Question

声明如下：

enum DrawBoldMode : unsigned
{
    DBM_NONE =              0,
    DBM_ITEM =              1<<0,   // bold just the nearest line
    DBM_SECTION =           1<<1,   // bold all lines in the same section
    DBM_LINETYPE =          1<<2,   // bold all lines of the same line type
    DBM_POINTAGE =          1<<3,   // bold all lines of the same line type
};

如何导出DrawBoldMode的基础类型（即无符号）？

Answer 1

std::underlying_type在GCC 4.7中可用，但在此之前，您可以得到一个近似的emulation with templates：

#include <tuple>
// This is a hack because GCC 4.6 does not support std::underlying_type yet.
// A specialization for each enum is preferred
namespace detail {
    template <typename T, typename Acc, typename... In>
    struct filter;

    template <typename T, typename Acc>
    struct filter<T, Acc> {
        typedef typename std::tuple_element<0, Acc>::type type;
    };

    template <typename T, typename... Acc, typename Head, typename... Tail>
    struct filter<T, std::tuple<Acc...>, Head, Tail...>
    : std::conditional<sizeof(T) == sizeof(Head) && (T(-1) < T(0)) == (Head(-1) < Head(0))
                      , filter<T, std::tuple<Acc...,Head>, Tail...>
                      , filter<T, std::tuple<Acc...>, Tail...>
                      >::type {};

    template <typename T, typename... In>
    struct find_best_match : filter<T, std::tuple<>, In...> {};
}

namespace std {
    template <typename E>
    struct underlying_type : detail::find_best_match<E,
                                signed short,
                                unsigned short,
                                signed int,
                                unsigned int,
                                signed long,
                                unsigned long,
                                signed long long,
                                unsigned long long,
                                bool,
                                char,
                                signed char,
                                unsigned char,
                                wchar_t,
                                char16_t,
                                char32_t> {};
}

它不会为您提供确切的类型，但它会为您提供具有相同大小和签名特征的类型。

Answer 2

它应该以{{1}}的形式提供。但是，我的编译器（GCC 4.6.1）似乎没有实现它。

我认为用模板实现它是不可能的，但我可能错了。

如何获得枚举的基本类型？

2 个答案: