How to get from a tuple to a tuple of references to elements in the tuple?

问题

I have a C++11 tuple, and I would like to get a tuple of std::reference_wrappers to the same elements of the tuple. Is there an easy way to do that?

Thanks

回答1:

Mapping a tuple is easy given a pack of indices, e.g.:

#include <tuple>
#include <functional>
#include <iostream>

template <int...>
struct Seq {};

template <int n, int... s>
struct Gens : Gens<n-1, n-1, s...> {};

template <int... s>
struct Gens<0, s...> {
  typedef Seq<s...> Type;
};

// The above are taken from https://stackoverflow.com/q/7858817

Using the indices we could apply std::ref to every element of the tuple using std::get:

template <int... s, typename Tuple>
auto ref_tuple_impl(Seq<s...> seq, Tuple& tup)
-> std::tuple<
    std::reference_wrapper<
        typename std::tuple_element<s, Tuple>::type
    >...
>
{
    return std::make_tuple(std::ref(std::get<s>(tup))...);
}

template <typename Tuple>
auto ref_tuple(Tuple& tup)
    -> decltype( ref_tuple_impl(typename Gens<std::tuple_size<Tuple>::value>::Type>(), tup) )
{
    return ref_tuple_impl(typename Gens<std::tuple_size<tuple>::value>::Type(), tup);
}

Usage demonstration:

int main() {
    auto t = std::make_tuple(1, 4.5, "66");
    auto rt = ref_tuple(t);

    std::get<0>(rt).get() = 123;
    std::get<1>(rt).get() = 5.67;
    std::get<2>(rt).get() = "34";

    std::cout << std::get<0>(t) << std::endl;
    std::cout << std::get<1>(t) << std::endl;
    std::cout << std::get<2>(t) << std::endl;
}

来源：https://stackoverflow.com/questions/13070218/how-to-get-from-a-tuple-to-a-tuple-of-references-to-elements-in-the-tuple

标签

c++

c++11

tuples