<T>, C<T>>::value);
};
// Catches std::array<T,S>
template<typename T, size_t S>
struct accepted_dense_vector<std::array<T,S>>
    { constexpr static bool value = std::is_arithmetic<T>::value; };
// Catches container_reference<C>
template<typename C>
struct accepted_dense_vector<container_reference<C>>
    { constexpr static bool value = accepted_dense_vector<C>::value; };
int main()
{ // Tests!
    cout << accepted_dense_vector<std::array<double, 5>>::value << endl;
    cout << accepted_dense_vector<uvector<double>>::value << endl;
    cout << accepted_dense_vector<A>::value << endl;
    cout << accepted_dense_vector<D>::value << endl;
    cout << accepted_dense_vector<B<int>>::value << endl;
    cout << accepted_dense_vector<container_reference<uvector<double>>>::value << endl;
    cout << accepted_dense_vector<int>::value << endl;
    return 0;
}

您可以进一步使用您的解决方案。首先，您不需要std::enable_if。代码将更具可读性。然后，不要使用std::true_type，而是使用一个自定义结构体来调用accepted_dense_vector:

template<typename T>
struct forwarder
{
    constexpr static bool value = accepted_dense_vector<T>::value;
};
template<typename T>
static forwarder<uvector<T>> inherit_uvector(uvector<T>*);
static std::false_type inherit_uvector(...);

然后，通过继承uvector<T>来了解类型是否满足accepted...，只需使用以下行:

constexpr static bool is_uvector = decltype(inherit_uvector(new C()))::value;
// C is the template parameter of accepted_dense_vector

这是如何工作的?

  • 如果C继承uvector<T>，则选择第一个静态函数，decltype将返回forwarder<uvector<T>>。然后，'调用' ::value将调用accepted...。
  • else，如果C没有继承uvector<T>，则选择第二个函数，decltype将返回std::false_type。

同样的技巧也可以用于其他继承。同样，B<T>被accepted...的第二个专门化捕获。因此，不再需要::parent类型定义。

下面是完整的代码:

#include <iostream>
#include <array>
#include <string>
using namespace std;
// Definition of used types.
template<typename T> struct uvector {};
template<typename T> struct dynamic_array {};
template<typename T> struct container_reference {};
struct A : public uvector<double> {};
template<typename T> struct B : public uvector<T> {};
struct C : public A {};
struct D : public std::array<double,5> {};
struct E : public uvector<string> {};
struct F : public std::array<string,16> {};
// Catches 'false', A, C, D  ======== HERE IT IS !!! ========
template<typename C>
struct accepted_dense_vector
{
   template<typename T>
   struct forwarder
   {
      constexpr static bool value = accepted_dense_vector<T>::value;
   };
   // uvector
   template<typename T>
   static forwarder<uvector<T>> inherit_uvector(uvector<T>*);
   static std::false_type inherit_uvector(...);
   // std::array
   template<typename T, size_t S>
   static forwarder<std::array<T,S>> inherit_stdarray(std::array<T,S>*);
   static std::false_type inherit_stdarray(...);
   // same for dynamic_array<T>
    constexpr static bool is_uvector = decltype(inherit_uvector(new C()))::value;
    constexpr static bool is_stdarray = decltype(inherit_stdarray(new C()))::value;
    constexpr static bool value = is_uvector || is_stdarray;
};

// Catches C<T>
// /! Also catches anything which is template<typename> class (for example, B)
// => B::parent is not needed, because of the use of std::is_base_of
template<template<typename> class C, typename T>
struct accepted_dense_vector<C<T>>
{
    constexpr static bool value = std::is_arithmetic<T>::value &&
        (std::is_base_of<uvector<T>, C<T>>::value || std::is_base_of<dynamic_array<T>, C<T>>::value);
};
// Catches std::array<T,S>
template<typename T, size_t S>
struct accepted_dense_vector<std::array<T,S>>
{
   constexpr static bool value = std::is_arithmetic<T>::value;
};
// Catches container_reference<C>
template<typename C>
struct accepted_dense_vector<container_reference<C>>
{
   constexpr static bool value = accepted_dense_vector<C>::value;
};
int main()
{ // Tests!
    cout << "array = " << accepted_dense_vector<std::array<double, 5>>::value << endl;
    cout << "uvector = " << accepted_dense_vector<uvector<double>>::value << endl;
    cout << "ref = " << accepted_dense_vector<container_reference<uvector<double>>>::value << endl;
    cout << "A = " << accepted_dense_vector<A>::value << endl;
    cout << "B = " << accepted_dense_vector<B<int>>::value << endl;
    cout << "C = " << accepted_dense_vector<C>::value << endl;
    cout << "D = " << accepted_dense_vector<D>::value << endl;
    cout << "E = " << accepted_dense_vector<E>::value << endl;
    cout << "F = " << accepted_dense_vector<F>::value << endl;
    cout << "int = " << accepted_dense_vector<int>::value << endl;
    return 0;
}
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