如何在C ++ 11中输出枚举类的值 [英] How can I output the value of an enum class in C++11

问题描述

如何在C ++ 11中输出枚举类的值？在C ++ 03中，它是这样：

  #include< iostream> 
 
 using namespace std; 
 
枚举A {
 a = 1，
 b = 69，
 c = 666 
}; 
 
 int main（）{
 A a = A :: c; 
 cout<< a<< endl; 
} 
  

在c ++ 0x中此代码不编译

  #include< iostream> 
 
 using namespace std; 
 
枚举类A {
 a = 1，
 b = 69，
 c = 666 
}; 
 
 int main（）{
 A a = A :: c; 
 cout<< a<< endl; 
} 
 
 
 prog.cpp：13：11：错误：无法将'std :: ostream'lvalue绑定到'std :: basic_ostream< char>& 
 /usr/lib/gcc/i686-pc-linux-gnu/4.5.1/../../../../include/c++/4.5.1/ostream:579:5：错误：初始化'std :: basic_ostream< _CharT，_Traits>& std :: operator<<<<<<（std :: basic_ostream< _CharT，_Traits&&&&& amp; const _Tp&在Ideone.com中编译的 
 
        / div> 
与无范围枚举不同，范围枚举不会隐式地转换为其整数值。您需要使用强制转型将其转换为整数：
  std :: cout< < static_cast< std :: underlying_type< A> :: type>（a）<< std :: endl; 
  
您可能想将逻辑封装到函数模板中：
  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）; 
} 
  
用作：
  std :: cout<< as_integer（a）<< std :: endl; 
  
 
How can I output the value of an enum class in C++11? In C++03 it's like this:
#include <iostream>

using namespace std;

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

int main () {
  A a = A::c;
  cout << a << endl;
}
in c++0x this code doesn't compile
#include <iostream>

using namespace std;

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

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


prog.cpp:13:11: error: cannot bind 'std::ostream' lvalue to 'std::basic_ostream<char>&&'
/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<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char, _Traits = std::char_traits<char>, _Tp = A]'
compiled at Ideone.com
解决方案
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;


                        
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