与其他类似的问题不同,这个问题是关于如何使用c++的新特性。

2008 c Is there a simple way to convert C++ enum to string? 2008 c Easy way to use variables of enum types as string in C? 2008 c++ How to easily map c++ enums to strings 2008 c++ Making something both a C identifier and a string? 2008 c++ Is there a simple script to convert C++ enum to string? 2009 c++ How to use enums as flags in C++? 2011 c++ How to convert an enum type variable to a string? 2011 c++ Enum to String C++ 2011 c++ How to convert an enum type variable to a string? 2012 c How to convert enum names to string in c 2013 c Stringifying an conditionally compiled enum in C

看了很多答案后,我还没有找到:

优雅的方式使用c++ 11、c++ 14或c++ 17的新特性 或者在Boost中使用一些现成的东西 还有一些东西计划在c++ 20中实现

例子

举例往往比冗长的解释更好。 您可以在Coliru上编译和运行这个代码片段。 (另一个前面的例子也可用)

#include <map>
#include <iostream>

struct MyClass
{
    enum class MyEnum : char {
        AAA = -8,
        BBB = '8',
        CCC = AAA + BBB
    };
};

// Replace magic() by some faster compile-time generated code
// (you're allowed to replace the return type with std::string
// if that's easier for you)
const char* magic (MyClass::MyEnum e)
{
    const std::map<MyClass::MyEnum,const char*> MyEnumStrings {
        { MyClass::MyEnum::AAA, "MyClass::MyEnum::AAA" },
        { MyClass::MyEnum::BBB, "MyClass::MyEnum::BBB" },
        { MyClass::MyEnum::CCC, "MyClass::MyEnum::CCC" }
    };
    auto   it  = MyEnumStrings.find(e);
    return it == MyEnumStrings.end() ? "Out of range" : it->second;
}

int main()
{
   std::cout << magic(MyClass::MyEnum::AAA) <<'\n';
   std::cout << magic(MyClass::MyEnum::BBB) <<'\n';
   std::cout << magic(MyClass::MyEnum::CCC) <<'\n';
}

约束

请不要无价值的重复其他答案或基本链接。 请避免基于宏的臃肿答案,或尽量减少#define开销。 请不要手动enum ->字符串映射。

很高兴有

支持从不同于零的数字开始的enum值 支持负enum值 支持碎片enum值 支持类枚举(c++ 11) 支持类枚举:<类型>有任何允许的<类型> (c++ 11) 编译时(不是运行时)到字符串的转换, 或者至少在运行时快速执行(例如std::map不是一个好主意…) constexpr (c++ 11,然后在c++ 14/17/20中放松) noexcept (C + + 11) c++ 17/ c++ 20友好的代码片段

一个可能的想法是使用c++编译器功能,在编译时使用基于可变参数模板类和constexpr函数的元编程技巧来生成c++代码……


当前回答

这个要点提供了一个基于c++可变参数模板的简单映射。

这是一个c++ 17简化版的基于类型的映射的要点:

#include <cstring> // http://stackoverflow.com/q/24520781

template<typename KeyValue, typename ... RestOfKeyValues>
struct map {
  static constexpr typename KeyValue::key_t get(const char* val) noexcept {
    if constexpr (sizeof...(RestOfKeyValues)==0)  // C++17 if constexpr
      return KeyValue::key; // Returns last element
    else {
      static_assert(KeyValue::val != nullptr,
                  "Only last element may have null name");
      return strcmp(val, KeyValue::val()) 
            ? map<RestOfKeyValues...>::get(val) : KeyValue::key;
    }
  }
  static constexpr const char* get(typename KeyValue::key_t key) noexcept {
    if constexpr (sizeof...(RestOfKeyValues)==0)
      return (KeyValue::val != nullptr) && (key == KeyValue::key)
            ? KeyValue::val() : "";
    else
      return (key == KeyValue::key) 
            ? KeyValue::val() : map<RestOfKeyValues...>::get(key);
  }
};

template<typename Enum, typename ... KeyValues>
class names {
  typedef map<KeyValues...> Map;
public:
  static constexpr Enum get(const char* nam) noexcept {
    return Map::get(nam);
  }
  static constexpr const char* get(Enum key) noexcept {
    return Map::get(key);
  }
};

