How to initialize std::array elegantly if T is not default constructible?

Question

How do I initialize std::array if T is not default constructible?

I know it\'s possible to initialize it like that:

Answer 1

Given N, you could generate a sequence-type calledseq<0,1,2,3,...N-1> using a generator called genseq_t<>, then do this:

template
void f(T value)
{
     //genseq_t is seq<0,1,...N-1>
     std::array items = repeat(value, genseq_t{});
}

where repeat is defined as:

template
auto repeat(T value, seq) -> std::array 
{
   //unpack N, repeating `value` sizeof...(N) times
   //note that (X, value) evaluates to value
   return {(N, value)...}; 
}

And the rest is defined as:

template
struct seq
{
   using type = seq;

   static const std::size_t size = sizeof ... (N);

   template
   struct push_back : seq {};
};

template
struct genseq : genseq::type::template push_back {};

template<>
struct genseq<0> : seq<> {};

template
using genseq_t = typename genseq::type;

Online demo

Hope that helps.