Question

是否可以多次打开参数包的包装？

例如：我想得到一个包含4个向量的元组 - 其中2个为int类型，2个为float类型。要创建这样的元组，我想使用如下语法：

ExampleClass<2, int, float> class;

是否可以创建这样的课程？我正在考虑这样的事情：

template <int numUnpacks, typename ... Types>
class ExampleClass
{
     using Types = std::tuple<std::vector<Types>...>; // here i don't know how to unpack "std::vector<Types>...>" as often as "numUnpacks">
}

Answer 1

通常的递归模板部分特化废话似乎很简单：

template<int n,class T,class... TT>
struct Impl;

template<int n,class... TT,class... UU>
struct Impl<n,std::tuple<TT...>,UU...> :
  Impl<n-1,std::tuple<TT...,UU...>,UU...> {};

template<class... TT,class... UU>  // just "class T" would be ambiguous
struct Impl<0,std::tuple<TT...>,UU...>
{using Types=std::tuple<std::vector<TT>...>;};

template<int n,class... TT>
struct ExampleClass : Impl<n,std::tuple<>,TT...> {};

诀窍是尽早创建tuple，以便我们一次可以保留多个模板参数包。

Answer 2

这可以使用C ++ 14＆＃39; integer_sequence来完成。如果您还没有，请implementation Jonathan Wakely。

template <int numUnpacks, typename... Types>
struct ExampleClass
{
    template<typename T, size_t>
    using T_alias = T;

    template<typename T, size_t N, typename I = std::make_index_sequence<N>>
    struct repeat;

    template<typename T, size_t N, size_t... I>
    struct repeat<T, N, std::index_sequence<I...>> {
        using type = decltype(std::tuple_cat(std::declval<T_alias<T, I>>()...));
    };

    using type = typename repeat<std::tuple<std::vector<Types>...>, numUnpacks>::type;
};

ExampleClass<2, int, float>::type是std::tuple<std::vector<int>, std::vector<float>, std::vector<int>, std::vector<float>>

Live demo

多次打开参数包的包装

2 个答案: