我更喜欢使用magic_enum，因为它有助于自动将字符串转换为枚举，反之亦然。 它是一个仅头文件的库，用c++ 17标准编写。

Magic_enum已经有了枚举位操作符的模板函数。 见文档。

用法:

#include <magic_enum.hpp>

enum Flag { ... };

Flag flag{};
Flag value{};

using namespace magic_enum::bitwise_operators;
flag |= value;

我想详细说明Uliwitness的回答，为c++ 98修复他的代码，并使用Safe Bool习语，因为在c++ 11以下的c++版本中缺少std::underlying_type<>模板和显式关键字。

我还修改了它，使枚举值可以是连续的，而不需要任何显式的赋值，因此您可以有

enum AnimalFlags_
{
    HasClaws,
    CanFly,
    EatsFish,
    Endangered
};
typedef FlagsEnum<AnimalFlags_> AnimalFlags;

seahawk.flags = AnimalFlags() | CanFly | EatsFish | Endangered;

然后，您可以获得原始标志值

seahawk.flags.value();

这是代码。

template <typename EnumType, typename Underlying = int>
class FlagsEnum
{
    typedef Underlying FlagsEnum::* RestrictedBool;

public:
    FlagsEnum() : m_flags(Underlying()) {}

    FlagsEnum(EnumType singleFlag):
        m_flags(1 << singleFlag)
    {}

    FlagsEnum(const FlagsEnum& original):
        m_flags(original.m_flags)
    {}

    FlagsEnum& operator |=(const FlagsEnum& f) {
        m_flags |= f.m_flags;
        return *this;
    }

    FlagsEnum& operator &=(const FlagsEnum& f) {
        m_flags &= f.m_flags;
        return *this;
    }

    friend FlagsEnum operator |(const FlagsEnum& f1, const FlagsEnum& f2) {
        return FlagsEnum(f1) |= f2;
    }

    friend FlagsEnum operator &(const FlagsEnum& f1, const FlagsEnum& f2) {
        return FlagsEnum(f1) &= f2;
    }

    FlagsEnum operator ~() const {
        FlagsEnum result(*this);
        result.m_flags = ~result.m_flags;
        return result;
    }

    operator RestrictedBool() const {
        return m_flags ? &FlagsEnum::m_flags : 0;
    }

    Underlying value() const {
        return m_flags;
    }

protected:
    Underlying  m_flags;
};

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.

在我看来，到目前为止没有一个答案是理想的。理想的解决方案是:

支持==，!=，=，&，&=，|，|=和~运算符 意义(即a和b) 类型安全，即不允许分配非枚举值，如字面量或整数类型(枚举值的按位组合除外)，或允许将枚举变量分配给整数类型 允许使用if (a & b)… 不需要邪恶的宏，实现特定的功能或其他hack

到目前为止，大多数解都停留在第2点或第3点上。WebDancer在我看来是封闭的，但在第3点失败了，需要在每个枚举中重复。

我提出的解决方案是WebDancer的一个广义版本，也解决了第3点:

#include <cstdint>
#include <type_traits>

template<typename T, typename = typename std::enable_if<std::is_enum<T>::value, T>::type>
class auto_bool
{
    T val_;
public:
    constexpr auto_bool(T val) : val_(val) {}
    constexpr operator T() const { return val_; }
    constexpr explicit operator bool() const
    {
        return static_cast<std::underlying_type_t<T>>(val_) != 0;
    }
};

template <typename T, typename = typename std::enable_if<std::is_enum<T>::value, T>::type>
constexpr auto_bool<T> operator&(T lhs, T rhs)
{
    return static_cast<T>(
        static_cast<typename std::underlying_type<T>::type>(lhs) &
        static_cast<typename std::underlying_type<T>::type>(rhs));
}

template <typename T, typename = typename std::enable_if<std::is_enum<T>::value, T>::type>
constexpr T operator|(T lhs, T rhs)
{
    return static_cast<T>(
        static_cast<typename std::underlying_type<T>::type>(lhs) |
        static_cast<typename std::underlying_type<T>::type>(rhs));
}

enum class AnimalFlags : uint8_t 
{
    HasClaws = 1,
    CanFly = 2,
    EatsFish = 4,
    Endangered = 8
};

