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 !
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
