c++ - 默认在std::variant中构造所有类型，并将它们放入std::vector中

Question

我将如何制作一个std::vector，其中包含std::variant中包含的所有类型的默认构造的实例？

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.解决方案:您可以将变体转换为元组，然后使用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{});
}

编辑:
2.解决方案:使用 make_index_sequence ， variant_alternative_t 和 variant_size_v (所有C++ 17)(感谢 @s通过将sant 指向__on_strong，将 _ont而不是递归):

#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>>{});
}