Is using a enum class inside of a class class un-necessary - c++

I should start by saying I'm fairly new to OOP. I am trying to write a portable class which will probably be used in multiple programs. The class requires multiple enumerated lists to store information which are only relevant to the class so I was planning on nesting the enum inside the class. My questions are
Is there any reason to use an enum class inside a class, or should I just use enum? Just curious. I posted sample code for reference below.
Is there any reason why it would be a really bad idea to make enums in a class public?
#include <iostream>
enum class Color {
red = 0,
green = 1,
blue = 2
};
class CodeLibray {
public:
enum Color2 {
red = 0,
green = 1,
blue = 2
};
enum class Color3 {
red = 0,
green = 1,
blue = 2
};
Color color;
Color2 color2;
Color3 color3;
};
int main(){
CodeLibray c;
c.color = Color::red;
c.color2 = CodeLibray::Color2::blue;
c.color3 = CodeLibray::Color3::blue;
}

Is there any reason to use an enum class inside a class, or should I just use enum? Just curious. I posted sample code for
reference below.
It's pretty obvious that the enum values defined inside CodeLibray would become ambigous if you omit the enum class there.
Also, If I want enums to be accessible for assigning public member values but I do not want values in the enum to be changed is this
possible?
I don't fully understand what you mean. Just a shot in the dark, something like the following?
enum Color2 {
red = ::red,
// ^^ Refer to the enum declaration in the global namespace
green = ::green,
blue = ::blue
};
enum class Color3 {
red = ::red,
green = ::green,
blue = ::blue
};

Related

Namespace enum vs enum class [duplicate]

