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The return type alias of a const overloaded function

I have the following overloaded function:

template<size_t N, typename T>
auto get(const T & _t) -> decltype(std::get<...>(_t)) {
    ...
}

template<size_t N, typename T>
auto get(T & _t) -> decltype(std::get<...>(_t)) {
    ...
}

First question:

the first one uses std::get(const tuple<_Elements...>& __t)

 and the second one std::get(tuple<_Elements...>& __t)

??

now I want to alias the return type of my new function get

:

using type = typename decltype(aux::get<I>(data))::type;

which is used here? const or not? and how can i choose? I would like to alias like !! data

is not constant

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1 answer


Yes, the former uses overload const

and the latter doesn't const

. This is because _t

const

in the first case and not const

in the second.

Which one is used in a type alias depends on the type data

. Is it const

? If so, the overload const

is done with an alias. If not, then not const

one.



To get any type of "virtual value" you can use std::declval

. This code will match the version const

:

using type = typename decltype(aux::get<I>(std::declval<const YourTypeHere>()))::type;
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