依赖非类型模板参数和可变参数模板 [英] Dependant non-type template parameter and variadic template
问题描述
我想扩展 std ::使用一个名为
。我的实现是基于我对此问题的回答,我试图适应 integer_range
的新类(它有意地在两个边界之间创建一个整数序列)来创建整数_序列 std: :integer_sequence
:
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). My implementation was based of my answer to this question that I tried to adapt to 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>;
我认为这个改变是微不足道的(添加一个 typename
template参数),但这实际上在 0
的专业化中引入了一个依赖非类型参数的问题。这里是编译器错误:
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
. Here is the compiler error:
error: type `Int` of template argument `0` depends on a template parameter
的基本问题有一些解决方案。但是,由于我使用可变参数模板,因此更难修复:此回答无法正常工作,因为我不是允许在可变参数模板之后具有默认模板参数。因此,我试图实现接受的答案中提到的修复程序,但似乎,如评论中提到的,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具有同等的专业性:
The basic problem already has some solutions. 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. 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:
-
struct increasing_integer_range< Int,std :: integral_constant< Int,C>,P,N ...> ;:/* * /
-
struct increasing_integer_range< Int,std :: integral_constant< Int,0>,P,N ...> ;:/* * /
struct increasing_integer_range<Int, std::integral_constant<Int, C>, P, N...>: /* */
struct increasing_integer_range<Int, std::integral_constant<Int, 0>, P, N...>: /* */
有没有其他方法来解决这个问题?
Is there any other way to fix this problem? I am out of ideas.
推荐答案
我会简单地减少你的 integer_range
到 std :: integer_sequence
的单个非递归调用:
I would simply reduce your integer_range
to a single, non-recursive call to std::integer_sequence
:
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>;
我测试过:
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 >() );
}
得到预期的链接器错误:
getting the expected linker errors:
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&)'
活动示例 (使用自己的实现 integer_sequence
),只需跳过第一部分)
Live example (with own implementation of integer_sequence
, just skip the first part)
这篇关于依赖非类型模板参数和可变参数模板的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!