std::experimental:: apply

From cppreference.com
Defined in header <experimental/tuple>
template < class F, class Tuple >
constexpr decltype ( auto ) apply ( F && f, Tuple && t ) ;
(library fundamentals TS)

Invoke the Callable object f with a tuple of arguments.

Parameters

f - Callable object to be invoked
t - tuple whose elements to be used as arguments to f

Return value

What returned by f .

Possible implementation

namespace detail
{
    template<class F, class Tuple, std::size_t... I>
    constexpr decltype(auto) apply_impl(F&& f, Tuple&& t, std::index_sequence<I...>)
    {
        return std::invoke(std::forward<F>(f), std::get<I>(std::forward<Tuple>(t))...);
        // Note: std::invoke is a C++17 feature
    }
} // namespace detail
 
template<class F, class Tuple>
constexpr decltype(auto) apply(F&& f, Tuple&& t)
{
    return detail::apply_impl(std::forward<F>(f), std::forward<Tuple>(t),
        std::make_index_sequence<std::tuple_size_v<std::decay_t<Tuple>>>{});
}

Example

#include <iostream>
#include <tuple>
 
template<typename... Ts>
void print_tuple(const std::tuple<Ts...> &tuple)
{
    std::apply([](const auto&... elem) 
    {
        ((std::cout << elem << '\n'), ...); 
    }, tuple);
}
 
int main()
{
    const std::tuple<int, char> t = std::make_tuple(5, 'a');
    print_tuple(t);
}

Output:

5
a

See also

(C++11)
creates a tuple object of the type defined by the argument types
(function template)
creates a tuple of forwarding references
(function template)