Question

假设我有一个嵌套的boost::variant - 类型TNested包含一些类型和一些其他boost::variant类型（它们本身不能再包含boost::variant types，所以会有没有递归）。

我正在寻找一个模板别名flatten<TNested>，评估为boost::variant类型没有嵌套boost::variant，例如TFlatten，而正在删除可能的重复类型，例如int只发生一次。

我真的不知道，如果可以某种方式完成。

#include <boost/variant.hpp>
#include <boost/any.hpp>
#include <iostream>

struct Person;

typedef boost::variant<int, double, boost::variant<std::string, int>, boost::variant<Person>> TNested;
typedef boost::variant<int, double, std::string, Person> TFlatten;

template<typename NestedVariant>
using flatten = //???

int main() {
    flatten<TNested> x; 
    std::cout << typeid(x) == typeid(TFlatten) << std::endl;
}

Answer 1

wandbox example

这是一个有效的C ++ 11递归模板解决方案，适用于您的示例：

using nested = variant<int, double, variant<std::string, int>, variant<Person>>;
using flattened = variant<int, double, std::string, int, Person>;

static_assert(std::is_same<flatten_variant_t<nested>, flattened>{}, "");

总体思路：

std::tuple用作泛型类型列表，std::tuple_cat用于连接类型列表。

定义了flatten_variant<TResult, Ts...>递归元函数，它执行以下操作：

TResult将填充扁平类型，并将在递归结束时返回。
- 当Ts...为空时，递归结束。
非变体类型会附加到TResult。
解压缩变体类型，并递归展平其所有内部类型，然后附加到TResult。

实现：

namespace impl
{
    // Type of the concatenation of all 'Ts...' tuples.
    template <typename... Ts>
    using cat = decltype(std::tuple_cat(std::declval<Ts>()...));

    template <typename TResult, typename... Ts>
    struct flatten_variant;

    // Base case: no more types to process.
    template <typename TResult>
    struct flatten_variant<TResult>
    {
        using type = TResult;
    };

    // Case: T is not a variant.
    // Return concatenation of previously processed types,
    // T, and the flattened remaining types.
    template <typename TResult, typename T, typename... TOther>
    struct flatten_variant<TResult, T, TOther...>
    {
        using type = cat<TResult, std::tuple<T>,
                         typename flatten_variant<TResult, TOther...>::type>;
    };

    // Case: T is a variant.
    // Return concatenation of previously processed types,
    // the types inside the variant, and the flattened remaining types.    
    // The types inside the variant are recursively flattened in a new
    // flatten_variant instantiation.
    template <typename TResult, typename... Ts, typename... TOther>
    struct flatten_variant<TResult, variant<Ts...>, TOther...>
    {
        using type =
            cat<TResult, typename flatten_variant<std::tuple<>, Ts...>::type,
                typename flatten_variant<TResult, TOther...>::type>;
    };

    template<typename T>
    struct to_variant;

    template<typename... Ts>
    struct to_variant<std::tuple<Ts...>>
    {
        using type = variant<Ts...>;
    };
}

template <typename T>
using flatten_variant_t =
    typename impl::to_variant<
        typename impl::flatten_variant<std::tuple<>, T>::type
    >::type;

Answer 2

boost::variant的问题在于，它不是真正的可变参数模板，只是通过添加一些默认值为boost::detail::variant::void_的模板参数来模拟一个。要像使用可变参数模板一样使用它，需要先将其更改为具有相关类型的包。完整的c ++ 11即用型方法可以实现您想要的功能，如下所示。该方法考虑了结果类型的平坦性和唯一性假设。如果您决定使用c ++ 14：

