Question

I was looking at this SO question and I couldn’t understand how the answer worked. I will post a copy of code in one of the answers for reference:

template<int ...> struct seq {};

// How does this line work?
template<int N, int ...S> struct gens : gens<N-1, N-1, S...> {};

template<int ...S> struct gens<0, S...>{ typedef seq<S...> type; };

double foo(int x, float y, double z)
{
    return x + y + z;
}

template <typename ...Args>
struct save_it_for_later
{
  std::tuple<Args...> params;
  double (*func)(Args...);

  double delayed_dispatch()
  {
     return callFunc(typename gens<sizeof...(Args)>::type());
  }

  template<int ...S>
  double callFunc(seq<S...>)
  {
     return func(std::get<S>(params) ...);
  }
};

int main(void)
{
  std::tuple<int, float, double> t = std::make_tuple(1, 1.2, 5);
  save_it_for_later<int,float, double> saved = {t, foo};
  cout << saved.delayed_dispatch() << endl;
}

The part I don’t understand is this:

template<int N, int ...S> struct gens : gens<N-1, N-1, S...> {};

In the example at return callFunc(typename gens<sizeof...(Args)>::type()); I’m assuming that sizeof..(Args) would be 3.
So,

template<int N, int ...S> struct gens : gens<N-1, N-1, S...> {};

becomes

template<3, {}> struct gens : gens<3-1, 3-1, {}> {};

Is this correct, and if so, what happens from there?

Answer 1

Let’s write down the recursion manually:

gens<3> : gens<2, 2>
gens<3> : gens<2, 2> : gens<1, 1, 2>
gens<3> : gens<2, 2> : gens<1, 1, 2> : gens<0, 0, 1, 2>

The recursion stops because of the partial specialization for 0:

struct gens<0, S...>{ typedef seq<S...> type; }; 

// first 0 consumed by the partial specialization
// S = 0,1,2
struct gens<0, 0, 1, 2> { 
    typedef seq<0, 1, 2> type;
}