Remove default-constructibility requirement in ChainRange

[Hazurl]

Remove default-constructibility requirement in ChainRange

[simonask]

That line exists solely to silence "no return statement in non-void function" warnings. I wonder if calling std::abort() and adding [[noreturn]] would be more elegant.

[Hazurl]

No. You just need to not create the _get method with an empty index_sequence, and never call it.
Nevermind you always get a branch with no return statement with my solution.

[Hazurl]

I end up using your solution as this snippet is ugly (returning the last element if not in range):

    template <size_t i, size_t... rest>
    constexpr output_type _get(std::index_sequence<i, rest...>) const {
        RX_ASSERT(current_input < sizeof...(Rs)); // chain was at end!
        if constexpr (sizeof...(rest) <= 0) {
            return std::get<i>(inputs).get();
        } else {
            if (i == current_input) {
                return std::get<i>(inputs).get();
            } else {
                return _get(std::index_sequence<rest...>{});
            }
        }
    }

[simonask]

And now I realize that the drawback of using std::abort() here is that ChainRange can no longer be used at all in a constexpr context. I wonder if throw can be used instead, but I'm not sure because I think the recursion base case will be compiled for any instantiation of get<size_t...>. This might have already been true with the assertion.

Alexandrescu has a solution here, but nobody seems to know why it works: https://akrzemi1.wordpress.com/2017/05/18/asserts-in-constexpr-functions/

Hmm. I think it's more important to lift the default-constructibility requirement than to remain constexpr. I'll have a look at ensuring constexpr-ness later.