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依赖的非类型模板参数和可变参数模板

[英]Dependant non-type template parameter and variadic template

I am trying to extend the possibilities offered by std::integer_sequence with a new class named integer_range (which obiously creates a sequence of integers between two bounds). 我试图用一个名为integer_range的新类扩展std::integer_sequence提供的可能性(它可以在两个边界之间创建一个整数序列)。 My implementation was based of my answer to this question that I tried to adapt to std::integer_sequence : 我的实现是基于我对这个问题的回答 ,我试图适应std::integer_sequence

namespace details
{
    template<typename Int, Int C, Int P, Int... N>
    struct increasing_integer_range:
        increasing_integer_range<Int, C-1, P+1, N..., P>
    {};

    template<typename Int, Int C, Int P, Int... N>
    struct decreasing_integer_range:
        decreasing_integer_range<Int, C+1, P-1, N..., P>
    {};

    template<typename Int, Int P, Int... N>
    struct increasing_integer_range<Int, 0, P, N...>:
        std::integer_sequence<Int, N...>
    {};

    template<typename Int, Int P, Int... N>
    struct decreasing_integer_range<Int, 0, P, N...>:
        std::integer_sequence<Int, N...>
    {};
}

template<typename Int, Int S, Int E,  bool Increasing=(S<E)>
struct integer_range;

template<typename Int, Int S, Int E>
struct integer_range<Int, S, E, true>:
    details::increasing_integer_range<Int, std::integral_constant<Int, E-S>, S>
{};

template<typename Int, Int S, Int E>
struct integer_range<Int, S, E, false>:
    details::decreasing_integer_range<Int, std::integral_constant<Int, E-S>, S>
{};

template<std::size_t S, std::size_t E>
using index_range = integer_range<std::size_t, S, E>;

I thought that the change would be trivial (adding a typename template parameter), but this actually introduces a problem with a dependant non-type parameter in the specialization for 0 . 我认为更改将是微不足道的(添加typename模板参数),但这实际上在0化中引入了依赖的非类型参数的问题。 Here is the compiler error: 这是编译器错误:

error: type `Int` of template argument `0` depends on a template parameter

The basic problem already has some solutions. 基本问题已经有了一些解决方案。 However, since I use variadic templates, it becomes even harder to fix: this answer can't work because I it is not allowed to have a default template parameter after a variadic template. 但是,由于我使用了可变参数模板,因此修复起来更加困难: 这个答案无法正常工作,因为我不允许在可变参数模板之后使用默认模板参数。 Therefore, I tried to implement the fix mentioned in the accepted answer but it seems that, as mentioned in the comments, my compiler (g++ 4.8.1) is unable to disambiguate and considers that the two following specialization are both equally specialized: 因此,我尝试实现接受的答案中提到的修复,但似乎正如评论中所提到的,我的编译器(g ++ 4.8.1)无法消除歧义并认为以下两个专业化都是同样专业的:

  • struct increasing_integer_range<Int, std::integral_constant<Int, C>, P, N...>: /* */
  • struct increasing_integer_range<Int, std::integral_constant<Int, 0>, P, N...>: /* */

Is there any other way to fix this problem? 有没有其他方法来解决这个问题? I am out of ideas. 我没有想法。

I'm just waiting for Yakk to suddenly appear out of nowhere and provide a non-hacky better solution ;) (I'm too tired to come up with something better..) but alas, this works for g++4.8.1: 我只是在等待Yakk突然冒出来并提供一个非hacky更好的解决方案;) (我太累了,无法想出更好的东西..)但是,这很适用于g ++ 4.8.1 :

#include <type_traits>

// C++1y Standard Library
// use this ONLY FOR TESTING PURPOSES
namespace std
{
    template<class T, T... vals> struct integer_sequence {};
}

namespace details
{
    template<typename T>
    struct minus_helper;
    template<typename T, T val>
    struct minus_helper< std::integral_constant<T, val> >
    {
        using type = std::integral_constant<T, val-1>;
    };

    template<typename T>
    using minus = typename minus_helper<T>::type;

    template<typename Int, typename C, Int P, Int... N>
    struct increasing_integer_range :
        increasing_integer_range<Int, minus<C>, P+1, N..., P>
    {};

