0
我想从另一个元组的前n个元素创建一个元组。这是我使用的递归(在这里不计算任何有用的东西)。这个模板专业为什么不在GCC中工作?
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,应该选择它。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;