如何获取指向std :: get &lt;0，tuple的指针 <int,int> &gt;）？

Question

I know that std::get is overloaded. And I know that to extract an overload I need to cast to a specific signature. Let say I need a pointer to std::get which returns non-const ref to 1st element from std::tuple&. Below is one of my many attempts (does not compile):

auto f = static_cast<
    int& (*)(std::tuple<int,int>&) noexcept
>(
    &std::get<(size_t)0u, std::tuple<int,int>>
);

How should I specify this static_cast?

Answer 1

The signature of the tuple get is (directly taken from libstdc++):

 template<std::size_t __i, typename... _Elements>
    constexpr typename __add_ref<     
                      typename tuple_element<__i, tuple<_Elements...>>::type 
                    >::type 
    get(tuple<_Elements...>& __t) noexcept

As such the template parameter to get is the different types of the tuple, not the tuple, so the function you take the address from should be:

&std::get<(size_t)0u,int,int> 

Answer 2

Do you need a pointer to an actual std::get or any function with the same behavior will do? If so you can just wrap it in a lambda:

auto f = [](std::tuple<int,int>& tuple) -> int&
{
    return std::get<0>(tuple);
}