用法示例:

enum class fasion {
    fancy,
    classic,
    sporty,
    emo,
    __last__ = emo,
    __unknown__ = -1
};

#define NAME(s) static inline constexpr const char* s() noexcept {return #s;}
namespace name {
    NAME(fancy)
    NAME(classic)
    NAME(sporty)
    NAME(emo)
}

template<auto K, const char* (*V)()>  // C++17 template<auto>
struct _ {
    typedef decltype(K) key_t;
    typedef decltype(V) name_t;
    static constexpr key_t  key = K; // enum id value
    static constexpr name_t val = V; // enum id name
};

typedef names<fasion,
    _<fasion::fancy, name::fancy>,
    _<fasion::classic, name::classic>,
    _<fasion::sporty, name::sporty>,
    _<fasion::emo, name::emo>,
    _<fasion::__unknown__, nullptr>
> fasion_names;

map < keyvalue…>可以双向使用:

fasion_names:把(in fashion: emo) fasion_names::把(“emo”)

这个例子可以在godbolt.org上找到

int main ()
{
  constexpr auto str = fasion_names::get(fasion::emo);
  constexpr auto fsn = fasion_names::get(str);
  return (int) fsn;
}

结果:gc -7 -std=c++1z -Ofast -S

main:
        mov     eax, 3
        ret

其他回答

我的解决方案是不使用宏。

优点:

你知道你在做什么 访问是通过哈希映射进行的,因此适用于许多有值枚举 不需要考虑顺序值或非连续值 既enum到字符串和字符串到enum转换,而添加的enum值必须添加在一个额外的地方

缺点:

您需要将所有枚举值复制为文本 哈希映射中的访问必须考虑字符串大小写 维护如果添加值是痛苦的-必须添加在enum和直接翻译映射

所以…直到c++实现c# Enum。解析功能,我将坚持这个:

            #include <unordered_map>

            enum class Language
            { unknown, 
                Chinese, 
                English, 
                French, 
                German
                // etc etc
            };

            class Enumerations
            {
            public:
                static void fnInit(void);

                static std::unordered_map <std::wstring, Language> m_Language;
                static std::unordered_map <Language, std::wstring> m_invLanguage;

            private:
                static void fnClear();
                static void fnSetValues(void);
                static void fnInvertValues(void);

                static bool m_init_done;
            };

            std::unordered_map <std::wstring, Language> Enumerations::m_Language = std::unordered_map <std::wstring, Language>();
            std::unordered_map <Language, std::wstring> Enumerations::m_invLanguage = std::unordered_map <Language, std::wstring>();

            void Enumerations::fnInit()
            {
                fnClear();
                fnSetValues();
                fnInvertValues();
            }

            void Enumerations::fnClear()
            {
                m_Language.clear();
                m_invLanguage.clear();
            }

            void Enumerations::fnSetValues(void)
            {   
                m_Language[L"unknown"] = Language::unknown;
                m_Language[L"Chinese"] = Language::Chinese;
                m_Language[L"English"] = Language::English;
                m_Language[L"French"] = Language::French;
                m_Language[L"German"] = Language::German;
                // and more etc etc
            }

            void Enumerations::fnInvertValues(void)
            {
                for (auto it = m_Language.begin(); it != m_Language.end(); it++)
                {
                    m_invLanguage[it->second] = it->first;
                }
            }

            // usage -
            //Language aLanguage = Language::English;
            //wstring sLanguage = Enumerations::m_invLanguage[aLanguage];

            //wstring sLanguage = L"French" ;
            //Language aLanguage = Enumerations::m_Language[sLanguage];

在类/struct (struct默认为public成员)和重载操作符中使用enum的解决方案:

struct Color
{
    enum Enum { RED, GREEN, BLUE };
    Enum e;

    Color() {}
    Color(Enum e) : e(e) {}

    Color operator=(Enum o) { e = o; return *this; }
    Color operator=(Color o) { e = o.e; return *this; }
    bool operator==(Enum o) { return e == o; }
    bool operator==(Color o) { return e == o.e; }
    operator Enum() const { return e; }

    std::string toString() const
    {
        switch (e)
        {
        case Color::RED:
            return "red";
        case Color::GREEN:
            return "green";
        case Color::BLUE:
            return "blue";
        default:
            return "unknown";
        }
    }
};

从外部看,它几乎完全像一个类枚举:

Color red;
red = Color::RED;
Color blue = Color::BLUE;

cout << red.toString() << " " << Color::GREEN << " " << blue << endl;

这将输出“red 12”。你可以重载<<使蓝色输出成为一个字符串(尽管这可能会导致歧义,所以不可能),但它不会与Color::GREEN一起工作,因为它不会自动转换为Color。

隐式转换为Enum(隐式转换为int或给定类型)的目的是能够做到:

Color color;
switch (color) ...