    template<typename Int, Int P, Int... N>
    struct increasing_integer_range<Int, std::integral_constant<Int, 0>,
                                    P, N...>
        : std::integer_sequence<Int, N...>
    {};
}

template<typename Int, Int S, Int E,  bool Increasing=(S<E)>
struct integer_range;

template<typename Int, Int S, Int E>
struct integer_range<Int, S, E, true>:
    details::increasing_integer_range<Int, std::integral_constant<Int, E-S>, S>
{};

template<std::size_t S, std::size_t E>
using index_range = integer_range<std::size_t, S, E>;

int main()
{
    index_range<1, 5> f;
}

The reason why I think this could be done better is that it seems the increasing and decreasing range implementations are redundant. 我认为可以做得更好的原因是, 似乎增加和减少范围的实现是多余的。 But I might have to consider corner cases (overflow) before providing a unifying solution. 但在提供统一解决方案之前,我可能不得不考虑角落情况(溢出)。

For example, you can easily make a shifted integer range by using std::make_integer_range plus partial specialization plus pack expansion. 例如,您可以使用std::make_integer_range加上部分特化加包扩展轻松创建一个移位的整数范围。 But this might overflow. 但这可能会溢出。

I would simply reduce your integer_range to a single, non-recursive call to std::integer_sequence : 我只是将你的integer_range减少为对std::integer_sequence的单个非递归调用:

namespace details
{
    template<typename Int, typename, Int S>
    struct increasing_integer_range;

    template<typename Int, Int... N, Int S>
    struct increasing_integer_range<Int, std::integer_sequence<Int, N...>, S>
        : std::integer_sequence<Int, N+S...>
    {};

    template<typename Int, typename, Int S>
    struct decreasing_integer_range;

    template<typename Int, Int... N, Int S>
    struct decreasing_integer_range<Int, std::integer_sequence<Int, N...>, S>
        : std::integer_sequence<Int, S-N...>
    {};
}

template<typename Int, Int S, Int E,  bool Increasing=(S<E)>
struct integer_range;

template<typename Int, Int S, Int E>
struct integer_range<Int, S, E, true>:
    details::increasing_integer_range<Int, std::make_integer_sequence<Int, E-S>, S>
{};

template<typename Int, Int S, Int E>
struct integer_range<Int, S, E, false>:
    details::decreasing_integer_range<Int, std::make_integer_sequence<Int, S-E>, S>
{};

template<std::size_t S, std::size_t E>
using index_range = integer_range<std::size_t, S, E>;

Which I tested with: 我测试过的是:

template<std::size_t... N>
void dummy( const std::integer_sequence< std::size_t, N... >& );

int main()
{
    dummy( index_range< 2, 5 >() );
    dummy( index_range< 5, 2 >() );
}

getting the expected linker errors: 获得预期的链接器错误:

main.cpp:(.text.startup+0xa): undefined reference to `void dummy<2ul, 3ul, 4ul>(detail::integer_sequence<unsigned long, 2ul, 3ul, 4ul> const&)'
main.cpp:(.text.startup+0x14): undefined reference to `void dummy<5ul, 4ul, 3ul>(detail::integer_sequence<unsigned long, 5ul, 4ul, 3ul> const&)'

Live example (with own implementation of integer_sequence , just skip the first part) 实例 (使用自己的integer_sequence实现,只跳过第一部分)

For example, you can easily make a shifted integer range by using std::make_integer_range plus partial specialization plus pack expansion. 例如,您可以使用std::make_integer_range加上部分特化加包扩展轻松创建一个移位的整数范围。

A brief demonstration. 简要演示。

template<typename T>
constexpr T abs(T t)
{ return t < static_cast<T>(0) ? -t : t; }

template<typename Sequence>
struct match_sequence {};

template<typename Int, Int... Ns>
struct match_sequence<std::integer_sequence<Int, Ns...>> {
    template<Int Base, Int Stride = static_cast<Int>(1)>
    using Offset = std::integer_sequence<Int, Base + Stride * Ns...>;
};

template<typename Int, Int Low, Int High>
struct integer_range {
private:
    static constexpr auto width = abs(High - Low);
    using base = std::make_integer_sequence<Int, width>;
    static constexpr bool increasing = High > Low;
    static constexpr auto stride = increasing ? static_cast<Int>(1) : static_cast<Int>(-1);

public:
    using type = typename match_sequence<base>::template Offset<Low, stride>;
};

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