I heard a few people recommending to use enum classes in C++ because of their type safety.
But what does that really mean?
C++ has two kinds of enum:
enum classes
Plain enums
Here are a couple of examples on how to declare them:
enum class Color { red, green, blue }; // enum class
enum Animal { dog, cat, bird, human }; // plain enum
What is the difference between the two?
enum classes - enumerator names are local to the enum and their values do not implicitly convert to other types (like another enum or int)
Plain enums - where enumerator names are in the same scope as the enum and their
values implicitly convert to integers and other types
Example:
enum Color { red, green, blue }; // plain enum
enum Card { red_card, green_card, yellow_card }; // another plain enum
enum class Animal { dog, deer, cat, bird, human }; // enum class
enum class Mammal { kangaroo, deer, human }; // another enum class
void fun() {
// examples of bad use of plain enums:
Color color = Color::red;
Card card = Card::green_card;
int num = color; // no problem
if (color == Card::red_card) // no problem (bad)
cout << "bad" << endl;
if (card == Color::green) // no problem (bad)
cout << "bad" << endl;
// examples of good use of enum classes (safe)
Animal a = Animal::deer;
Mammal m = Mammal::deer;
int num2 = a; // error
if (m == a) // error (good)
cout << "bad" << endl;
if (a == Mammal::deer) // error (good)
cout << "bad" << endl;
}
Conclusion:
enum classes should be preferred because they cause fewer surprises that could potentially lead to bugs.
From Bjarne Stroustrup's C++11 FAQ:
The enum classes ("new enums", "strong enums") address three problems
with traditional C++ enumerations:
conventional enums implicitly convert to int, causing errors when someone does not want an enumeration to act as an integer.
conventional enums export their enumerators to the surrounding scope, causing name clashes.
the underlying type of an enum cannot be specified, causing confusion, compatibility problems, and makes forward declaration
impossible.
The new enums are "enum class" because they combine aspects of traditional enumerations (names values) with aspects of classes (scoped members and absence of conversions).
So, as mentioned by other users, the "strong enums" would make the code safer.
The underlying type of a "classic" enum shall be an integer type large enough to fit all the values of the enum; this is usually an int. Also each enumerated type shall be compatible with char or a signed/unsigned integer type.
This is a wide description of what an enum underlying type must be, so each compiler will take decisions on its own about the underlying type of the classic enum and sometimes the result could be surprising.
For example, I've seen code like this a bunch of times:
enum E_MY_FAVOURITE_FRUITS
{
E_APPLE = 0x01,
E_WATERMELON = 0x02,
E_COCONUT = 0x04,
E_STRAWBERRY = 0x08,
E_CHERRY = 0x10,
E_PINEAPPLE = 0x20,
E_BANANA = 0x40,
E_MANGO = 0x80,
E_MY_FAVOURITE_FRUITS_FORCE8 = 0xFF // 'Force' 8bits, how can you tell?
};
In the code above, some naive coder is thinking that the compiler will store the E_MY_FAVOURITE_FRUITS values into an unsigned 8bit type... but there's no warranty about it: the compiler may choose unsigned char or int or short, any of those types are large enough to fit all the values seen in the enum. Adding the field E_MY_FAVOURITE_FRUITS_FORCE8 is a burden and doesn't forces the compiler to make any kind of choice about the underlying type of the enum.
If there's some piece of code that rely on the type size and/or assumes that E_MY_FAVOURITE_FRUITS would be of some width (e.g: serialization routines) this code could behave in some weird ways depending on the compiler thoughts.
And to make matters worse, if some workmate adds carelessly a new value to our enum:
E_DEVIL_FRUIT = 0x100, // New fruit, with value greater than 8bits
The compiler doesn't complain about it! It just resizes the type to fit all the values of the enum (assuming that the compiler were using the smallest type possible, which is an assumption that we cannot do). This simple and careless addition to the enum could subtlety break related code.
Since C++11 is possible to specify the underlying type for enum and enum class (thanks rdb) so this issue is neatly addressed:
enum class E_MY_FAVOURITE_FRUITS : unsigned char
{
E_APPLE = 0x01,
E_WATERMELON = 0x02,
E_COCONUT = 0x04,
E_STRAWBERRY = 0x08,
E_CHERRY = 0x10,
E_PINEAPPLE = 0x20,
E_BANANA = 0x40,
E_MANGO = 0x80,
E_DEVIL_FRUIT = 0x100, // Warning!: constant value truncated
};
Specifying the underlying type if a field have an expression out of the range of this type the compiler will complain instead of changing the underlying type.
I think that this is a good safety improvement.
So Why is enum class preferred over plain enum?, if we can choose the underlying type for scoped(enum class) and unscoped (enum) enums what else makes enum class a better choice?:
They don't convert implicitly to int.
They don't pollute the surrounding namespace.
They can be forward-declared.
The basic advantage of using enum class over normal enums is that you may have same enum variables for 2 different enums and still can resolve them(which has been mentioned as type safe by OP)
For eg:
enum class Color1 { red, green, blue }; //this will compile
enum class Color2 { red, green, blue };
enum Color1 { red, green, blue }; //this will not compile
enum Color2 { red, green, blue };
As for the basic enums, compiler will not be able to distinguish whether red is refering to the type Color1 or Color2 as in hte below statement.
enum Color1 { red, green, blue };
enum Color2 { red, green, blue };
int x = red; //Compile time error(which red are you refering to??)
Enumerations are used to represent a set of integer values.
The class keyword after the enum specifies that the enumeration is strongly typed and its enumerators are scoped. This way enum classes prevents accidental misuse of constants.
For Example:
enum class Animal{Dog, Cat, Tiger};
enum class Pets{Dog, Parrot};
Here we can not mix Animal and Pets values.
Animal a = Dog; // Error: which DOG?
Animal a = Pets::Dog // Pets::Dog is not an Animal
It's worth noting, on top of these other answers, that C++20 solves one of the problems that enum class has: verbosity. Imagining a hypothetical enum class, Color.
void foo(Color c)
switch (c) {
case Color::Red: ...;
case Color::Green: ...;
case Color::Blue: ...;
// etc
}
}
This is verbose compared to the plain enum variation, where the names are in the global scope and therefore don't need to be prefixed with Color::.
However, in C++20 we can use using enum to introduce all of the names in an enum to the current scope, solving the problem.
void foo(Color c)
using enum Color;
switch (c) {
case Red: ...;
case Green: ...;
case Blue: ...;
// etc
}
}
So now, there is no reason not to use enum class.
do not implicitly convert to int
can choose which type underlie
ENUM namespace to avoid polluting happen
Compared with normal class, can be declared forward, but do not have methods
C++11 FAQ mentions below points:
conventional enums implicitly convert to int, causing errors when someone does not want an enumeration to act as an integer.
enum color
{
Red,
Green,
Yellow
};
enum class NewColor
{
Red_1,
Green_1,
Yellow_1
};
int main()
{
//! Implicit conversion is possible
int i = Red;
//! Need enum class name followed by access specifier. Ex: NewColor::Red_1
int j = Red_1; // error C2065: 'Red_1': undeclared identifier
//! Implicit converison is not possible. Solution Ex: int k = (int)NewColor::Red_1;
int k = NewColor::Red_1; // error C2440: 'initializing': cannot convert from 'NewColor' to 'int'
return 0;
}
conventional enums export their enumerators to the surrounding scope, causing name clashes.
// Header.h
enum vehicle
{
Car,
Bus,
Bike,
Autorickshow
};
enum FourWheeler
{
Car, // error C2365: 'Car': redefinition; previous definition was 'enumerator'
SmallBus
};
enum class Editor
{
vim,
eclipes,
VisualStudio
};
enum class CppEditor
{
eclipes, // No error of redefinitions
VisualStudio, // No error of redefinitions
QtCreator
};
The underlying type of an enum cannot be specified, causing confusion, compatibility problems, and makes forward declaration impossible.
// Header1.h
#include <iostream>
using namespace std;
enum class Port : unsigned char; // Forward declare
class MyClass
{
public:
void PrintPort(enum class Port p);
};
void MyClass::PrintPort(enum class Port p)
{
cout << (int)p << endl;
}
.
// Header.h
enum class Port : unsigned char // Declare enum type explicitly
{
PORT_1 = 0x01,
PORT_2 = 0x02,
PORT_3 = 0x04
};
.
// Source.cpp
#include "Header1.h"
#include "Header.h"
using namespace std;
int main()
{
MyClass m;
m.PrintPort(Port::PORT_1);
return 0;
}
One thing that hasn't been explicitly mentioned - the scope feature gives you an option to have the same name for an enum and class method. For instance:
class Test
{
public:
// these call ProcessCommand() internally
void TakeSnapshot();
void RestoreSnapshot();
private:
enum class Command // wouldn't be possible without 'class'
{
TakeSnapshot,
RestoreSnapshot
};
void ProcessCommand(Command cmd); // signal the other thread or whatever
};
Because, as said in other answers, class enum are not implicitly convertible to int/bool, it also helps to avoid buggy code like:
enum MyEnum {
Value1,
Value2,
};
...
if (var == Value1 || Value2) // Should be "var == Value2" no error/warning

