C++11-元组数组中的元组

C++ 11 - Tuple of Arrays from Array of Tuple

本文关键字：元组 数组 C++11-      更新时间：2023-10-16

给定一个std:：元组，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>>

正如您所看到的，是原始类型的向量的元组。以下是标准场景：

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

场景1的访问方式如下：

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

场景2的访问方式如下：

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

场景3的访问方式如下：

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.
}

HereIsTheQuestion:：Type必须如何编写以类似于原始Struct类型的数组元组？

谢谢，m.

以下是应该如何实现HereIsTheQuestion。

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>>

希望能有所帮助。

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;
}

在那之后，事情就变得简单了。这是您的程序：

#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);
}

演示

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