在c#中,通过[flags]属性将枚举视为标志,但在c++中实现这一点的最佳方法是什么?

例如,我想写:

enum AnimalFlags
{
    HasClaws = 1,
    CanFly =2,
    EatsFish = 4,
    Endangered = 8
};

seahawk.flags = CanFly | EatsFish | Endangered;

然而,我得到编译器错误关于int/enum转换。除了生硬的角色转换,还有更好的表达方式吗?最好,我不想依赖第三方库(如boost或Qt)的构造。

编辑:如答案中所示,我可以通过声明seahawk来避免编译器错误。标记为int。但是,我希望有某种机制来执行类型安全,这样就不能编写seahawk了。flags = HasMaximizeButton。


当前回答

我使用以下宏:

#define ENUM_FLAG_OPERATORS(T)                                                                                                                                            \
    inline T operator~ (T a) { return static_cast<T>( ~static_cast<std::underlying_type<T>::type>(a) ); }                                                                       \
    inline T operator| (T a, T b) { return static_cast<T>( static_cast<std::underlying_type<T>::type>(a) | static_cast<std::underlying_type<T>::type>(b) ); }                   \
    inline T operator& (T a, T b) { return static_cast<T>( static_cast<std::underlying_type<T>::type>(a) & static_cast<std::underlying_type<T>::type>(b) ); }                   \
    inline T operator^ (T a, T b) { return static_cast<T>( static_cast<std::underlying_type<T>::type>(a) ^ static_cast<std::underlying_type<T>::type>(b) ); }                   \
    inline T& operator|= (T& a, T b) { return reinterpret_cast<T&>( reinterpret_cast<std::underlying_type<T>::type&>(a) |= static_cast<std::underlying_type<T>::type>(b) ); }   \
    inline T& operator&= (T& a, T b) { return reinterpret_cast<T&>( reinterpret_cast<std::underlying_type<T>::type&>(a) &= static_cast<std::underlying_type<T>::type>(b) ); }   \
    inline T& operator^= (T& a, T b) { return reinterpret_cast<T&>( reinterpret_cast<std::underlying_type<T>::type&>(a) ^= static_cast<std::underlying_type<T>::type>(b) ); }

它类似于上面提到的那些,但有几个改进:

它是类型安全的(它不假设基础类型是int型) 它不需要手动指定底层类型(与@LunarEclipse的答案相反)

它需要包含type_traits:

#include <type_traits>

其他回答

对于像我这样的懒人来说,下面是复制粘贴的模板解决方案:

template<class T> inline T operator~ (T a) { return (T)~(int)a; }
template<class T> inline T operator| (T a, T b) { return (T)((int)a | (int)b); }
template<class T> inline T operator& (T a, T b) { return (T)((int)a & (int)b); }
template<class T> inline T operator^ (T a, T b) { return (T)((int)a ^ (int)b); }
template<class T> inline T& operator|= (T& a, T b) { return (T&)((int&)a |= (int)b); }
template<class T> inline T& operator&= (T& a, T b) { return (T&)((int&)a &= (int)b); }
template<class T> inline T& operator^= (T& a, T b) { return (T&)((int&)a ^= (int)b); }

注意(也有点离题):另一种制作唯一标志的方法可以使用位移位。我自己觉得这更容易理解。

enum Flags
{
    A = 1 << 0, // binary 0001
    B = 1 << 1, // binary 0010
    C = 1 << 2, // binary 0100
    D = 1 << 3  // binary 1000
};

它可以保存不超过int的值,也就是说,大多数情况下,32个标志清楚地反映在移位量中。

“正确”的方法是为枚举定义位操作符,如下所示:

enum AnimalFlags
{
    HasClaws   = 1,
    CanFly     = 2,
    EatsFish   = 4,
    Endangered = 8
};

inline AnimalFlags operator|(AnimalFlags a, AnimalFlags b)
{
    return static_cast<AnimalFlags>(static_cast<int>(a) | static_cast<int>(b));
}

等等,其余的位操作符。如果枚举范围超过int range,则根据需要修改。

也许像Objective-C的NS_OPTIONS。

#define ENUM(T1, T2) \
enum class T1 : T2; \
inline T1 operator~ (T1 a) { return (T1)~(int)a; } \
inline T1 operator| (T1 a, T1 b) { return static_cast<T1>((static_cast<T2>(a) | static_cast<T2>(b))); } \
inline T1 operator& (T1 a, T1 b) { return static_cast<T1>((static_cast<T2>(a) & static_cast<T2>(b))); } \
inline T1 operator^ (T1 a, T1 b) { return static_cast<T1>((static_cast<T2>(a) ^ static_cast<T2>(b))); } \
inline T1& operator|= (T1& a, T1 b) { return reinterpret_cast<T1&>((reinterpret_cast<T2&>(a) |= static_cast<T2>(b))); } \
inline T1& operator&= (T1& a, T1 b) { return reinterpret_cast<T1&>((reinterpret_cast<T2&>(a) &= static_cast<T2>(b))); } \
inline T1& operator^= (T1& a, T1 b) { return reinterpret_cast<T1&>((reinterpret_cast<T2&>(a) ^= static_cast<T2>(b))); } \
enum class T1 : T2

ENUM(Options, short) {
    FIRST  = 1 << 0,
    SECOND = 1 << 1,
    THIRD  = 1 << 2,
    FOURTH = 1 << 3
};

auto options = Options::FIRST | Options::SECOND;
options |= Options::THIRD;
if ((options & Options::SECOND) == Options::SECOND)
    cout << "Contains second option." << endl;
if ((options & Options::THIRD) == Options::THIRD)
    cout << "Contains third option." << endl;
return 0;

// Output:
// Contains second option. 
// Contains third option.

如上(启)或做以下。实际上枚举是“枚举”,你想做的是有一个集合,因此你应该使用stl::set

enum AnimalFlags
{
    HasClaws = 1,
    CanFly =2,
    EatsFish = 4,
    Endangered = 8
};

int main(void)
{
    AnimalFlags seahawk;
    //seahawk= CanFly | EatsFish | Endangered;
    seahawk= static_cast<AnimalFlags>(CanFly | EatsFish | Endangered);
}