Yes. std::optional<T> allocates the T in-place. Just checked compiler explorer, and on g++ 8.1, sizeof(std::string) == 32, sizeof(std::optional<std::string>) ==

Yes. std::optional<T> allocates the T in-place. Just checked compiler explorer, and on g++ 8.1, sizeof(std::string) == 32, sizeof(std::optional<std::string>) == 40. (of course sizeof(char* ) == 8)