265

I have an enum class with two values, and I want to create a method which receives a value and returns the other one. I also want to maintain type safety(that's why I use enum class instead of enums).

http://www.cplusplus.com/doc/tutorial/other_data_types/ doesn't mention anything about methods However, I was under the impression that any type of class can have methods.

6
  • 5
    No, it cannot. See here. Commented Jan 22, 2014 at 23:05
  • @octavian Note my answer and rethink about your use cases please! Commented Jan 22, 2014 at 23:41
  • @πάνταῥεῖ you're totally right, I've read enum but thought union, killed the comment. Commented Jan 23, 2014 at 1:10
  • @octavian Are you even asking for a particular use case at all, or did you just want to have the standards restrictions on c++11 enum class/struct confirmed? Commented Jan 23, 2014 at 1:19
  • 1
    In the 4th Ed of Stroustrup's book The C++ Programming Language, he says, "By default, an enum class has only assignment, initialization, and comparisons (e.g. == and < ) defined. However, an enumeration is a user-defined type so we can define operators for it." He then defines the operator++ for an enum class So you can define operators for enum classes at least. Commented Sep 6, 2017 at 13:11

10 Answers 10

232

While the answer that "you can't" is technically correct, I believe you may be able to achieve the behavior you're looking for using the following idea:

I imagine that you want to write something like:

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

And you were hoping that the code looks something like this:

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

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

But of course, it doesn't work, because enums can't have methods (and 'this' doesn't mean anything in the above context)

However, if you use the idea of a normal class containing a non-class enum and a single member variable that contains a value of that type, you can get extremely close to the syntax/behavior/type safety that you want. i.e.:

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;
};

Now you can write:

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

And the compiler will prevent things like:

Fruit f = 1;  // Compile time error.

You could easily add methods such that:

Fruit f("Apple");

and

f.ToString();

can be supported.

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11 Comments

Shouldn't be also IsYellow(), operator==, != marked as constexpr?
I'm getting " error: missing binary operator before token "switch" "
Could you elaborate on the macro if-else part?
It should be explicit operator bool() const = delete;, otherwise if(fruit) is possible if your fruit is of type const Fruit.
The precompiler #if is because you can make a choice on if you want to enable the usage or not. Having automatic conversions to Value() if you don't need it is imho a chance to start using the enum in a way that it shouldn't be.
|
186

No, they can't.

I can understand that the enum class part for strongly typed enums in C++11 might seem to imply that your enum has class traits too, but it's not the case. My educated guess is that the choice of the keywords was inspired by the pattern we used before C++11 to get scoped enums:

class Foo {
public:
  enum {BAR, BAZ};
};

However, that's just syntax. Again, enum class is not a class.

4 Comments

On ##C++ I was told that "c++ aims to be as confusing and expert friendly as possible". Obviously it's a joke, but you get the idea :)
A union is not what John Doe would consider a class, too. Yet they can have member functions. And classes are really not mandatory for member functions. Using a designator like value or this, something like enum Size { Huge, Mega, Apocalypse; bool operator<(X rhs) const { return *this < rhs; } (here also allowing ;), it can make just as much sense as other forms of functions.
Why wasn't this struct Foo{ enum {BAR, BAZ}; }; and therefore enum struct?
Oh, enum struct is actually a valid way to create a scoped enum! Who knew!
22

Concentrating on the description of the question instead of the title a possible answer is

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 "";
    }
};

Comments

8

There is a pretty compatible ability(§) to refactor an enum into a class without having to rewrite your code, which means that effectively you can do what you were asking to do without too much editing.

(§) as ElementW points out in a comment, type_traits dependent code will not work, so e.g. one cannot use auto, etc. There may be some way of handling such stuff, but in the end one is converting an enum into a class, and it is always a mistake to subvert C++

the enum struct and enum class specifications are about scoping so not part of this.

Your original enum is e.g. 'pet' (this is as an example only!).

enum pet { 
    fish, cat, dog, bird, rabbit, other 
};

(1) You modify that to eg petEnum (so as to hide it from your existing code).

enum petEnum { 
    fish, cat, dog, bird, rabbit, other 
};

(2) You add a new class declaration below it (named with the original enum)

class pet {
    private:
        petEnum value;
        pet() {}

    public:
        pet(const petEnum& v) : value{v} {} //not explicit here.
        operator petEnum() const { return value; }
        pet& operator=(petEnum v) { value = v; return *this;}
        bool operator==(const petEnum v) const { return value == v; }
        bool operator!=(const petEnum v) const { return value != v; }
 //     operator std::string() const;

};

(3) You can now add whatever class methods you like to your pet class. eg. a string operator

    pet::operator std::string() const {
        switch (value) {
            case fish: return "fish";
            case cat:  return "cat";
            case dog:  return "dog";
            case bird: return "bird";
            case rabbit: return "rabbit";
            case other: return "Wow. How exotic of you!";
        }
    }

Now you can use eg std::cout...

int main() {
    pet myPet = rabbit;
    if(myPet != fish) {
        cout << "No splashing! ";
    }
    std::cout << "I have a " << std::string(myPet) << std::endl;
    return 0;
}

5 Comments

It is not fully compatible: if you use the enum values with any kind of type deduction where it is expected to get a pet typename/instance, be it templates, auto, or decltype, this breaks, as you get a petEnum instead.
By wrapping it in a class you lose the ability to use it for switch-case statements - one of the things were enums normally would really shine.
@ABaumstumpf, sure. But the Q is about enums with methods. Needless to say if one is wishing to switch on an instance, the switch would be handled as a method within the class, cf. my example 3 above.
@Konchog It is an alternative that allows you to have names and functions, but it lacks many of the traits of enums - implicit/explicit conversions to underlying type and operators need to be implemented, switch-statements need Constant-expressions etc. If you don#t need those features of enums then a class works just fine.
@ABaumstumpf, there are use cases for this - for instance when needing to update legacy code - but I don't disagree. It is better to be wise when one can be.
5

It may not fulfill all your needs, but with non-member operators you can still have a lot of fun. For example:

#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";
    }
}

This allows code like

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"

Comments

4

As mentioned in the other answer, no. Even enum class isn't a class.


Usually the need to have methods for an enum results from the reason that it's not a regular (just incrementing) enum, but kind of bitwise definition of values to be masked or need other bit-arithmetic operations:

enum class Flags : unsigned char {
    Flag1 = 0x01 , // Bit #0
    Flag2 = 0x02 , // Bit #1
    Flag3 = 0x04 , // Bit #3
    // aso ...
}

// Sets both lower bits
unsigned char flags = (unsigned char)(Flags::Flag1 | Flags::Flag2);

// Set Flag3
flags |= Flags::Flag3;

// Reset Flag2
flags &= ~Flags::Flag2;

Obviously one thinks of encapsulating the necessary operations to re-/set single/group of bits, by e.g. bit mask value or even bit index driven operations would be useful for manipulation of such a set of 'flags'.

The struct/class specification just supports better scoping of enum values for access. No more, no less!

Ways to get out of the restriction you cannot declare methods for enum (classes) are , either to use a std::bitset (wrapper class), or a bitfield union.

unions, and such bitfield unions can have methods (see here for the restrictions!).

I have a sample, how to convert bit mask values (as shown above) to their corresponding bit indices, that can be used along a std::bitset here: BitIndexConverter.hpp
I've found this pretty useful for enhancing readability of some 'flag' decison based algorithms.

2 Comments

There are more use cases that warrant methods on enum classe, e.g. toString() and fromString(). Every (even not so) modern major language has this (e.g. C#, Java, Swift), just not C++.
Let's hope for unified call syntax next time around... open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4165.pdf
2

Based on jtlim's answer

Idea (Solution)

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;
};

Usage

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"); }

Comments

1

I know this is an old question but something that I don't see here is using an std::map<enum, std::string> for a to_string() equivalent function and, if necessary, another std::map<std::string, enum> for a from_string() equivalent function.

