result_of，make_tuple，参数包

Question

我想获取std::make_tuple为给定参数包返回的类型。到目前为止，我已经编写了以下代码：

#include <tuple>
#include <functional>

template <class T>
struct unwrap_refwrapper
{
    using type = T;
};

template <class T>
struct unwrap_refwrapper<std::reference_wrapper<T>>
{
    using type = T&;
};

template <class T>
using special_decay_t = typename unwrap_refwrapper<typename std::decay<T>::type>::type;

template<class ... Types>
struct foo
{
    typedef std::tuple<special_decay_t<Types>...> tuple_t;
};

int main()
{
    short s;
    // t should be std::tuple<int, double&, short&>
    typedef foo<int, double&, decltype(std::ref(s))>::tuple_t t;
}

但是我觉得复制部分possible implementation of std::make_tuple非常难看，我在这里做了。

我想使用std::result_of或类似的东西来实现给定的效果。

我的尝试如下：

#include <tuple>
#include <functional>

template<class ... Types>
struct foo
{
    typedef typename std::result_of<
        std::make_tuple(Types...)>::type tuple_t;
};

int main()
{
    short s;
    // t should be std::tuple<int, double&, short&>
    typedef foo<int, double&, decltype(std::ref(s))>::tuple_t t;
}

但确实not compile。

怎么做？

Answer 1

template<class... Ts>
struct foo
{
    using tuple_t = decltype(std::make_tuple(std::declval<Ts>()...));
};