具有独特类型的变体

Question

在编写 cpp 库时，我最终发现std::variants 可能包含重复类型，例如std::variant<int, double, int, std::string, int>. 它在某些情况下可能很有用，但对于我的图书馆，我发现这是不必要的。所以我想像这样过滤类型重复：

std::variant<int, double, int, std::string, int>->std::variant<int, double, std::string>

std::variant<std::string, std::string, int>->std::variant<std::string, int>

如何才能做到这一点？

Answer 1

#include <type_traits>
#include <variant>

template <typename T, typename... Ts>
struct filter_duplicates { using type = T; };

template <template <typename...> class C, typename... Ts, typename U, typename... Us>
struct filter_duplicates<C<Ts...>, U, Us...>
    : std::conditional_t<(std::is_same_v<U, Ts> || ...)
                       , filter_duplicates<C<Ts...>, Us...>
                       , filter_duplicates<C<Ts..., U>, Us...>> {};

template <typename T>
struct unique_variant;

template <typename... Ts>
struct unique_variant<std::variant<Ts...>> : filter_duplicates<std::variant<>, Ts...> {};

template <typename T>
using unique_variant_t = typename unique_variant<T>::type;

演示版

Answer 2

对于Boost.Mp11，这是一个简短的单行代码（一如既往）：

using V1 = mp_unique<std::variant<int, double, int, std::string, int>>;
// V1 is std::variant<int, double, std::string>

using V2 = mp_unique<std::variant<std::string, std::string, int>>;
// V2 is std::variant<std::string, int>