But the question is about an enum class with only two values, like a bool. Then I think it can be reduced to:

enum class TwoValues {
    value1,
    value2
};

static TwoValues operator!(TwoValues tv)
{
    return tv == TwoValues::value1 ? TwoValues::value2 : TwoValues::value1;
}

int main() {
    TwoValues result = TwoValues::value1;
    result = !result;
}

5 Comments

If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
It's not a new question. I was giving another possible solution to the original question.
Then the wording is vague at best. 1- you do not stated the fact enum class cannot have methods. 2- the flipflop table map is not a memeber of the enum class neither. btw, I understand your idea, you may improve the answer by a new question about how to generalize the lack of enum class methods (better visibility than enum class method original question). I would suggest to implement an operator++ on the enum class using the flipflop map ; enum class operator is allowed if I remember well.
Thanks for the suggestion, @tomrtc. The std::map isn't needed if there is a not operator (in this case).
nice the final result fit the question imo.
0

Here is what I do. Hope it helps.

In addition to the enum, I create another calls extension with static members.

/// @brief Current type selector for Nano only (Merlin)
enum class CurrentTypeSelector {
    /// Force DC (Direct current)
    currentTypeSelectorForceDc = 0,

    /// Force PC (pulsed current)
    currentTypeSelectorForcePc = 1,

    /// Automatic mode: select the one sent by the batteries
    currentTypeSelectorAutomatic = 2,

    /// Force shunt current
    currentTypeSelectorShunt = 3,

    /// Pick either DC or PC, aka what the battery has chosen
    currentTypeSelectorBatteryAutomatic = 4,
};

// CurrentTypeSelector enum extension
struct CurrentTypeSelectorExtension {

    static CurrentTypeSelector fromInt(int v) {
        switch (v) {
        case (unsigned)CurrentTypeSelector::currentTypeSelectorForceDc:             return CurrentTypeSelector::currentTypeSelectorForceDc;
        case (unsigned)CurrentTypeSelector::currentTypeSelectorForcePc:             return CurrentTypeSelector::currentTypeSelectorForcePc;
        case (unsigned)CurrentTypeSelector::currentTypeSelectorShunt:               return CurrentTypeSelector::currentTypeSelectorShunt;
        case (unsigned)CurrentTypeSelector::currentTypeSelectorBatteryAutomatic:    return CurrentTypeSelector::currentTypeSelectorBatteryAutomatic;
        case (unsigned)CurrentTypeSelector::currentTypeSelectorAutomatic:           return CurrentTypeSelector::currentTypeSelectorAutomatic;
        default: 
            pepsr.alert("CurrentTypeSelectorEnum::fromValue: Unknown value %d, returns DC", v);
            return CurrentTypeSelector::currentTypeSelectorForceDc;
        }
    }

    static String toString(CurrentTypeSelector v) {

        // if (v >= std::numeric_limits<enum CurrentTypeSelector>::max()) {
        //  pepsr.alert("CurrentTypeSelectorEnum::toString: CurrentTypeSelectorEnum has more values");
        // }
        switch (v) {
        case CurrentTypeSelector::currentTypeSelectorForceDc: return "dc";
        case CurrentTypeSelector::currentTypeSelectorForcePc: return "pc";
        case CurrentTypeSelector::currentTypeSelectorShunt: return "shunt";
        case CurrentTypeSelector::currentTypeSelectorBatteryAutomatic: return "battery";
        case CurrentTypeSelector::currentTypeSelectorAutomatic: return "auto";

        default: 
            pepsr.alert("CurrentTypeSelectorEnum::toString: Unknown. CurrentTypeSelectorEnum has more values npow. You must upgrade toString()");
            return "pc";
        }
    }
};

Comments

-3

Yes, they can, but you need to make a wrapper class, for example:

#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;
}

1 Comment

This was already suggested 4 years ago.

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