Default-construct all the types in a std::variant and put them in a std::vector

Question

How would I make a std::vector that contains a default-constructed instance of all the types contained in a std::variant?

using TaskVariant = std::variant<TypeA, TypeB, TypeC>;
std::vector<TaskVariant> variants;

// I'd like to do this in a loop
variants.push_back(TypeA());
variants.push_back(TypeB());
variants.push_back(TypeC());

Answer 1

1.Solution: You can convert variant to tuple and then use C++17 std::apply:

#include <variant>
#include <iostream>
#include <string>
#include <vector>
#include <tuple>

using namespace std;

template <typename... Types>
struct ToTuple;

template <typename... Types>
struct ToTuple<variant<Types...>>
{
    using type = tuple<Types...>;
};

struct TypeA {};
struct TypeB {};
struct TypeC {};

using TaskVariant = std::variant<TypeA, TypeB, TypeC>;

int main()
{
    vector<TaskVariant> vec;
    apply([&vec](auto&&... args) {(vec.push_back(args), ...); }, ToTuple<TaskVariant>::type{});
}

EDIT:

2.Solution: Using make_index_sequence, variant_alternative_t and variant_size_v (all C++17) (thanks to @s d by pointing out to variant_alternative_t, but I have modified example by using make_index_sequence instead of recursion):

#include <variant>
#include <iostream>
#include <string>
#include <vector>
using namespace std;

struct TypeA {};
struct TypeB {};
struct TypeC {};

using TaskVariant = std::variant<TypeA, TypeB, TypeC>;

template <size_t... Idx>
void fill_vec(vector<TaskVariant>& vec, index_sequence<Idx...> idx)
{
    (vec.push_back(variant_alternative_t<Idx, TaskVariant>{}), ...);
}

int main()
{
    vector<TaskVariant> vec;
    fill_vec(vec, make_index_sequence<variant_size_v<TaskVariant>>{});
}