enum class PlantFlags : uint8_t
{
    HasLeaves = 1,
    HasFlowers = 2,
    HasFruit = 4,
    HasThorns = 8
};

int main()
{
    AnimalFlags seahawk = AnimalFlags::CanFly;        // Compiles, as expected
    AnimalFlags lion = AnimalFlags::HasClaws;         // Compiles, as expected
    PlantFlags rose = PlantFlags::HasFlowers;         // Compiles, as expected
//  rose = 1;                                         // Won't compile, as expected
    if (seahawk != lion) {}                           // Compiles, as expected
//  if (seahawk == rose) {}                           // Won't compile, as expected
//  seahawk = PlantFlags::HasThorns;                  // Won't compile, as expected
    seahawk = seahawk | AnimalFlags::EatsFish;        // Compiles, as expected
    lion = AnimalFlags::HasClaws |                    // Compiles, as expected
           AnimalFlags::Endangered;
//  int eagle = AnimalFlags::CanFly |                 // Won't compile, as expected
//              AnimalFlags::HasClaws;
//  int has_claws = seahawk & AnimalFlags::CanFly;    // Won't compile, as expected
    if (seahawk & AnimalFlags::CanFly) {}             // Compiles, as expected
    seahawk = seahawk & AnimalFlags::CanFly;          // Compiles, as expected

    return 0;
}

This creates overloads of the necessary operators but uses SFINAE to limit them to enumerated types. Note that in the interests of brevity I haven't defined all of the operators but the only one that is any different is the &. The operators are currently global (i.e. apply to all enumerated types) but this could be reduced either by placing the overloads in a namespace (what I do), or by adding additional SFINAE conditions (perhaps using particular underlying types, or specially created type aliases). The underlying_type_t is a C++14 feature but it seems to be well supported and is easy to emulate for C++11 with a simple template<typename T> using underlying_type_t = underlying_type<T>::type;

编辑:我纳入了弗拉基米尔·阿菲内洛建议的变化。用GCC 10、CLANG 13和Visual Studio 2022测试。

c++ 20类型安全Enum操作符

博士TL;

template<typename T>
requires std::is_enum_v<T> and
         requires (std::underlying_type_t<T> x) {
             { x | x } -> std::same_as<std::underlying_type_t<T>>;
             T(x);
         }
T operator|(T left, T right)
{
    using U = std::underlying_type_t<T>;
    return T( U(left) | U(right) );
}

template<typename T>
requires std::is_enum_v<T> and
         requires (std::underlying_type_t<T> x) {
             { x | x } -> std::same_as<std::underlying_type_t<T>>;
             T(x);
         }
T operator&(T left, T right)
{
    using U = std::underlying_type_t<T>;
    return T( U(left) & U(right) );
}

template<typename T>
requires std::is_enum_v<T> and requires (T x) { { x | x } -> std::same_as<T>; }
T & operator|=(T &left, T right)
{
    return left = left | right;
}

template<typename T>
requires std::is_enum_v<T> and requires (T x) { { x & x } -> std::same_as<T>; }
T & operator&=(T &left, T right)
{
    return left = left & right;
}

基本原理

使用类型特征std::is_enum，我们可以测试一些类型T是否为枚举类型。 这包括无作用域和有作用域的枚举(即enum和enum类)。 使用类型trait std::underlying_type，我们可以得到枚举的底层类型。 使用c++ 20的概念和约束，很容易为按位操作提供重载。

有作用域和无作用域

如果操作只应该重载有作用域或无作用域的枚举，std::is_scoped_enum可以用于相应地扩展模板约束。

c++ 23

在c++ 23中，我们使用std::to_underlying来更容易地将枚举值转换为其底层类型。

移动语义和完善转发

如果你遇到了一种奇怪的情况，即你的底层类型对于copy和move有不同的语义，或者它不提供复制c'tor，那么你应该使用std::forward对操作数进行完美的转发。

目前还没有语言支持枚举标志，如果它将成为c++标准的一部分，元类可能会固有地添加这个特性。

我的解决方案是创建仅枚举实例化的模板函数，为枚举类使用其底层类型添加类型安全的按位操作支持:

