How can I output the value of an enum class in C 11

Question

How can I output the value of an enum class in C  11  In C  03 it s like this    include  lt iostream gt   using namespace std   enum A     a   1    b   69    c  666     int main        A a   A  c    cout  lt  lt  a  lt  lt  endl      in c  0x this code doesn t compile   include  lt iostream gt   using namespace std   enum class A     a   1    b   69    c  666     int main        A a   A  c    cout  lt  lt  a  lt  lt  endl      prog cpp 13 11  error  cannot bind  std  ostream  lvalue to  std  basic ostream lt char gt  amp  amp    usr lib gcc i686-pc-linux-gnu 4 5 1             include c   4 5 1 ostream 579 5  error    initializing argument 1 of  std  basic ostream lt  CharT   Traits gt  amp  std  operator lt  lt  std  basic ostream lt  CharT   Traits gt  amp  amp   const  Tp amp    with  CharT   char   Traits   std  char traits lt char gt    Tp   A     compiled at Ideone com

User · Answer

To write simpler   enum class Color       Red   1      Green   11      Blue   111     int value   static cast lt int gt  Color  Blue      111

User · Answer

include  lt iostream gt   include  lt type traits gt   using namespace std   enum class A     a   1    b   69    c  666     std  ostream amp  operator  lt  lt   std  ostream amp  os  const A amp  obj       os  lt  lt  static cast lt std  underlying type lt A gt   type gt  obj      return os     int main        A a   A  c    cout  lt  lt  a  lt  lt  endl

User · Answer

You could do something like this     outside of main namespace A       enum A               a   0          b   69          c   666              in main   A  A a   A  c  std  cout  lt  lt  a  lt  lt  std  endl

User · Answer

Following worked for me in C  11   template  lt typename Enum gt  constexpr typename std  enable if lt std  is enum lt Enum gt   value                                    typename std  underlying type lt Enum gt   type gt   type to integral Enum const amp  value        return static cast lt typename std  underlying type lt Enum gt   type gt  value

User · Answer

It is possible to get your second example  i e   the one using a scoped enum  to work using the same syntax as unscoped enums  Furthermore  the solution is generic and will work for all scoped enums  versus writing code for each scoped enum  as shown in the answer provided by  ForEveR    The solution is to write a generic operator lt  lt  function which will work for any scoped enum  The solution employs SFINAE via std  enable if and is as follows    include  lt iostream gt   include  lt type traits gt      Scoped enum enum class Color       Red      Green      Blue        Unscoped enum enum Orientation       Horizontal      Vertical        Another scoped enum enum class ExecStatus       Idle      Started      Running     template lt typename T gt  std  ostream amp  operator lt  lt  typename std  enable if lt std  is enum lt T gt   value  std  ostream gt   type amp  stream  const T amp  e        return stream  lt  lt  static cast lt typename std  underlying type lt T gt   type gt  e      int main         std  cout  lt  lt  Color  Blue  lt  lt    n       std  cout  lt  lt  Vertical  lt  lt    n       std  cout  lt  lt  ExecStatus  Running  lt  lt    n       return 0

User · Answer

Unlike an unscoped enumeration  a scoped enumeration is not implicitly convertible to its integer value   You need to explicitly convert it to an integer using a cast   std  cout  lt  lt  static cast lt std  underlying type lt A gt   type gt  a   lt  lt  std  endl    You may want to encapsulate the logic into a function template   template  lt typename Enumeration gt  auto as integer Enumeration const value      - gt  typename std  underlying type lt Enumeration gt   type       return static cast lt typename std  underlying type lt Enumeration gt   type gt  value       used as   std  cout  lt  lt  as integer a   lt  lt  std  endl

User · Answer

I m not allowed to comment yet   I would suggest the following improvements to the already great answer of James McNellis   template  lt typename Enumeration gt  constexpr auto as integer Enumeration const value      - gt  typename std  underlying type lt Enumeration gt   type       static assert std  is enum lt Enumeration gt   value   parameter is not of type enum or enum class        return static cast lt typename std  underlying type lt Enumeration gt   type gt  value       with   constexpr  allowing me to use an enum member value as compile-time array size static assert is enum  to  ensure  compile-time that the function does sth  with enumerations only  as suggested   By the way I m asking myself  Why should I ever use enum class when I would like to assign number values to my enum members   Considering the conversion effort   Perhaps I would then go back to ordinary enum as I suggested here  How to use enums as flags in C       Yet another  better  flavor of it without static assert  based on a suggestion of  TobySpeight   template  lt typename Enumeration gt  constexpr std  enable if t lt std  is enum lt Enumeration gt   value  std  underlying type t lt Enumeration gt  gt  as number const Enumeration value        return static cast lt std  underlying type t lt Enumeration gt  gt  value

[c++] How can I output the value of an enum class in C++11

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