Question

我定义了一个类array_view和一个类strided_view（想想array_view和strided_array_view http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0122r0.pdf），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，它试图专门研究不同的情况。

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);
}

然后我定义了不同的专业化

// 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主

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 
}

一切都很好并正确切换

但是当我尝试使用const时，事情变得奇怪了例如

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 https://stackoverflow.com/questions/14926482/const-and-non-const-template-specialization) 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，但在我的情况下它没有任何改变。

我缺少什么？此致

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!std::is_same_v<T1 const, T2 const>案例测试T1（而不是测试T2和"both view != type"是否属于同一类型）（2）在T1中使用T2和T（而不是"both view same type"）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，因此我使用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：{而不是强加View1<T1>和View2<T2>与view<T1>和view<T2>的类型相同，您可以（在您的专业化中）避免使用{ {1}}和View1并直接使用View2。

您可以按如下方式简化您的专业化

view

视图和跨步视图以及const噩梦的功能模板专业化

1 个答案: