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

 1 #include <iostream>
 2 
 3 namespace myenum {
 4   // 一般的枚举
 5   enum PieceType {
 6     PieceTypeKing, PieceTypeQueen, PieceTypeRook, PieceTypePawn
 7   };
 8 
 9   // 设置了值得枚举
10   enum PieceTypeAnother {
11     PieceTypeKingA=1, PieceTypeQueenA, PieceTypeRookA=10, PieceTypePawnA
12   };
13   // 强类型的枚举
14   enum class MyEnum {
15     EnumValue1,
16       EnumValue2 = 10,
17       EnumValue3
18       };
19   // 改变了类型的强类型枚举
20   enum class MyEnumLong : unsigned long {
21     EnumValueLong1,
22       EnumValueLong2 = 10,
23       EnumValueLong3
24       };
25 
26   template<typename T> std::ostream& operator<<(typename std::enable_if<std::is_enum<T>::value, std::ostream>::type& stream, const T& e) {
27     return stream << static_cast<typename std::underlying_type<T>::type>(e);
28   }
29 }

main函数：main.cpp

 1 #include "enum.h"
 2 int main() {
 3 
 4   std::cout << "test enum" << std::endl;
 5   std::cout << "一般的枚举" << myenum::PieceType::PieceTypeKing << std::endl;
 6   std::cout << "设置了值的枚举" << myenum::PieceTypeAnother::PieceTypeQueenA << std::endl;
 7   std::cout << "强类型的枚举" << myenum::MyEnum::EnumValue3 << std::endl;
 8   std::cout << "改变了类型的强类型枚举" << myenum::MyEnumLong::EnumValueLong1 << std::endl;
 9 
10   std::cout << "end of this program" << std::endl;
11 }

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：

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