为什么这个模板专业化在GCC中不起作用？

Question

I want to create a tuple from the first n elements of another tuple. 我想从另一个元组的前n个元素创建一个元组。 This is the recursion I use (doesn't calculate anything useful here). 这是我使用的递归（不计算任何有用的东西）。

Weirdly, it doesn't work using g++ 4.6.1, even though as I understand it, front<0,…> is more specialized than front<n,…> and it should be selected. 奇怪的是，它不能使用g ++ 4.6.1，尽管据我所知， front<0,…>比front<n,…>更专业front<n,…>应该选择它。 Is this a bug or am I confused about specialization? 这是一个错误还是我对专业化感到困惑？

#include <tuple>

template<std::size_t, typename>
struct front;

#if 0
// This works, but I only want one case.
template<typename Head, typename... Tail>
struct front<0, std::tuple<Head, Tail...>> {
    typedef std::tuple<> type;
};
template<typename Head>
struct front<0, std::tuple<Head>> {
    typedef std::tuple<> type;
};
template<>
struct front<0, std::tuple<>> {
    typedef std::tuple<> type;
};
#elseif 0
// this doesn't work, but I don't understand why:
// ambiguous class template instantiation, candidates are:
//  struct front<0u, std::tuple<_Elements ...> >
//  struct front<n, std::tuple<_Head, _Tail ...> >
template<typename... Tail>
struct front<0, std::tuple<Tail...>> {
    typedef std::tuple<> type;
};
#else
// neither does this:
// ambiguous class template instantiation, candidates are:
//  struct front<0u, T>
//  struct front<n, std::tuple<_Head, _Tail ...> >
template<typename T>
struct front<0, T> {
    typedef std::tuple<> type;
};
#endif
// this makes no sense, but it's short.
template<std::size_t n, typename Head, typename... Tail>
struct front<n, std::tuple<Head, Tail...>> {
    typedef typename front<n - 1, std::tuple<Tail...>>::type type;
};
// check all cases, error includes calculated type:
front<0, std::tuple<int, float, double, long>>::type x0 = 0;
front<2, std::tuple<int, float, double, long>>::type x2 = 0;
front<4, std::tuple<int, float, double, long>>::type x4 = 0;

Answer 1

Template arguments aren't picked from left to right. 不从左到右挑选模板参数。 Yes, 0 is more specialized than n , but std::tuple<Head, Tail...> is more specialized than T . 是的， 0比n更专业，但是std::tuple<Head, Tail...>比T更专业。 You can add an extra specialization, which is the most specialized for both arguments: 您可以添加额外的特化，这是两个参数最专业的：

#include <tuple>

template<typename A, typename B>
struct front_helper;

template<typename... A, typename... B>
struct front_helper<std::tuple<A...>, std::tuple<B...>> {
    typedef std::tuple<A..., B...> type;
};

template<std::size_t, typename>
struct front;

template<>
struct front<0, std::tuple<>> {
    typedef std::tuple<> type;
};
template<typename Head, typename... Tail>
struct front<0, std::tuple<Head, Tail...>> {
    typedef std::tuple<> type;
};
template<std::size_t n, typename Head, typename... Tail>
struct front<n, std::tuple<Head, Tail...>> {
    typedef typename front_helper<std::tuple<Head>, typename front<n-1, std::tuple<Tail...>>::type>::type type;
};

void x0(front<0, std::tuple<int, float, double, long>>::type) { }
void x1(front<1, std::tuple<int, float, double, long>>::type) { }
void x2(front<2, std::tuple<int, float, double, long>>::type) { }
void x3(front<3, std::tuple<int, float, double, long>>::type) { }
void x4(front<4, std::tuple<int, float, double, long>>::type) { }

$ g++ test.cc -c -std=c++0x && nm -C test.o
00000000 b .bss
00000000 d .data
00000000 t .text
00000000 T x0(std::tuple<>)
00000005 T x1(std::tuple<int>)
0000000a T x2(std::tuple<int, float>)
0000000f T x3(std::tuple<int, float, double>)
00000014 T x4(std::tuple<int, float, double, long>)
00000000 b std::(anonymous namespace)::ignore

为什么这个模板专业化在GCC中不起作用？

问题描述

1 个解决方案

解决方案1
2

为什么这个模板专业化在GCC中不起作用？

问题描述

1 个解决方案

解决方案1 2

解决方案1
2