Enum methods in Java and C++11?

Consider the following Enum Class in Java:
package com.gousslegend.deepov;
public enum Color
{
WHITE, BLACK;
private Color opposite;
static
{
WHITE.opposite = BLACK;
BLACK.opposite = WHITE;
}
public Color getOppositeColor()
{
return opposite;
}
}
How can I achieve the same thing with C++11?
Is is possible to have enum methods with C++11?
Should I put my enum in a separate class/file with C++?
Thanks
How can I achieve the same thing with C++11?
In C++ you would be better off defining an enum, along with a free-standing function:
enum Color {BLACK, WHITE};
Color getOppositeColor(Color c) {
static Color opposite[] = {WHITE, BLACK};
return opposite[c];
}
If you are set on modeling Java exactly, make a class with two static members BLACK and WHITE, and use it instead of enum. The use would be limited, though: for instance, you wouldn't be able to use your enum values in switch cases.
Is is possible to have enum methods with C++11?
No
Should I put my enum in a separate class/file with C++?
You should put enum definition and a function prototype in a header; implementation of the function goes in a separate C++ file.
C++'s enums are simple enumerated values, they're not full-fledged classes (enum classes in C++ only provide some additional type-safety).
I would take the following approach to implement the equivalent to how Java's enums are used here:
// Header file
class Color {
Color *opposite_p;
public:
Color(Color *opposite_pArg) : opposite_p(opposite_pArg) {}
Color &opposite() { return *opposite_p; }
static Color WHITE, BLACK;
};
// .C file
Color Color::WHITE(&BLACK);
Color Color::BLACK(&WHITE);
Then, Color::WHITE.opposite() returns Color::BLACK, and COLOR::BLACK.oppsoite() returns Color::WHITE.