，则integer_sequence部分将是不必要的

#include <boost/variant.hpp>
#include <tuple>
#include <type_traits>

template <class T, T... Vs>
struct integer_sequence { };

template <class T, class, class, class = integer_sequence<T>, class = integer_sequence<T, 0>, class = void>
struct make_integer_sequence_impl;

template <class T, T ICV1, T... Res, T... Pow>
struct make_integer_sequence_impl<T, std::integral_constant<T, ICV1>, std::integral_constant<T, 0>, integer_sequence<T, Res...>, integer_sequence<T, Pow...>, typename std::enable_if<(ICV1 > 0)>::type>: make_integer_sequence_impl<T, std::integral_constant<T, ICV1/2>, std::integral_constant<T, ICV1%2>, integer_sequence<T, Res...>, integer_sequence<T, Pow..., (Pow + sizeof...(Pow))...>> { };

template <class T, T ICV1, T... Res, T... Pow>
struct make_integer_sequence_impl<T, std::integral_constant<T, ICV1>, std::integral_constant<T, 1>, integer_sequence<T, Res...>, integer_sequence<T, Pow...>, void>: make_integer_sequence_impl<T, std::integral_constant<T, ICV1/2>, std::integral_constant<T, ICV1%2>, integer_sequence<T, Pow..., (Res + sizeof...(Pow))...>, integer_sequence<T, Pow..., (Pow + sizeof...(Pow))...>> { };

template <class T, class Res, class Pow>
struct make_integer_sequence_impl<T, std::integral_constant<T, 0>, std::integral_constant<T, 0>, Res, Pow, void> {
   using type = Res;
};

template <class T, T V>
using make_integer_sequence = typename make_integer_sequence_impl<T, std::integral_constant<T, V/2>, std::integral_constant<T, V%2>>::type;

template <class>
struct is_variant_void {
   static constexpr size_t value = 0;
};

template <>
struct is_variant_void<boost::detail::variant::void_> {
   static constexpr size_t value = 1;
};

constexpr size_t sum(size_t v){
    return v;
}

template <class... Args>
constexpr size_t sum(size_t v, Args... vs){
    return v + sum(vs...);
}

template <class Variant>
struct variant_size;

template <class... Ts>
struct variant_size<boost::variant<Ts...>> {
   static constexpr size_t value = sizeof...(Ts) - sum(is_variant_void<Ts>::value...);
};

template <class...>
struct pack { };

template <class V, class=make_integer_sequence<size_t, variant_size<V>::value>>
struct variant_to_pack;

template <class... Ts, size_t... Is>
struct variant_to_pack<boost::variant<Ts...>, integer_sequence<size_t, Is...>> {
    using type = pack<typename std::tuple_element<Is, std::tuple<Ts...>>::type...>;
};

template <class>
struct unique_opaque { };

template <class... Ts>
struct unique_pack: unique_opaque<Ts>... { };

template <class, class, class = void>
struct unique;

template <class... Res, class T, class... Ts>
struct unique<unique_pack<Res...>, pack<T, Ts...>, typename std::enable_if<std::is_base_of<unique_opaque<T>, unique_pack<Res...>>::value>::type>:
       unique<unique_pack<Res...>, pack<Ts...>> { };

template <class... Res, class T, class... Ts>
struct unique<unique_pack<Res...>, pack<T, Ts...>, typename std::enable_if<!std::is_base_of<unique_opaque<T>, unique_pack<Res...>>::value>::type>:
       unique<unique_pack<Res..., T>, pack<Ts...>> { };

template <class... Res>
struct unique<unique_pack<Res...>, pack<>, void> {
    using type = boost::variant<Res...>;
};

template <class...>
struct flatten;

template <class... Res, class T, class... Rest>
struct flatten<pack<Res...>, T, Rest...>: flatten<pack<Res..., T>, Rest...> { };

template <class... Res, class... Vs, class... Rest>
struct flatten<pack<Res...>, boost::variant<Vs...>, Rest...>: flatten<pack<Res...>, typename variant_to_pack<boost::variant<Vs...>>::type, Rest...> {};

template <class... Res, class... Vs, class... Rest>
struct flatten<pack<Res...>, pack<Vs...>, Rest...>: flatten<pack<Res...>, Vs..., Rest...> { };

template <class Res>
struct flatten<Res>:unique<unique_pack<>, Res> { };

int main() {
   boost::variant<int, boost::variant<float, int, boost::variant<char, bool>, bool>> vif;
   static_assert(std::is_same<flatten<pack<>, decltype(vif)>::type, boost::variant<int, float, char, bool>>::value, "Test");
}

从给定的嵌套boost-variant类型创建新的boost-variant类型

2 个答案: