Overload operator >> for all enum types

c++, operator-overloading

Solution

Make the operator a template and enable it only if the template parameter is an enum (using `enable_if`):

#include <type_traits>

template<typename T>
typename std::enable_if<std::is_enum<T>::value, std::istream&>::type
operator >>(std::istream &is, T& enumVar)
{
  std::cout << "enum\n"; // just to see it is this one that gets used
  int intVal;
  is >> intVal;
  enumVar = static_cast<T>(intVal); // note the explicit cast to make it legal
  return is;
}

If you don't have C++11 avaliable, you can use boost's `type_traits` library.

Seeing your comment that you don't know exactly how `enable_if` works, here is a link with a detailed explanation.

In short, `enable_if` templates come in pairs - one with a member typedef of second template parameter to `type` when the condition (first template parameter) is true, and one without that typedef when condition is false. Template instantiations with false condition are invalid, because the member typedef does not exist. Such instantiation is then (instead of being a hard compiler error) discarded from overload set for later overload resolution. Read also about SFINAE.

EDIT: boost's `enable_if` works slightly differently than the standard one. Use either:

boost::enable_if < boost::is_enum<T>, std::istream& >::type
                // ^^^^^^^^^^^^^^^^^
                // expects a type

or

boost::enable_if_c < boost::is_enum<T>::value, std::istream& >::type
                  // ^^^^^^^^^^^^^^^^^^^^^^^^
                  // expects a value

Problem

I have enumeration type , for which I overload the `operator>>`. ``` std::istream& operator >>(std::istream &is,MyEnum& enumVar) { int intVal; is>>intVal; enumVar = intVal; return is; } ``` How to avoid writing this code for all future enum types, i.e how to write the function so it is applicable to all enum types?

Original source