Question

Is it possible to use the C++11 initializer_list to assemble a recursively defined class such as Foo, below, using constexpr constructors:

template <size_t N>
struct Foo {
  constexpr Foo(int x, Foo<N-1> f) : x(x), xs(xs) {}
  int x;
  Foo<N-1> xs;
};

template <> struct Foo<0> {};

I can initialise a Foo<3> using:

int main(int argc, char *argv[])
{
  Foo<3> a = Foo<3>(1,Foo<2>(2,Foo<1>(3,Foo<0>())));
  return 0;
}

It would be nice to use Foo<3> a = {1,2,3} instead. If there was a constexpr tail function in initializer_list I think it should work.

Answer 1

Yes, in a kind of round-about way, effectively unpacking and repacking the initializer list to a more suited format. However, there is a better (imho) way: Variadic templates.

#include <stddef.h>
#include <iostream>

template <size_t N>
struct Foo {
  template<class... Tail>
  constexpr Foo(int i, Tail... t) : x(i), xs(t...) {}

  void print(){
    std::cout << "(" << x << ", ";
    xs.print();
    std::cout << ")";
  }

  int x;
  Foo<N-1> xs;
};

template <> 
struct Foo<1> {
  constexpr Foo(int i) : x(i) {}
  void print(){ std::cout << "(" << x << ")"; }
  int x;
};

int main(){
 Foo<3> x = {1, 2, 3};
 x.print();
 std::cout << "\n";
}

Output as expected:

(1, (2, (3)))

Note that I chose 1 as the base case, as it simply makes more sense.