最简单的方法如下所示，使用标准库类bitset。

为了以一种类型安全的方式模拟c#特性，您必须编写一个模板包装器来封装bitset，将int参数替换为模板的类型参数枚举。喜欢的东西:

    template <class T, int N>
class FlagSet
{

    bitset<N> bits;

    FlagSet(T enumVal)
    {
        bits.set(enumVal);
    }

    // etc.
};

enum MyFlags
{
    FLAG_ONE,
    FLAG_TWO
};

FlagSet<MyFlags, 2> myFlag;

下面是我的解决方案，不需要任何一堆重载或强制转换:

namespace EFoobar
{
    enum
    {
        FB_A    = 0x1,
        FB_B    = 0x2,
        FB_C    = 0x4,
    };
    typedef long Flags;
}

void Foobar(EFoobar::Flags flags)
{
    if (flags & EFoobar::FB_A)
        // do sth
        ;
    if (flags & EFoobar::FB_B)
        // do sth
        ;
}

void ExampleUsage()
{
    Foobar(EFoobar::FB_A | EFoobar::FB_B);
    EFoobar::Flags otherflags = 0;
    otherflags|= EFoobar::FB_B;
    otherflags&= ~EFoobar::FB_B;
    Foobar(otherflags);
}

我认为这是可以的，因为我们无论如何都会识别(非强类型)枚举和整数。

只是作为一个(较长的)边注，如果你

要使用强类型枚举和 不需要重一点摆弄你的旗帜 性能不是问题

我会想到这个:

#include <set>

enum class EFoobarFlags
{
    FB_A = 1,
    FB_B,
    FB_C,
};

void Foobar(const std::set<EFoobarFlags>& flags)
{
    if (flags.find(EFoobarFlags::FB_A) != flags.end())
        // do sth
        ;
    if (flags.find(EFoobarFlags::FB_B) != flags.end())
        // do sth
        ;
}

void ExampleUsage()
{
    Foobar({EFoobarFlags::FB_A, EFoobarFlags::FB_B});
    std::set<EFoobarFlags> otherflags{};
    otherflags.insert(EFoobarFlags::FB_B);
    otherflags.erase(EFoobarFlags::FB_B);
    Foobar(otherflags);
}

使用c++ 11初始化列表和枚举类。

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>

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;

这里有一个位掩码的选项，如果你实际上不需要使用单个枚举值(例如，你不需要关闭它们)…如果你不担心保持二进制兼容性，即:你不关心你的位在哪里…你可能就是这样。此外，您最好不要过于关注范围和访问控制。嗯，枚举对于位域有一些不错的属性…不知道是否有人尝试过:)

struct AnimalProperties
{
    bool HasClaws : 1;
    bool CanFly : 1;
    bool EatsFish : 1;
    bool Endangered : 1;
};

union AnimalDescription
{
    AnimalProperties Properties;
    int Flags;
};

void TestUnionFlags()
{
    AnimalDescription propertiesA;
    propertiesA.Properties.CanFly = true;

    AnimalDescription propertiesB = propertiesA;
    propertiesB.Properties.EatsFish = true;

    if( propertiesA.Flags == propertiesB.Flags )
    {
        cout << "Life is terrible :(";
    }
    else
    {
        cout << "Life is great!";
    }

    AnimalDescription propertiesC = propertiesA;
    if( propertiesA.Flags == propertiesC.Flags )
    {
        cout << "Life is great!";
    }
    else
    {
        cout << "Life is terrible :(";
    }
}

我们可以看到生命是伟大的，我们有离散的值，我们有一个漂亮的int到&和|到我们的心内容，它仍然有它的位的含义的上下文。一切都是一致的和可预测的……对我来说……只要我继续使用微软的vc++编译器w/ Update 3在Win10 x64上，不碰我的编译器标志:)

Even though everything is great... we have some context as to the meaning of flags now, since its in a union w/ the bitfield in the terrible real world where your program may be be responsible for more than a single discrete task you could still accidentally (quite easily) smash two flags fields of different unions together (say, AnimalProperties and ObjectProperties, since they're both ints), mixing up all yours bits, which is a horrible bug to trace down... and how I know many people on this post don't work with bitmasks very often, since building them is easy and maintaining them is hard.

class AnimalDefinition {
public:
    static AnimalDefinition *GetAnimalDefinition( AnimalFlags flags );   //A little too obvious for my taste... NEXT!
    static AnimalDefinition *GetAnimalDefinition( AnimalProperties properties );   //Oh I see how to use this! BORING, NEXT!
    static AnimalDefinition *GetAnimalDefinition( int flags ); //hmm, wish I could see how to construct a valid "flags" int without CrossFingers+Ctrl+Shift+F("Animal*"). Maybe just hard-code 16 or something?

    AnimalFlags animalFlags;  //Well this is *way* too hard to break unintentionally, screw this!
    int flags; //PERFECT! Nothing will ever go wrong here... 
    //wait, what values are used for this particular flags field? Is this AnimalFlags or ObjectFlags? Or is it RuntimePlatformFlags? Does it matter? Where's the documentation? 
    //Well luckily anyone in the code base and get confused and destroy the whole program! At least I don't need to static_cast anymore, phew!

    private:
    AnimalDescription m_description; //Oh I know what this is. All of the mystery and excitement of life has been stolen away :(
}

因此，然后你让你的联合声明私有，以防止直接访问“Flags”，并必须添加getter /setter和操作符重载，然后为所有这些做一个宏，你基本上回到你开始的地方，当你试图用Enum来做这件事。

不幸的是，如果你想要你的代码是可移植的，我不认为有任何方法可以A)保证位布局或B)在编译时确定位布局(这样你就可以跟踪它，至少纠正跨版本/平台等的变化) 带位字段的结构中的偏移量

在运行时，你可以玩一些技巧，设置字段和XORing标志，看看哪些位发生了变化，听起来对我来说很糟糕，尽管有一个100%一致的，平台独立的，完全确定的解决方案，即:ENUM。

TL;DR: Don't listen to the haters. C++ is not English. Just because the literal definition of an abbreviated keyword inherited from C might not fit your usage doesn't mean you shouldn't use it when the C and C++ definition of the keyword absolutely includes your use case. You can also use structs to model things other than structures, and classes for things other than school and social caste. You may use float for values which are grounded. You may use char for variables which are neither un-burnt nor a person in a novel, play, or movie. Any programmer who goes to the dictionary to determine the meaning of a keyword before the language spec is a... well I'll hold my tongue there.

如果你确实希望你的代码模仿口语，你最好使用Objective-C编写，顺便说一句，它也大量使用枚举作为位域。

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;
}