c++ - 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

此结构化绑定等效于：

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

既然tuple有类型std::tuple<int&, int>，那也是的类型e。

结构化绑定语法没有从tuple. 但是，您可以像在这个问题中那样创建一个辅助函数来执行此操作。这是一个工作示例：

#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';
}

输出：

5 6
5 6