我想要我的课
template <class T, unsigned int n>
class X;
创建一个 std::tuple
,其中包含 n
倍的 T
类型。有没有特别巧妙的方法?对于任意可变参数模板类,是否还有一种很好的方法来做到这一点?
这是我首先做的:
#include <tuple>
template <class, unsigned int, class>
struct simple_repeat_helper;
template <class T, unsigned int n, class... Args>
struct simple_repeat_helper<T, n, std::tuple<Args...>>
{
typedef typename simple_repeat_helper<T, n-1, std::tuple<Args..., T>>::type type;
};
template <class T, class... Args>
struct simple_repeat_helper<T, 0, std::tuple<Args...>>
{
typedef std::tuple<Args...> type;
};
template <class T, unsigned int n>
struct simple_repeat
{
using type = typename simple_repeat_helper<T, n, std::tuple<>>::type;
};
但实际上,对于 std::tuple
我不需要这个,但是对于另一个行为类似的类。所以我想我会创建一个更通用的版本:
template <class, unsigned int, template <class...> class, class>
struct repeat_helper;
template <class T, template <class...> class M, class... Args>
struct repeat_helper<T, 0, M, M<Args...>>
{
typedef M<Args...> type;
};
template <class T, unsigned int n, template <class...> class M, class... Args>
struct repeat_helper<T, n, M, M<Args...>>
{
typedef typename repeat_helper<T, n-1, M, M<Args..., T>>::type type;
};
template <class T, unsigned int n, template <class...> class M = std::tuple>
struct repeat
{
using type = typename repeat_helper<T, n, M, M<>>::type;
};
我以为我可以这样使用它:
repeat<double, 5, std::tuple>::type x = std::make_tuple( 1., 2., 3., 4., 5. );
但不幸的是它未能compile由于:
ambiguous class template instantiation for ‘struct repeat_helper<double, 0u, std::tuple, std::tuple<double, double, double, double, double> >’
如有任何关于此错误的帮助,我们将不胜感激!
最佳答案
我会这样做:
template<typename, typename>
struct append_to_type_seq { };
template<typename T, typename... Ts, template<typename...> class TT>
struct append_to_type_seq<T, TT<Ts...>>
{
using type = TT<Ts..., T>;
};
template<typename T, unsigned int N, template<typename...> class TT>
struct repeat
{
using type = typename
append_to_type_seq<
T,
typename repeat<T, N-1, TT>::type
>::type;
};
template<typename T, template<typename...> class TT>
struct repeat<T, 0, TT>
{
using type = TT<>;
};
作为一个小测试:
#include <type_traits>
#include <tuple>
template<typename... Ts>
struct X { };
int main()
{
repeat<double, 5, std::tuple>::type t = std::make_tuple(1., 2., 3., 4., 5.);
static_assert(
std::is_same<
decltype(t),
std::tuple<double, double, double, double, double>
>::value, "!");
repeat<double, 3, X>::type y;
static_assert(
std::is_same<decltype(y), X<double, double, double>>::value, "!");
}
最后,一个 live example .
关于c++ - 如何创建包含 n 次相同类型的类型列表(用于可变参数模板)?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23985879/