Is it possible for the compiler to deduce a parameter pack from a function's default aargument? Particularly, I have the following code:
template <int ... Is> struct seq {};
template <int ... Is> struct make_seq;
template <int head, int ... tail>
struct make_seq<head, tail...>
{
using type = typename make_seq<head - 1, head - 1, tail...>::type;
};
template <int ... Is>
struct make_seq<0, Is...>
{
using type = seq<Is...>;
};
template <int N>
using make_seq_t = typename make_seq<N>::type;
template<int N, int ...Is>
int deduceParamPack(seq<Is...> s = make_seq_t<N>{})
{
return sizeof...(Is);
}
int main()
{
return deduceParamPack<5>();
}
And the compiler deduces the parameter pack as empty, and tries to cast the default argument to it. Instead, I would like to achieve a similar behaviour to:
int main()
{
return deduceParamPack<5>(make_seq_t<5>{});
}
where the deduced parameter pack is 0,1,2,3,4, without explicitly passing in this argument.
Is it possible for the compiler to deduce a parameter pack from a function's default aargument?
No, as far I know.
But... not exactly what you asked... but maybe you can find useful the following solution based in partial specialization of structs
template <std::size_t N, typename = std::make_index_sequence<N*N>>
struct deduceParamPackStruct;
template <std::size_t N, std::size_t ... Is>
struct deduceParamPackStruct<N, std::index_sequence<Is...>>
{
static constexpr std::size_t func ()
{ return sizeof...(Is); }
};
You can use it as follows
static_assert( 25 == deduceParamPackStruct<5>::func() );