Question

给定std :: tuple，例如：

std::tuple<int, float, char>

我想生成这样的类型：

std::tuple<std::vector<int>, std::vector<float>, std::vector<char>>

你可以看到，它是原始类型向量的元组。这是标准方案：

typedef std::tuple<int, float, char>    Struct;          // scenario 1
typedef std::vector<Struct>             ArrayOfStructs;  // scenario 2
typedef HereIsTheQuestion<Struct>::Type StructOfArrays;  // scenario 3

方案1旨在像这样访问：

Struct x = ...; // single tuple
std::get<0>(x) = 11;
// etc.

方案2旨在像这样访问：

ArrayOfStructs xs = ...; // array of tuples
for (size_t i=0; i<xs.size(); ++i) {
    std::get<0>(xs[i]) = 11;
    // etc.
}

方案3旨在像这样访问：

StructsOfArrays xs = ...; // single tuple of arrays
size_t n = std::get<0>(xs).size(); // first tuple array size
for (size_t i=0; i<n; ++i) {
    std::get<0>(xs)[i] = 11;
    // etc.
}

如何将HereIsTheQuestion :: Type写成类似于原始Struct类型的数组元组？

谢谢，米。

Answer 1

以下是HereIsTheQuestion的实施方式。

template<typename T>       //primary template
struct HereIsTheQuestion;  //leave it undefined

template<typename ...T>
struct HereIsTheQuestion<std::tuple<T...>>  //partial specialization
{
    using Type = std::tuple<std::vector<T>...>;
};

现在

HereIsTheQuestion<std::tuple<int, float, char>>::Type

是

std::tuple<std::vector<int>,std::vector<float>, std::vector<char>>

希望有所帮助。

Answer 2

您可以使用此模板创建类型：

namespace detail
{
    template <typename... Ts>
    struct tuple_change { };

    template <typename... Ts>
    struct tuple_change<std::tuple<Ts...>>
    {
        using type = std::tuple<std::vector<Ts>...>;
    };
}

并创建一个这样的索引序列：

namespace detail
{
    template <int... Is>
    struct index { };

    template <int N, int... Is>
    struct gen_seq : gen_seq<N - 1, N - 1, Is...> { };

    template <int... Is>
    struct gen_seq<0, Is...> : index<Is...> { };
}

您还需要一个模板来允许打印元组：

template <typename... Ts, int... Is>
static void print(std::tuple<Ts...>& var, detail::index<Is...>)
{
    auto l = { (print(std::get<Is>(var)), 0)... };
    (void)l;
}

template <typename... Ts>
static void print(std::tuple<Ts...>& var)
{
    print(var, detail::gen_seq<sizeof...(Ts)>{});
}

template <typename T>
static void print(std::vector<T>& v)
{
    for (auto a : v)
    {
        std::cout << std::boolalpha << a << std::endl;
    }
    std::cout << std::endl;
}

之后变得简单。这是你的计划：

#include <iostream>
#include <tuple>
#include <vector>

namespace detail
{
    template <typename... Ts>
    struct tuple_change { };

    template <typename... Ts>
    struct tuple_change<std::tuple<Ts...>>
    {
        using type = std::tuple<std::vector<Ts>...>;
    };

    template <int... Is>
    struct index { };

    template <int N, int... Is>
    struct gen_seq : gen_seq<N - 1, N - 1, Is...> { };

    template <int... Is>
    struct gen_seq<0, Is...> : index<Is...> { };
}

template <typename... Args, int... Is>
void fill(std::tuple<Args...>& var, detail::index<Is...>)
{
    auto l = { (std::get<Is>(var).assign(5, 11), 0)... };
    // here I just decided to make the size 5
    (void)l;
}

template <typename... Args>
void fill(std::tuple<Args...>& var)
{
    fill(var, detail::gen_seq<sizeof...(Args)>{});
}

template <typename T>
static void print(std::vector<T>& v)
{
    for (auto a : v)
    {
        std::cout << std::boolalpha << a << std::endl;
    }
    std::cout << std::endl;
}

template <typename... Ts, int... Is>
static void print(std::tuple<Ts...>& var, detail::index<Is...>)
{
    auto l = { (print(std::get<Is>(var)), 0)... };
    (void)l;
}

template <typename... Ts>
static void print(std::tuple<Ts...>& var)
{
    print(var, detail::gen_seq<sizeof...(Ts)>{});
}

using result_type = detail::tuple_change<std::tuple<int, bool>>::type;

int main()
{
    result_type r;

    fill(r);
    print(r);
}

的 Demo 的

C ++ 11 - 元组数组中的数组元组

2 个答案: