C++11中如何输出enum class的值

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;

示例：

头文件：enum.h

#include <iostream>

namespace myenum {
  // 一般的枚举
  enum PieceType {
    PieceTypeKing, PieceTypeQueen, PieceTypeRook, PieceTypePawn
  };

  // 设置了值得枚举
  enum PieceTypeAnother {
    PieceTypeKingA=1, PieceTypeQueenA, PieceTypeRookA=10, PieceTypePawnA
  };
  // 强类型的枚举
  enum class MyEnum {
    EnumValue1,
      EnumValue2 = 10,
      EnumValue3
      };
  // 改变了类型的强类型枚举
  enum class MyEnumLong : unsigned long {
    EnumValueLong1,
      EnumValueLong2 = 10,
      EnumValueLong3
      };

  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);
  }
}

main函数：main.cpp

#include "enum.h"
int main() {

  std::cout << "test enum" << std::endl;
  std::cout << "一般的枚举" << myenum::PieceType::PieceTypeKing << std::endl;
  std::cout << "设置了值的枚举" << myenum::PieceTypeAnother::PieceTypeQueenA << std::endl;
  std::cout << "强类型的枚举" << myenum::MyEnum::EnumValue3 << std::endl;
  std::cout << "改变了类型的强类型枚举" << myenum::MyEnumLong::EnumValueLong1 << std::endl;

  std::cout << "end of this program" << std::endl;
}

重写了之后，就不用再对强类型的进行强转了，呵呵

编译的时候要加上c++11的选项：

g++ main.cpp  -std=c++11