Why is it possible to get this enum member value without having to delve in the nested enum first?

(under Visual Studio 2012)
Consider the following class :
class pe_exception : public std::runtime_error
{
public:
//Exception IDs
enum exception_id
{
unknown_error,
bad_pe_file,
bad_dos_header,
image_nt_headers_not_found,
error_reading_image_nt_headers,
error_reading_data_directories,
error_reading_file,
pe_signature_incorrect,
...
Can you tell why is it possible to directly get the enum member with such code :
bool b = e.get_id() == pe_exception::error_reading_image_nt_headers;
Instead of having to go into the enum first ?
bool b = e.get_id() == pe_exception::exception_id::error_reading_image_nt_headers;
(library where this is taken from : https://code.google.com/p/portable-executable-library/)
That is just the way old-school enums work: they have no scope. This comes directly from C.
For example:
enum Colour { red, black, orange };
Colour c = orange; // OK
enum Fruit { banana, apple, orange }; // ERROR: orange has already been defined
C++11 class enums or scoped enumerations fix this (see scoped enumerations here for more information).
enum class Colour { red, black, orange };
Colour c = orange; // ERROR
Colour c = Color::red; // OK
enum class Fruit { banana, apple, orange }; // OK
An enum is absolutly equivalent to a list of static int declaration, except the compiler generates initialization value starting at 0, and incrementing by one between each value.
In your case
enum exception_id
{
unknown_error,
bad_pe_file,
bad_dos_header
};
int val = exception_id::unknown_error; //Not ok (except >MSVC 2010)
Note that on visual since VS 2010 (maybe before) you can add the nested scope to be more precise, but that was not standard (and not compiling on GCC and CLang).
is equivalent to:
static int unknown_error = 0;
static int bad_pe_file= 1;
static int bad_dos_header= 2;
The solution before C++11 was
struct exception_id{
enum exception_id_enum
{
unknown_error,
bad_pe_file,
bad_dos_header
};};
int val = exception_id::unknown_error; //ok
Now you can write
enum class exception_id{
{
unknown_error,
bad_pe_file,
bad_dos_header
};
auto val = exception_id::unknown_error; //ok
Last but very important point The typing changes, now you have strong typing on this.
I let you read the FAQ of bjarne
See here a nice explaination on MSDN