文件:EnumClassBitwise.h

#pragma once
#ifndef _ENUM_CLASS_BITWISE_H_
#define _ENUM_CLASS_BITWISE_H_

#include <type_traits>

//unary ~operator    
template <typename Enum, typename std::enable_if_t<std::is_enum<Enum>::value, int> = 0>
constexpr inline Enum& operator~ (Enum& val)
{
    val = static_cast<Enum>(~static_cast<std::underlying_type_t<Enum>>(val));
    return val;
}

// & operator
template <typename Enum, typename std::enable_if_t<std::is_enum<Enum>::value, int> = 0>
constexpr inline Enum operator& (Enum lhs, Enum rhs)
{
    return static_cast<Enum>(static_cast<std::underlying_type_t<Enum>>(lhs) & static_cast<std::underlying_type_t<Enum>>(rhs));
}

// &= operator
template <typename Enum, typename std::enable_if_t<std::is_enum<Enum>::value, int> = 0>
constexpr inline Enum operator&= (Enum& lhs, Enum rhs)
{
    lhs = static_cast<Enum>(static_cast<std::underlying_type_t<Enum>>(lhs) & static_cast<std::underlying_type_t<Enum>>(rhs));
    return lhs;
}

//| operator

template <typename Enum, typename std::enable_if_t<std::is_enum<Enum>::value, int> = 0>
constexpr inline Enum operator| (Enum lhs, Enum rhs)
{
    return static_cast<Enum>(static_cast<std::underlying_type_t<Enum>>(lhs) | static_cast<std::underlying_type_t<Enum>>(rhs));
}
//|= operator

template <typename Enum, typename std::enable_if_t<std::is_enum<Enum>::value, int> = 0>
constexpr inline Enum& operator|= (Enum& lhs, Enum rhs)
{
    lhs = static_cast<Enum>(static_cast<std::underlying_type_t<Enum>>(lhs) | static_cast<std::underlying_type_t<Enum>>(rhs));
    return lhs;
}

#endif // _ENUM_CLASS_BITWISE_H_

为了方便和减少错误，你可能想要包装你的枚举和整数的位标志操作:

文件:BitFlags.h

#pragma once
#ifndef _BIT_FLAGS_H_
#define _BIT_FLAGS_H_

#include "EnumClassBitwise.h"

 template<typename T>
 class BitFlags
 {
 public:

     constexpr inline BitFlags() = default;
     constexpr inline BitFlags(T value) { mValue = value; }
     constexpr inline BitFlags operator| (T rhs) const { return mValue | rhs; }
     constexpr inline BitFlags operator& (T rhs) const { return mValue & rhs; }
     constexpr inline BitFlags operator~ () const { return ~mValue; }
     constexpr inline operator T() const { return mValue; }
     constexpr inline BitFlags& operator|=(T rhs) { mValue |= rhs; return *this; }
     constexpr inline BitFlags& operator&=(T rhs) { mValue &= rhs; return *this; }
     constexpr inline bool test(T rhs) const { return (mValue & rhs) == rhs; }
     constexpr inline void set(T rhs) { mValue |= rhs; }
     constexpr inline void clear(T rhs) { mValue &= ~rhs; }

 private:
     T mValue;
 };
#endif //#define _BIT_FLAGS_H_

可能的用法:

#include <cstdint>
#include <BitFlags.h>
void main()
{
    enum class Options : uint32_t
    { 
          NoOption = 0 << 0
        , Option1  = 1 << 0
        , Option2  = 1 << 1
        , Option3  = 1 << 2
        , Option4  = 1 << 3
    };

    const uint32_t Option1 = 1 << 0;
    const uint32_t Option2 = 1 << 1;
    const uint32_t Option3 = 1 << 2;
    const uint32_t Option4 = 1 << 3;

   //Enum BitFlags
    BitFlags<Options> optionsEnum(Options::NoOption);
    optionsEnum.set(Options::Option1 | Options::Option3);

   //Standard integer BitFlags
    BitFlags<uint32_t> optionsUint32(0);
    optionsUint32.set(Option1 | Option3); 

    return 0;
}