我有一个有两个值的枚举类,我想创建一个接收值的方法 然后返回另一个。我还想维护类型安全(这就是为什么我使用枚举类而不是enum)。

http://www.cplusplus.com/doc/tutorial/other_data_types/没有提到任何关于方法的内容 然而,在我的印象中,任何类型的类都可以有方法。


当前回答

把注意力集中在问题的描述上,而不是标题,一个可能的答案是

struct LowLevelMouseEvent {
    enum Enum {
        mouse_event_uninitialized = -2000000000, // generate crash if try to use it uninitialized.
        mouse_event_unknown = 0,
        mouse_event_unimplemented,
        mouse_event_unnecessary,
        mouse_event_move,
        mouse_event_left_down,
        mouse_event_left_up,
        mouse_event_right_down,
        mouse_event_right_up,
        mouse_event_middle_down,
        mouse_event_middle_up,
        mouse_event_wheel
    };
    static const char* ToStr (const type::LowLevelMouseEvent::Enum& event)
    {
        switch (event) {
            case mouse_event_unknown:         return "unknown";
            case mouse_event_unimplemented:   return "unimplemented";
            case mouse_event_unnecessary:     return "unnecessary";
            case mouse_event_move:            return "move";
            case mouse_event_left_down:       return "left down";
            case mouse_event_left_up:         return "left up";
            case mouse_event_right_down:      return "right down";
            case mouse_event_right_up:        return "right up";
            case mouse_event_middle_down:     return "middle down";
            case mouse_event_middle_up:       return "middle up";
            case mouse_event_wheel:           return "wheel";
            default:
                Assert (false);
                break;
        }
        return "";
    }
};

其他回答

它可能不能满足您的所有需求,但使用非成员操作符仍然可以获得很多乐趣。例如:

#include <iostream>

enum class security_level
{
    none, low, medium, high
};

static bool operator!(security_level s) { return s == security_level::none; }

static security_level& operator++(security_level& s)
{
    switch(s)
    {
        case security_level::none: s = security_level::low; break;
        case security_level::low: s = security_level::medium; break;
        case security_level::medium: s = security_level::high; break;
        case security_level::high: break;
    }
    return s;
}

static std::ostream & operator<<(std::ostream &o, security_level s)
{
    switch(s)
    {
        case security_level::none: return o << "none";
        case security_level::low: return o << "low";
        case security_level::medium: return o << "medium";
        case security_level::high: return o << "high";
    }
}

这允许如下代码

security_level l = security_level::none;   
if(!!l) { std::cout << "has a security level: " << l << std::endl; } // not reached
++++l;
if(!!l) { std::cout << "has a security level: " << l << std::endl; } // reached: "medium"

根据jtlim的回答

想法(解决方案)

enum ErrorType: int {
  noConnection,
  noMemory
};

class Error {
public:
  Error() = default;
  constexpr Error(ErrorType type) : type(type) { }

  operator ErrorType() const { return type; }
  constexpr bool operator == (Error error) const { return type == error.type; }
  constexpr bool operator != (Error error) const { return type != error.type; }    
  constexpr bool operator == (ErrorType errorType) const { return type == errorType; }
  constexpr bool operator != (ErrorType errorType) const { return type != errorType; }

  String description() { 
    switch (type) {
    case noConnection: return "no connection";
    case noMemory: return "no memory";
    default: return "undefined error";
    }
 }

private:
  ErrorType type;
};

使用

Error err = Error(noConnection);
err = noMemory;
print("1 " + err.description());

switch (err) {
  case noConnection: 
    print("2 bad connection");
    break;
  case noMemory:
    print("2 disk is full");
    break;
  default: 
    print("2 oops");
    break;
}

if (err == noMemory) { print("3 Errors match"); }
if (err != noConnection) { print("4 Errors don't match"); }

把注意力集中在问题的描述上,而不是标题,一个可能的答案是

struct LowLevelMouseEvent {
    enum Enum {
        mouse_event_uninitialized = -2000000000, // generate crash if try to use it uninitialized.
        mouse_event_unknown = 0,
        mouse_event_unimplemented,
        mouse_event_unnecessary,
        mouse_event_move,
        mouse_event_left_down,
        mouse_event_left_up,
        mouse_event_right_down,
        mouse_event_right_up,
        mouse_event_middle_down,
        mouse_event_middle_up,
        mouse_event_wheel
    };
    static const char* ToStr (const type::LowLevelMouseEvent::Enum& event)
    {
        switch (event) {
            case mouse_event_unknown:         return "unknown";
            case mouse_event_unimplemented:   return "unimplemented";
            case mouse_event_unnecessary:     return "unnecessary";
            case mouse_event_move:            return "move";
            case mouse_event_left_down:       return "left down";
            case mouse_event_left_up:         return "left up";
            case mouse_event_right_down:      return "right down";
            case mouse_event_right_up:        return "right up";
            case mouse_event_middle_down:     return "middle down";
            case mouse_event_middle_up:       return "middle up";
            case mouse_event_wheel:           return "wheel";
            default:
                Assert (false);
                break;
        }
        return "";
    }
};

虽然“你不能”这个答案在技术上是正确的,但我相信你可以用下面的想法来实现你想要的行为:

我猜你想写这样的东西:

Fruit f = Fruit::Strawberry;
f.IsYellow();

你希望代码看起来像这样:

enum class Fruit : uint8_t
{
  Apple, 
  Pear,
  Banana,
  Strawberry,

  bool IsYellow() { return this == Banana; }
};

但是,当然,它不起作用,因为枚举不能有方法(并且'this'在上面的上下文中没有任何意义)

但是,如果您使用包含非类枚举和包含该类型值的单个成员变量的普通类的思想,则可以非常接近您想要的语法/行为/类型安全。例如:

class Fruit
{
public:
  enum Value : uint8_t
  {
    Apple,
    Pear,
    Banana,
    Strawberry
  };

  Fruit() = default;
  constexpr Fruit(Value aFruit) : value(aFruit) { }

#if Enable switch(fruit) use case:
  // Allow switch and comparisons.
  constexpr operator Value() const { return value; }

  // Prevent usage: if(fruit)
  explicit operator bool() const = delete;        
#else
  constexpr bool operator==(Fruit a) const { return value == a.value; }
  constexpr bool operator!=(Fruit a) const { return value != a.value; }
#endif

  constexpr bool IsYellow() const { return value == Banana; }

private:
  Value value;
};

现在你可以这样写:

Fruit f = Fruit::Strawberry;
f.IsYellow();

编译器会阻止如下事情:

Fruit f = 1;  // Compile time error.

你可以很容易地添加这样的方法:

Fruit f("Apple");

and

f.ToString();

可支持。

是的,它们可以,但是你需要创建一个包装器类,例如:

#include <iostream>
using namespace std;

class Selection {
    public: 
        enum SelectionEnum {
            yes,
            maybe,
            iDontKnow,
            canYouRepeatTheQuestion
        };
        Selection(SelectionEnum selection){value=selection;};
        string toString() {
            string selectionToString[4]={
                "Yes",
                "Maybe",
                "I don't know",
                "Can you repeat the question?"
            };
            return selectionToString[value];
        };
    private:
        SelectionEnum value;
};


int main(){
    Selection s=Selection(Selection::yes);
    cout<<s.toString()<<endl;
    return 0;
}