Would it be possible to generate a variant from an integer sequence like in this pseudo C++ sample:
#include <utility>
#include <variant>
template <int MyInt>
struct MyStruct{};
using MyIntegerSequence = std::make_integer_sequence<int, 3>;
using MyVariant = std::variant<MyStruct<MyIntegerSequence>...>; // Does not compile
int main()
{
return sizeof(MyVariant);
}
On coliru: http://coliru.stacked-crooked.com/a/2439048b107642c2
It's possible, sure. The key is that one must use a pack expansion of some sort. Your struct takes an int non-type template parameter. So we must expand the integer sequence's pack to feed them into MyStruct one at a time.
MyStruct<MyIntegerSequence>... is not a pack expansion, since there is no template definition and no pack in that context.
For example, with a class template (and partial specialization):
template<typename Seq> struct MyVariantHelper;
template<int... Is> struct MyVariantHelper<std::integer_sequence<int, Is...>> {
using type = std::variant<MyStruct<Is>...>;
};
using MyVariant = MyVariantHelper<MyIntegerSequence>::type;
Or with a function template and decltype:
template<int... Is>
auto MyVariantHelper(std::integer_sequence<int, Is...>)
-> std::variant<MyStruct<Is>...>;
using MyVariant = decltype(MyVariantHelper(MyIntegerSequence{}));