Produce std::tuple of same type in compile time given its length by a template argument

Question

In c++, how can I implement a function with an int template argument indicating the tuple length and produce a std::tuple with that length?

E.g.

func<         


        

Answer 1

Using an index_sequence and a helper type alias you can generate the type you want:

// Just something to take a size_t and give the type `int`
template 
using Integer = int;

// will get a sequence of Is = 0, 1, ..., N
template 
auto func_impl(std::index_sequence) {
    // Integer... becomes one `int` for each element in Is...
    return std::tuple...>{};
}

template 
auto func() {
    return func_impl(std::make_index_sequence{});
}

It is worth calling out that in the general case you would probably be better with a std::array, (in your case you can't use one), but a std::array can behave like a tuple, similarly to a std::pair.

Update: since you've made it clear you're working with c++11 and not 14+, you'll need to get an implementation of index_sequence and related from somewhere (here is libc++'s). Here is the C++11 version of func and func_impl with explicit return types:

template 
auto func_impl(std::index_sequence) -> std::tuple...> {
  return std::tuple...>{};
}

template 
auto func() -> decltype(func_impl(std::make_index_sequence{})) {
  return func_impl(std::make_index_sequence{});
}