视图和跨步视图的函数模板专业化和 const 噩梦

Question

我定义了一个classarray_view和一个classstrided_view（想想array_view和strided_array_viewhttp://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0122r0.pdf ) 并且我想专注于我可以迭代它以提高效率的方式假设我有一个函数调度程序，它尝试专门处理不同的情况。

让我们从一些简单的代码开始

template <class T>
    class view
    {};
template <class T>
    class sview
    {};
template<typename T1, typename T2, template <typename> class View1,   template <typename> class View2>
void PixelWiseUnary(const View1<T1>& i_vin, View2<T2>& o_vout)
{       
    PixelWiseUnaryDispatch<T1, T2, View1, View2> dispatcher;
    dispatcher(i_vin, o_vout);
}

然后我定义不同的专业化

// primary use strided view 
template<typename T1, typename T2, template <typename> class View1, template <typename> class View2, typename = void>
struct PixelWiseUnaryDispatch
{
    void operator()(const View1<T1>& i_vin, View2<T2>& o_vout) const
    {
        std::cout << "***************" << std::endl;
        std::cout << "primary template" << std::endl;
        std::cout << "***************" << std::endl;

    }
};

template<typename T1, typename T2, template <typename> class View1, template <typename> class View2>
struct PixelWiseUnaryDispatch<T1,T2, View1, View2,
    std::enable_if_t<(!std::is_same_v<T1,T2> && std::is_same_v<View1<T1>,view<T1>> )&& std::is_same_v<View2<T2>, view<T2>>>
>
{
    void operator()(const View1<T1>& i_vin, View2<T2>& o_vout) const
    {
        std::cout << "***************" << std::endl;
        std::cout << "both view != type" << std::endl;
        std::cout << "***************" << std::endl;

    }
};

template<typename T,template <typename> class View1, template <typename> class View2>
struct PixelWiseUnaryDispatch<T,T, View1, View2,
    std::enable_if_t<(std::is_arithmetic_v<T> && std::is_same_v<View1<T>, view<T>>) && std::is_same_v<View2<T>, view<T>>>
>
{
    void operator()(const View1<T>& i_vin, View2<T>& o_vout) const
    {
        std::cout << "***************" << std::endl;
        std::cout << "both view same type" << std::endl;
        std::cout << "***************" << std::endl;

    }
};

然后定义一个简单的main

void main(void)
{
view<int> vin;
view<float> vinf;
view<int> vout;
sview<int> vsout;
PixelWiseUnary(vin, vsout); //primary template
PixelWiseUnary(vinf, vout); //both view != type
PixelWiseUnary(vin, vout);  //both view same type 
}

一切正常，正常切换

但是当我尝试使用 const 时，事情变得很奇怪例如

void main(void)
{
view<const int> vin;
view<const float> vinf;
view<int> vout;
sview<int> vsout;
PixelWiseUnary(vin, vsout); //primary template as expected
PixelWiseUnary(vinf, vout); //both view != type WTF i don't provide specialisation for const (cf  so i expected primary template
PixelWiseUnary(vin, vout); //both view != type WTF i don't provide specialisation for const and i loose the same type specialization
}
}

我尝试添加来自 Const and non const template specialization 的建议，但它对我的情况没有任何改变。

我错过了什么？问候

注意：我使用最新的 Visual2017 社区版

Answer 1

我想你可以解决问题 (1) 针对 "both view != type" 案例测试 !std::is_same_v<T1 const, T2 const>（而不是测试 T1 和 T2 不是同一类型） (2) 在 "both view same type" 中使用 T1 和 T2（而不是 T）并添加测试 std::is_same_v<T1 const, T2 const>

我的意思是（我只有一个 C++14 编译器，所以我使用 std::is_same<>::value 而不是 std::is_same_v<>）

template <typename T1, typename T2,
          template <typename> class View1,
          template <typename> class View2>
struct PixelWiseUnaryDispatch <T1, T2, View1, View2,
   std::enable_if_t<
         !std::is_same<T1 const, T2 const>::value
      && std::is_same<View1<T1>, view<T1>>::value
      && std::is_same<View2<T2>, view<T2>>::value>
>
 {
   void operator()(View1<T1> const & i_vin, View2<T2> & o_vout) const
    {
      std::cout << "***************" << std::endl;
      std::cout << "both view != type" << std::endl;
      std::cout << "***************" << std::endl;
    }
 };

template <typename T1, typename T2, template <typename> class View1,
          template <typename> class View2>
struct PixelWiseUnaryDispatch<T1, T2, View1, View2,
   std::enable_if_t<
         std::is_same<T1 const, T2 const>::value
      && std::is_arithmetic<T1>::value
      && std::is_same<View1<T1>, view<T1>>::value
      && std::is_same<View2<T2>, view<T2>>::value>
>
 {
   void operator()(View1<T1> const & i_vin, View2<T2> & o_vout) const
    {
      std::cout << "***************" << std::endl;
      std::cout << "both view same type" << std::endl;
      std::cout << "***************" << std::endl;

    }
 };

En passant: 不要强加 View1<T1> 和 View2<T2> 与 view<T1> 和 view<T2> 是同一类型，你可以（在你的专业中）避免使用View1 和 View2 并直接使用 view。

您可以按如下方式简化您的专业化

template <typename T1, typename T2>
struct PixelWiseUnaryDispatch <T1, T2, view, view,
   std::enable_if_t<!std::is_same<T1 const, T2 const>::value>>
 {
   void operator()(view<T1> const & i_vin, view<T2> & o_vout) const
    {
      std::cout << "***************" << std::endl;
      std::cout << "both view != type" << std::endl;
      std::cout << "***************" << std::endl;
    }
 };
template <typename T1, typename T2>
struct PixelWiseUnaryDispatch<T1, T2, view, view,
   std::enable_if_t<
         std::is_same<T1 const, T2 const>::value
      && std::is_arithmetic<T1>::value>>
 {
   void operator()(view<T1> const & i_vin, view<T2> & o_vout) const
    {
      std::cout << "***************" << std::endl;
      std::cout << "both view same type" << std::endl;
      std::cout << "***************" << std::endl;

    }
 };

视图和跨步视图的函数模板专业化和 const 噩梦

Function template specialization for both view and strided view and const nightmare

c++

templates

constants

array-view

c++17