structured bindings, reference ? Is it possible to remove them

Question

Let's say I have such a tuple.

std::tuple<int &, int> tuple{};

I want to do something like that :

auto [i1, i2] = tuple; // Here i1 is lvalue reference, i2 is int

i1 is a lvalue reference because the first value on the tuple is one lvalue reference. However, I did not write auto &[i1, i2]. So, is there a possibility to "remove" the reference in this case? So that I got i1 and i2 as "simple" int. Thanks !

Answer 1

This structured binding is equivalent to:

auto e = tuple;  
auto&& i1 = e.get<1>();
auto&& i2 = e.get<2>();

Since tuple has type std::tuple<int&, int>, that is also the type of e.

The structured binding syntax doesn't have a flavour to remove references from inside the tuple. However you can make a helper function to do that as in this question. Here is a working example:

#include <tuple>
#include <iostream>

template <typename... T>
using tuple_with_removed_refs = std::tuple<typename std::remove_reference<T>::type...>;

template <typename... T>
tuple_with_removed_refs<T...> remove_ref_from_tuple_members(std::tuple<T...> const& t) {
    return tuple_with_removed_refs<T...> { t };
}

int main()
{
    int x{5}, y{6};

    std::tuple<int& , int> t(x, y);

    auto [i1, i2] = remove_ref_from_tuple_members(t);

    std::cout << i1 << ' ' << i2 << '\n';
    i1 = 7; i2 = 8;
    std::cout << x << ' ' << y << '\n';
}

Output:

5 6
5 6