这是可行的,但这也意味着这也是可行的:

int i = color;

对于枚举类,它不会编译。 如果重载两个函数,接受枚举和整数,或者删除隐式转换…

另一个解决方案将涉及使用实际的枚举类和静态成员:

struct Color
{
    enum class Enum { RED, GREEN, BLUE };
    static const Enum RED = Enum::RED, GREEN = Enum::GREEN, BLUE = Enum::BLUE;

    //same as previous...
};

它可能会占用更多的空间,并且花费更长的时间,但会导致隐式int转换的编译错误。我就会用这个!

虽然这样做肯定有开销,但我认为它比我见过的其他代码更简单,看起来更好。还可以添加功能,这些功能都可以在类中进行范围限定。

编辑:这是有效的,大多数可以在执行前编译:

class Color
{
public:
    enum class Enum { RED, GREEN, BLUE };
    static const Enum RED = Enum::RED, GREEN = Enum::GREEN, BLUE = Enum::BLUE;

    constexpr Color() : e(Enum::RED) {}
    constexpr Color(Enum e) : e(e) {}

    constexpr bool operator==(Enum o) const { return e == o; }
    constexpr bool operator==(Color o) const { return e == o.e; }
    constexpr operator Enum() const { return e; }

    Color& operator=(Enum o) { const_cast<Enum>(this->e) = o; return *this; }
    Color& operator=(Color o) { const_cast<Enum>(this->e) = o.e; return *this; }

    std::string toString() const
    {
        switch (e)
        {
        case Enum::RED:
            return "red";
        case Enum::GREEN:
            return "green";
        case Enum::BLUE:
            return "blue";
        default:
            return "unknown";
        }
    }
private:
    const Enum e;
};

很长一段时间以来,我也一直为这个问题感到沮丧,还有以适当的方式将类型转换为字符串的问题。然而,对于最后一个问题,我对在标准c++中打印变量类型是否可能解释的解决方案感到惊讶?,使用的思想从Can I obtain c++ type name in a constexpr way?使用这种技术,可以构造一个类似的函数来获取枚举值为string:

#include <iostream>
using namespace std;

class static_string
{
    const char* const p_;
    const std::size_t sz_;

public:
    typedef const char* const_iterator;

    template <std::size_t N>
    constexpr static_string(const char(&a)[N]) noexcept
        : p_(a)
        , sz_(N - 1)
    {}

    constexpr static_string(const char* p, std::size_t N) noexcept
        : p_(p)
        , sz_(N)
    {}

    constexpr const char* data() const noexcept { return p_; }
    constexpr std::size_t size() const noexcept { return sz_; }

    constexpr const_iterator begin() const noexcept { return p_; }
    constexpr const_iterator end()   const noexcept { return p_ + sz_; }

    constexpr char operator[](std::size_t n) const
    {
        return n < sz_ ? p_[n] : throw std::out_of_range("static_string");
    }
};

inline std::ostream& operator<<(std::ostream& os, static_string const& s)
{
    return os.write(s.data(), s.size());
}

/// \brief Get the name of a type
template <class T>
static_string typeName()
{
#ifdef __clang__
    static_string p = __PRETTY_FUNCTION__;
    return static_string(p.data() + 30, p.size() - 30 - 1);
#elif defined(_MSC_VER)
    static_string p = __FUNCSIG__;
    return static_string(p.data() + 37, p.size() - 37 - 7);
#endif

}

namespace details
{
    template <class Enum>
    struct EnumWrapper
    {
        template < Enum enu >
        static static_string name()
        {
#ifdef __clang__
            static_string p = __PRETTY_FUNCTION__;
            static_string enumType = typeName<Enum>();
            return static_string(p.data() + 73 + enumType.size(), p.size() - 73 - enumType.size() - 1);
#elif defined(_MSC_VER)
            static_string p = __FUNCSIG__;
            static_string enumType = typeName<Enum>();
            return static_string(p.data() + 57 + enumType.size(), p.size() - 57 - enumType.size() - 7);
#endif
        }
    };
}

/// \brief Get the name of an enum value
template <typename Enum, Enum enu>
static_string enumName()
{
    return details::EnumWrapper<Enum>::template name<enu>();
}

enum class Color
{
    Blue = 0,
    Yellow = 1
};


int main() 
{
    std::cout << "_" << typeName<Color>() << "_"  << std::endl;
    std::cout << "_" << enumName<Color, Color::Blue>() << "_"  << std::endl;
    return 0;
}

上面的代码只在Clang(参见https://ideone.com/je5Quv)和VS2015上进行了测试,但是应该可以通过对整数常量进行一些调整来适应其他编译器。当然,它仍然在底层使用宏,但至少有一个宏不需要访问枚举实现。

#define ENUM_MAKE(TYPE, ...) \
        enum class TYPE {__VA_ARGS__};\
        struct Helper_ ## TYPE { \
            static const String& toName(TYPE type) {\
                int index = static_cast<int>(type);\
                return splitStringVec()[index];}\
            static const TYPE toType(const String& name){\
                static std::unordered_map<String,TYPE> typeNameMap;\
                if( typeNameMap.empty() )\
                {\
                    const StringVector& ssVec = splitStringVec();\
                    for (size_t i = 0; i < ssVec.size(); ++i)\
                        typeNameMap.insert(std::make_pair(ssVec[i], static_cast<TYPE>(i)));\
                }\
                return typeNameMap[name];}\
            static const StringVector& splitStringVec() {\
                static StringVector typeNameVector;\
                if(typeNameVector.empty()) \
                {\
                    typeNameVector = StringUtil::split(#__VA_ARGS__, ",");\
                    for (auto& name : typeNameVector)\
                    {\
                        name.erase(std::remove(name.begin(), name.end(), ' '),name.end()); \
                        name = String(#TYPE) + "::" + name;\
                    }\
                }\
                return typeNameVector;\
            }\
        };


using String = std::string;
using StringVector = std::vector<String>;

   StringVector StringUtil::split( const String& str, const String& delims, unsigned int maxSplits, bool preserveDelims)
    {
        StringVector ret;
        // Pre-allocate some space for performance
        ret.reserve(maxSplits ? maxSplits+1 : 10);    // 10 is guessed capacity for most case

        unsigned int numSplits = 0;

        // Use STL methods 
        size_t start, pos;
        start = 0;
        do 
        {
            pos = str.find_first_of(delims, start);
            if (pos == start)
            {
                // Do nothing
                start = pos + 1;
            }
            else if (pos == String::npos || (maxSplits && numSplits == maxSplits))
            {
                // Copy the rest of the string
                ret.push_back( str.substr(start) );
                break;
            }
            else
            {
                // Copy up to delimiter
                ret.push_back( str.substr(start, pos - start) );

                if(preserveDelims)
                {
                    // Sometimes there could be more than one delimiter in a row.
                    // Loop until we don't find any more delims
                    size_t delimStart = pos, delimPos;
                    delimPos = str.find_first_not_of(delims, delimStart);
                    if (delimPos == String::npos)
                    {
                        // Copy the rest of the string
                        ret.push_back( str.substr(delimStart) );
                    }
                    else
                    {
                        ret.push_back( str.substr(delimStart, delimPos - delimStart) );
                    }
                }

                start = pos + 1;
            }
            // parse up to next real data
            start = str.find_first_not_of(delims, start);
            ++numSplits;

        } while (pos != String::npos);



        return ret;
    }

例子

ENUM_MAKE(MY_TEST, MY_1, MY_2, MY_3)


    MY_TEST s1 = MY_TEST::MY_1;
    MY_TEST s2 = MY_TEST::MY_2;
    MY_TEST s3 = MY_TEST::MY_3;

