How to unpack a variadic template parameter with a numeric sequence?

Question

How to (or is it possible to) unpack a parameter pack with a numeric sequence? For example,

template <typename C, typename... T>
C* init_from_tuple(bp::tuple tpl)
{
   return new C{bp::extract<T>("magic"(tpl))...}; // <--
}

which the <-- line should expand to

   return new C{bp::extract<T0>(tpl[0]),
                bp::extract<T1>(tpl[1]),
                .....
                bp::extract<Tn>(tpl[n])};

where n == sizeof...(T) - 1.

The purpose is to create a __init__ function for Boost.Python which accepts a tuple with predefined types.

score 3 · Accepted Answer · answered Nov 06 '11 at 16:11

Actually, it is possible for the unpack operations to target two different packs of parameters at once (I think they need be of equal length). Here we would like a pack of types, and a pack of numbers.

Something akin to:

template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl) {
  return new C{ bp::extract<T>(tpl[N])... };
}

We "just" need to generate the pack of indices:

template <size_t... N> struct Collection {};

template <typename C> struct ExtendCollection;

template <size_t... N>
struct ExtendCollection< Collection<N...> > {
  typedef Collection<N..., sizeof...(N)> type;
};

template <typename... T>
struct GenerateCollection;

template <>
struct GenerateCollection<> { typedef Collection<> type; };

template <typename T0, typename... T>
struct GenerateCollection<T0, T...> {
  typedef typename ExtendCollection<
    typename GenerateCollection<T...>::type
  >::type type;
};

And then use it:

template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl, Collection<N...>) {
  return new C { bp::extract<T>(tpl[N])... };
}

template <typename C, typename... T>
C* init_from_tuple(bp::tuple tpl) {
  typename GenerateCollection<T...>::type collection;
  return init_from_tuple_impl<C, T...>(tpl, collection);
}

In action at Ideone.

We can witness the correctness by making a "mistake" in the implementation of init_from_tuple_impl (remove the new for example):

template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl, Collection<N...>) {
  return C { bp::extract<T>(tpl[N])... };
}

In action at Ideone:

prog.cpp: In function 'C* init_from_tuple_impl(bp::tuple, Collection<N ...>)
[with
    C = bp::Object,
    T = {int, float, char},
    unsigned int ...N = {0u, 1u, 2u},
    bp::tuple = std::basic_string<char>
]':

Exactly what we wanted :)

score 1 · Answer 2 · answered Nov 06 '11 at 15:23

Its possible if you first extract the parameters into their own pack and then call the constructor. its far from finished, but you get the general idea:

template <typename C, int N, typename... T>
C* init_from_tuple(bp::tuple tpl, T... args) // enable_if N == sizeof...(T)
{
    return new C{args...};
}

template <typename C, int N, typename T0, typename... T>
C* init_from_tuple(bp::tuple tpl, T... args) // enable_if N != sizeof...(T)
{
    return init_from_tuple<C, N + 1>(tpl, args, bp::extract<T0>(tpl[N]));
}

template <typename C, typename... T>
C* init_from_tuple(bp::tuple tpl, T... args)
{
    return init_from_tuple<C, 0, T...>(tpl, args);
}

Use boost's enable_if to make the places indicated only enabled in some cases, and probably the template arguments need some changes, but this is a start.

How to unpack a variadic template parameter with a numeric sequence?

2 Answers2

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