C++ enums and inline header functions

When using an enum in C++ how does one create getter and setters?
Example:
enum class FieldLayerStates
{
kEmpty = 0, // new to the game.
kActive = 1, // has fields already in the same.
kAdd = 2 // field layer is ready to have a field added.
};
FieldLayerStates _fieldLayerState;
inline FieldLayerStates getFieldLayerState() { return _fieldLayerState; };
inline void setFieldLayerState(FieldLayerStates i) { _fieldLayerState = i; };
I am getting errors at the inline functions:
: Unknown type name 'FieldLayerStates'
: Cannot initialize return object of type 'int' with an lvalue of type 'FieldLayer::FieldLayerStates'
and when I go to use it:
// check the status of the layer to see if it has fields in it already (loaded from CCUserDefaults
if (getFields().empty())
{
// nothing, this must be our first time in.
setFieldLayerStatus(kEmpty);
}
It says kEmpty is undeclared.
Can someone help me clear up my confusion?
You're using an enum class, are you sure that's what you want?
If you stop doing that your code will work.
otherwise refer to FieldLayerStates::kEmpty (because enumerators of an enum class must be qualified by their type name)
I don't know why you get the Unknown type name 'FieldLayerStates' error because you haven't shown enough context to understand the code, at a guess I'd say you're trying to define the functions outside the class and you need to say FieldLayer::FieldLayerStates
Please show the full code so we have a chance of seeing what you're really compiling.
enum class Foo
is a new C++11 language feature that means "strongly typed, and scoped" enumerations. It is significantly different than just
enum Foo
When you use strongly typed enums, you have to qualify them with the scope they are contained in.
enum class Colors { Red, Green, Blue };
enum class Moods { Happy, Bored, Blue };
Without the "class", this won't compile, because you have defined "Blue" twice. With the "Class" you have actually defined two scopes with their own, private, scoped enums, that require the "::" operator to access.
The "class" are also strongly typed, which means that they won't cast - e.g. to an integer type - without you explicitly casting them.
enum COLORS { RED, GREEN, BLUE };
enum class Colors { Red, Green Blue };
int i = RED; // Legal
Colors j = Colors::Red; // Legal
int i = Colors::Red; // Illegal: Colors::Red is not an integer, it's a Colors.
Colors j = Red; // Illegal: You didn't specify the scope.
Colors j = RED; // Illegal: RED is an integer, not a Colors.
for (int i = RED; i <= BLUE; ++i) { cout << i << endl; } // Legal
// But Colors is not a numeric type, you can't do math or increment on it,
// so the following is illegal:
for (auto j = Colors::Red; j <= Colors::Blue; ++j)
enum class Flags = { Herp = 1, Derp = 2 };
Flags combo = Flags::Herp;
combo |= Flags::Derp; // Illegal, no '|' operator for Flags, casting required.
I think you just want this
enum FieldLayerStates
{
kEmpty = 0, // new to the game.
kActive = 1, // has fields already in the same.
kAdd = 2 // field layer is ready to have a field added.
};

Accessing to enum values by '::' in C++

I have class like following:
class Car
{
public:
Car();
// Some functions and members and <b>enums</b>
enum Color
{
Red,
Blue,
Black
};
Color getColor();
void setColor(Color);
private:
Color myColor;
}
I want to:
access to Color values as Color::Red. It is really hardly to understand code when Car::Red is used, when class have a lot enums, subclasses etc.
use type Color as function argument or return value
use variable type Color in switch
I know 3 partial solutions:
Using embedded class Color and enum in it
Using embedded namespace Color and enum in it
Using enum class
1 and 2 solutions solves a Color::Red accession problem, but I can't use functions like Color getColor() and void setColor(Color).
3 solution has a problem: VS2010 doen't support enum class. GCC v.4.1.2 doesn't support it too. I don't know about later versions of gcc.
Yes, I'm working on cross-platform project.
I have found this solution, but it seems ... heavy.
I hope somebody can help me here :)
In current C++ (i.e. C++11 and beyond), you can already access enum values like that:
enum Color { Red };
Color c = Color::Red;
Color d = Red;
You can go further and enforce the use of this notation:
enum class Color { Red };
Color c = Color::Red;
// Color d = Red; <-- error now
And on a sidenote, you now define the underlying type, which was previously only possible with hacky code (FORCEDWORD or so anyone?):
enum class Color : char { Red };
Name the enum inside the nested class (as example one):
class Car
{
public:
struct Color
{
enum Type
{
Red,
Blue,
Black
};
};
Color::Type getColor();
void setColor(Color::Type);
};
When I want to do something like this I tend to use a namespace and a typedef outside of th namespace (though usually I'm doing this globally rather than inside a class). Something like this:
namespace colors
{
enum Color
{
Red,
Blue
...
}
}
typedef colors::Color Color;
This way you use the namespace to get at the actual colors, but the Color type itself is still globally accessible:
Color myFav = colors::Red;