std tuple apply Revisited

[Remi123]

Here is some code I wrote on CompilerExplorer that work on tuple-like structure

There is some work to be done to make it work with lambda that "return void" and others cool things. This is the direction I want to go with te::tup<...>.

template<std::size_t ... Is,class F,typename Ts>
constexpr auto apply(F&& f,Ts&& t)
-> typename std::enable_if<sizeof...(Is) >= 1,decltype(f(std::get<Is>(std::forward<Ts>(t))...))>::type
{
    return f(std::get<Is>(std::forward<Ts>(t))...);
}
template <class F, class  Tuple>
constexpr auto apply(F&& f, Tuple&& t)
-> decltype(apply(std::forward<F>(f),std::forward<Tuple>(t),std::make_index_sequence<std::tuple_size<Tuple>::value>{}))
{
    using Iseq = std::make_index_sequence<std::tuple_size<Tuple>::value>; 
    return apply(std::forward<F>(f),std::forward<Tuple>(t),Iseq{});
}
template<class F,typename Tuple, std::size_t ... Is>
constexpr auto apply(F&& f,Tuple&& t, std::integer_sequence<std::size_t,Is...>)
-> decltype(f(std::get<Is>(std::forward<Tuple>(t))...))
{
    return f(std::get<Is>(std::forward<Tuple>(t))...);
}


int main()
{
    apply([](int i,int j){ std::cout <<  i << " " << j; return true;} , std::tuple<int,int>{1,2}); // print 1 2
    apply<0>([](int i){ std::cout <<  i  return true;} , std::tuple<int,int>{1,2}); // print 1
    apply<1>([](int i){ std::cout <<  i  return true;} , std::tuple<int,int>{1,2}); // print 2
    apply<1,0>([](int i,int j){ std::cout <<  i << " " << j; return true;} , std::tuple<int,int>{1,2}); // print 2 1
    apply<1,1>([](int i,int j){ std::cout <<  i << " " << j; return true;} , std::tuple<int,int>{1,2}); // print 2 2
}