使用枚举类的C++ 11标准符合位掩码

Question

您可以使用枚举类实现标准符合(如n3242草案的17.5.2.1.3所述)类型安全位掩码吗？我读它的方式,类型T是一个位掩码,如果它支持|,&,^,〜,| =,&=和^ =运算符,你还可以做if(l&r)其中l和r是T类型.列表中缺少的是运算符!=和==并允许排序一个也可能想要重载<.

让操作员工作只是烦人的样板代码,但我不知道如何(l&r).至少下面的代码不能用GCC编译(除了极其危险,因为它允许错误的隐式转换为int):

enum class Foo{
    operator bool(){
        return (unsigned)*this;
    }
};

编辑:我现在肯定知道枚举类不能有成员.实际问题如果(l&r)仍然存在.

Answer 1

我想你可以......你必须为bitmasky事情添加运算符.我没有在这里做,但你可以轻松地重载任何关系运算符.

  /**
   *
   */
  // NOTE: I changed to a more descriptive and consistent name
  //       This needs to be a real bitmask type.
  enum class file_permissions : int
  {
    no_perms        = 0,

    owner_read      =  0400,
    owner_write     =  0200,
    owner_exe       =  0100,
    owner_all       =  0700,

    group_read      =   040,
    group_write     =   020,
    group_exe       =   010,
    group_all       =   070,

    others_read     =    04,
    others_write    =    02,
    others_exe      =    01,
    others_all      =    07,

    all_all     = owner_all | group_all | others_all, // 0777

    set_uid_on_exe  = 04000,
    set_gid_on_exe  = 02000,
    sticky_bit      = 01000,

    perms_mask      = all_all | set_uid_on_exe | set_gid_on_exe | sticky_bit, // 07777

    perms_not_known = 0xffff,

    add_perms       = 0x1000,
    remove_perms    = 0x2000,
    symlink_perms   = 0x4000
  };

  inline constexpr file_permissions
  operator&(file_permissions x, file_permissions y)
  {
    return static_cast<file_permissions>
      (static_cast<int>(x) & static_cast<int>(y));
  }

  inline constexpr file_permissions
  operator|(file_permissions x, file_permissions y)
  {
    return static_cast<file_permissions>
      (static_cast<int>(x) | static_cast<int>(y));
  }

  inline constexpr file_permissions
  operator^(file_permissions x, file_permissions y)
  {
    return static_cast<file_permissions>
      (static_cast<int>(x) ^ static_cast<int>(y));
  }

  inline constexpr file_permissions
  operator~(file_permissions x)
  {
    return static_cast<file_permissions>(~static_cast<int>(x));
  }

  inline file_permissions &
  operator&=(file_permissions & x, file_permissions y)
  {
    x = x & y;
    return x;
  }

  inline file_permissions &
  operator|=(file_permissions & x, file_permissions y)
  {
    x = x | y;
    return x;
  }

  inline file_permissions &
  operator^=(file_permissions & x, file_permissions y)
  {
    x = x ^ y;
    return x;
  }

Answer 2

我不完全确定你的接受标准是什么,但你可以operator &通过适当的转换返回一个包装类,并且explicit operator bool:

#include <type_traits>

template<typename T> using Underlying = typename std::underlying_type<T>::type;
template<typename T> constexpr Underlying<T>
underlying(T t) { return Underlying<T>(t); }

template<typename T> struct TruthValue {
    T t;
    constexpr TruthValue(T t): t(t) { }
    constexpr operator T() const { return t; }
    constexpr explicit operator bool() const { return underlying(t); }
};

enum class Color { RED = 0xff0000, GREEN = 0x00ff00, BLUE = 0x0000ff };
constexpr TruthValue<Color>
operator&(Color l, Color r) { return Color(underlying(l) & underlying(r)); }

当然,所有其他运营商都可以继续返回Color:

constexpr Color
operator|(Color l, Color r) { return Color(underlying(l) | underlying(r)); }
constexpr Color operator~(Color c) { return Color(~underlying(c)); }

int main() {
    constexpr Color YELLOW = Color::RED | Color::GREEN;
    constexpr Color WHITE = Color::RED | Color::GREEN | Color::BLUE;
    static_assert(YELLOW == (WHITE & ~Color::BLUE), "color subtraction");
    return (YELLOW & Color::BLUE) ? 1 : 0;
}

Answer 3

作用域枚举（使用 或 创建的枚举enum class）enum struct不是类。它们不能有成员函数，它们只是提供封闭的枚举器（在命名空间级别不可见）。