    String z1 = Helper_MY_TEST::toName(s1);
    String z2 = Helper_MY_TEST::toName(s2);
    String z3 = Helper_MY_TEST::toName(s3);

    MY_TEST q1 = Helper_MY_TEST::toType(z1);
    MY_TEST q2 = Helper_MY_TEST::toType(z2);
    MY_TEST q3 = Helper_MY_TEST::toType(z3);

自动ENUM_MAKE宏生成“枚举类”和“枚举反射函数”辅助类。

为了减少错误,Everything只定义一个ENUM_MAKE。

这种代码的优点是自动创建用于反射和查看宏代码,易于理解的代码。'enum to string', 'string to enum'性能都是算法O(1)。

缺点是第一次使用时,枚举relection的string vector和map的helper类被初始化。 但是如果你想,你也会被预初始化。- - - - - -

几天前我也遇到了同样的问题。没有一些奇怪的宏魔法,我找不到任何c++解决方案,所以我决定写一个CMake代码生成器来生成简单的开关case语句。

用法:

enum2str_generate(
  PATH          <path to place the files in>
  CLASS_NAME    <name of the class (also prefix for the files)>
  FUNC_NAME     <name of the (static) member function>
  NAMESPACE     <the class will be inside this namespace>
  INCLUDES      <LIST of files where the enums are defined>
  ENUMS         <LIST of enums to process>
  BLACKLIST     <LIST of constants to ignore>
  USE_CONSTEXPR <whether to use constexpr or not (default: off)>
  USE_C_STRINGS <whether to use c strings instead of std::string or not (default: off)>
)

该函数搜索文件系统中的include文件(使用include_directories命令提供的include目录),读取它们并执行一些regex来生成类和函数。

注意:constexpr在c++中意味着内联,所以使用USE_CONSTEXPR选项将只生成一个头类!

例子:

- includes a。h:

enum AAA : char { A1, A2 };

typedef enum {
   VAL1          = 0,
   VAL2          = 1,
   VAL3          = 2,
   VAL_FIRST     = VAL1,    // Ignored
   VAL_LAST      = VAL3,    // Ignored
   VAL_DUPLICATE = 1,       // Ignored
   VAL_STRANGE   = VAL2 + 1 // Must be blacklisted
} BBB;

/ CMakeLists.txt:

include_directories( ${PROJECT_SOURCE_DIR}/includes ...)

enum2str_generate(
   PATH       "${PROJECT_SOURCE_DIR}"
   CLASS_NAME "enum2Str"
   NAMESPACE  "abc"
   FUNC_NAME  "toStr"
   INCLUDES   "a.h" # WITHOUT directory
   ENUMS      "AAA" "BBB"
   BLACKLIST  "VAL_STRANGE")

生成:

/ enum2Str.hpp:

/*!
  * \file enum2Str.hpp
  * \warning This is an automatically generated file!
  */

#ifndef ENUM2STR_HPP
#define ENUM2STR_HPP

#include <string>
#include <a.h>

namespace abc {

class enum2Str {
 public:
   static std::string toStr( AAA _var ) noexcept;
   static std::string toStr( BBB _var ) noexcept;
};

}

#endif // ENUM2STR_HPP

/ enum2Str.cpp:

/*!
  * \file enum2Str.cpp
  * \warning This is an automatically generated file!
  */

#include "enum2Str.hpp"

namespace abc {

/*!
 * \brief Converts the enum AAA to a std::string
 * \param _var The enum value to convert
 * \returns _var converted to a std::string
 */
std::string enum2Str::toStr( AAA _var ) noexcept {
   switch ( _var ) {
      case A1: return "A1";
      case A2: return "A2";
      default: return "<UNKNOWN>";
   }
}

/*!
 * \brief Converts the enum BBB to a std::string
 * \param _var The enum value to convert
 * \returns _var converted to a std::string
 */
std::string enum2Str::toStr( BBB _var ) noexcept {
   switch ( _var ) {
      case VAL1: return "VAL1";
      case VAL2: return "VAL2";
      case VAL3: return "VAL3";
      default: return "<UNKNOWN>";
   }
}
}

更新:

脚本现在还支持作用域枚举(枚举类|struct)和 我将它与一些我经常使用的其他脚本一起移动到一个单独的repo: https://github.com/mensinda/cmakeBuildTools