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

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 << static_cast<std::underlying_type<A>::type>(a) << std::endl;

You may want to encapsulate the logic into a function template:

template <typename Enumeration>
auto as_integer(Enumeration const value)
    -> typename std::underlying_type<Enumeration>::type
{
    return static_cast<typename std::underlying_type<Enumeration>::type>(value);
}

used as:

std::cout << as_integer(a) << std::endl;

#include <iostream>
#include <type_traits>

using namespace std;

enum class A {
  a = 1,
  b = 69,
  c= 666
};

std::ostream& operator << (std::ostream& os, const A& obj)
{
   os << static_cast<std::underlying_type<A>::type>(obj);
   return os;
}

int main () {
  A a = A::c;
  cout << a << endl;
}

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<< function which will work for any scoped enum. The solution employs SFINAE via std::enable_if and is as follows.

#include <iostream>
#include <type_traits>

// Scoped enum
enum class Color
{
    Red,
    Green,
    Blue
};

// Unscoped enum
enum Orientation
{
    Horizontal,
    Vertical
};

// Another scoped enum
enum class ExecStatus
{
    Idle,
    Started,
    Running
};

template<typename T>
std::ostream& operator<<(typename std::enable_if<std::is_enum<T>::value, std::ostream>::type& stream, const T& e)
{
    return stream << static_cast<typename std::underlying_type<T>::type>(e);
}

int main()
{
    std::cout << Color::Blue << "\n";
    std::cout << Vertical << "\n";
    std::cout << ExecStatus::Running << "\n";
    return 0;
}