在 C++ 中，是否可以枚举枚举(运行时或编译时(首选))并为每次迭代调用函数/生成代码？

示例用例:

enum abc
{    
    start
    a,
    b,
    c,
    end
}    
for each (__enum__member__ in abc)
{    
    function_call(__enum__member__);    
}

可能的重复:

• C++: Iterate through an enum
• Enum in C++ like Enum in Ada?

最佳答案

要添加到 @StackedCrooked 答案，您可以重载 operator++ , operator--和 operator*并具有类似迭代器的功能。

enum Color {
    Color_Begin,
    Color_Red = Color_Begin,
    Color_Orange,
    Color_Yellow,
    Color_Green,
    Color_Blue,
    Color_Indigo,
    Color_Violet,
    Color_End
};

namespace std {
template<>
struct iterator_traits<Color>  {
  typedef Color  value_type;
  typedef int    difference_type;
  typedef Color *pointer;
  typedef Color &reference;
  typedef std::bidirectional_iterator_tag
    iterator_category;
};
}

Color &operator++(Color &c) {
  assert(c != Color_End);
  c = static_cast<Color>(c + 1);
  return c;
}

Color operator++(Color &c, int) {
  assert(c != Color_End); 
  ++c;
  return static_cast<Color>(c - 1);
}

Color &operator--(Color &c) {
  assert(c != Color_Begin);
  return c = static_cast<Color>(c - 1);
}

Color operator--(Color &c, int) {
  assert(c != Color_Begin); 
  --c;
  return static_cast<Color>(c + 1);
}

Color operator*(Color c) {
  assert(c != Color_End);
  return c;
}

让我们测试一下 <algorithm>模板

void print(Color c) {
  std::cout << c << std::endl;
}

int main() {
  std::for_each(Color_Begin, Color_End, &print);
}

现在，Color是一个恒定的双向迭代器。这是我在上面手动编写时编写的可重用类。我注意到它可以用于更多的枚举，所以一遍遍重复相同的代码是相当乏味的

// Code for testing enum_iterator
// --------------------------------

namespace color_test {
enum Color {
  Color_Begin,
  Color_Red = Color_Begin,
  Color_Orange,
  Color_Yellow,
  Color_Green,
  Color_Blue,
  Color_Indigo,
  Color_Violet,
  Color_End
};

Color begin(enum_identity<Color>) {
  return Color_Begin;
}

Color end(enum_identity<Color>) {
  return Color_End;
}
}

void print(color_test::Color c) {
  std::cout << c << std::endl;
}

int main() {
  enum_iterator<color_test::Color> b = color_test::Color_Begin, e;
  while(b != e)
    print(*b++);
}

实现如下。

template<typename T>
struct enum_identity { 
  typedef T type; 
};

namespace details {
void begin();
void end();
}

template<typename Enum>
struct enum_iterator 
  : std::iterator<std::bidirectional_iterator_tag, 
                  Enum> {
  enum_iterator():c(end()) { }

  enum_iterator(Enum c):c(c) { 
    assert(c >= begin() && c <= end());
  }

  enum_iterator &operator=(Enum c) {
    assert(c >= begin() && c <= end());
    this->c = c; 
    return *this;
  }

  static Enum begin() {
    using details::begin; // re-enable ADL
    return begin(enum_identity<Enum>());
  }

  static Enum end() {
    using details::end; // re-enable ADL
    return end(enum_identity<Enum>());
  }

  enum_iterator &operator++() {
    assert(c != end() && "incrementing past end?");
    c = static_cast<Enum>(c + 1);
    return *this;
  }

  enum_iterator operator++(int) {
    assert(c != end() && "incrementing past end?");
    enum_iterator cpy(*this);
    ++*this;
    return cpy;
  }

  enum_iterator &operator--() {
    assert(c != begin() && "decrementing beyond begin?");
    c = static_cast<Enum>(c - 1);
    return *this;
  }

  enum_iterator operator--(int) {
    assert(c != begin() && "decrementing beyond begin?");
    enum_iterator cpy(*this);
    --*this;
    return cpy;
  }

  Enum operator*() {
    assert(c != end() && "cannot dereference end iterator");
    return c;
  }

  Enum get_enum() const {
    return c;
  }

private:
  Enum c;
};

template<typename Enum>
bool operator==(enum_iterator<Enum> e1, enum_iterator<Enum> e2) {
  return e1.get_enum() == e2.get_enum();
}

template<typename Enum>
bool operator!=(enum_iterator<Enum> e1, enum_iterator<Enum> e2) {
  return !(e1 == e2);
}

关于c++ - 枚举 C++ 中的枚举，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1390703/