c 11 – Fusion adaped std_tuple视图,转换为另一个元组

2023年9月27日 298次阅读

Boost Fusion的设计方式使得大多数转换都是“懒惰的”,因为它们都生成“视图”但不生成实际(Fusion)容器(
http://www.boost.org/doc/libs/1_58_0/libs/fusion/doc/html/fusion/algorithm.html).因此,例如,为了实际反转向量,需要使用转换函数as_vector(
http://www.boost.org/doc/libs/1_58_0/libs/fusion/doc/html/fusion/container/conversion/functions.html).

boost::fusion::vector<int, double, std::string> vec;
auto view_rev = boost::fusion::reverse(vec); // view object
auto vec_rev = boost::fusion::as_vector(view_rev);

现在,我想用改编的std :: tuple做到这一点：

#include<boost/fusion/adapted/std_tuple.hpp>
...
std::tuple<int, double, std::string> tup;
auto view_rev = boost::fusion::reverse(tup);
auto tup_rev = boost::fusion::???(view_rev); // type should be of type std::tuple<std::string, double, int>

如何将生成的视图转换回元组？

我期待这个？函数被称为as_std_tuple(类似于boost :: fusion :: as_vector,但它不存在(但是？).

有一些用于反转元组的解决方案,在这种情况下,我只想使用Boost Fusion中已有的内容.

最佳答案 我不知道将Boost Fusion序列转换为std :: tuple的任何内置方法,但使用
indices trick它可以很容易地实现：

template <std::size_t... Is>
struct indices {};

template <std::size_t N, std::size_t... Is>
struct build_indices
  : build_indices<N-1, N-1, Is...> {};

template <std::size_t... Is>
struct build_indices<0, Is...> : indices<Is...> {};

template<typename Sequence, std::size_t ...Is>
auto as_std_tuple_impl(const Sequence& s, indices<Is...>&&) -> decltype(std::tie(boost::fusion::at_c<Is>(s)...))
{
    return std::tie(boost::fusion::at_c<Is>(s)...);
}

template <typename Sequence, typename Indices = build_indices<boost::fusion::result_of::size<Sequence>::value>>
auto as_std_tuple(const Sequence& s) -> decltype(as_std_tuple_impl(s, Indices()))
{
    return as_std_tuple_impl(s, Indices());
}

下面是一个完整的示例,它使用boost :: fusion :: reverse来反转一个改编的std :: tuple并将其转换回std :: tuple并打印两个元组：

#include <tuple>
#include <utility>

#include<boost/fusion/adapted/std_tuple.hpp>
#include <boost/fusion/algorithm/transformation/reverse.hpp>
#include <boost/fusion/include/reverse.hpp>

#include <boost/fusion/sequence/intrinsic/size.hpp>
#include <boost/fusion/include/size.hpp>

#include <iostream>

template <std::size_t... Is>
struct indices {};

template <std::size_t N, std::size_t... Is>
struct build_indices
  : build_indices<N-1, N-1, Is...> {};

template <std::size_t... Is>
struct build_indices<0, Is...> : indices<Is...> {};

template<typename Sequence, std::size_t ...Is>
auto as_std_tuple_impl(const Sequence& s, indices<Is...>&&) -> decltype(std::tie(boost::fusion::at_c<Is>(s)...))
{
    return std::tie(boost::fusion::at_c<Is>(s)...);
}

template <typename Sequence, typename Indices = build_indices<boost::fusion::result_of::size<Sequence>::value>>
auto as_std_tuple(const Sequence& s) -> decltype(as_std_tuple_impl(s, Indices()))
{
    return as_std_tuple_impl(s, Indices());
}


template<class Tuple, std::size_t N>
struct TuplePrinter
{
    static void print(const Tuple& t) 
    {
        TuplePrinter<Tuple, N-1>::print(t);
        std::cout << ", " << std::get<N-1>(t);
    }
};

template<class Tuple>
struct TuplePrinter<Tuple, 1> 
{
    static void print(const Tuple& t) 
    {
        std::cout << std::get<0>(t);
    }
};

template<class... Args>
void print(const std::tuple<Args...>& t) 
{
    std::cout << "(";
    TuplePrinter<decltype(t), sizeof...(Args)>::print(t);
    std::cout << ")\n";
}

int main()
{
    std::tuple<int, double, std::string> tup(1,2.5,"hello");
    auto view_rev = boost::fusion::reverse(tup);
    auto reversed_tup = as_std_tuple(view_rev);

    print(tup);
    print(reversed_tup);
    return 0;
}

输出：

(1, 2.5, hello)
(hello, 2.5, 1)

Live example on ideone

点赞