枚举类是否可以转换为底层类型？ [英] Can an enum class be converted to the underlying type?

问题描述

有没有办法将枚举类字段转换为底层类型？我认为这是自动的，但显然不是。

Is there a way to convert an enum class field to the underlying type? I thought this would be automatic, but apparently not.

enum class my_fields : unsigned { field = 1 };

unsigned a = my_fields::field;

该作业被GCC拒绝。 错误：无法在赋值中将'my_fields'转换为'unsigned int'。

That assignment is being rejected by GCC. error: cannot convert 'my_fields' to 'unsigned int' in assignment.

推荐答案

我认为您可以使用 std :: underlying_type 了解底层类型，然后使用cast ：

I think you can use std::underlying_type to know the underlying type, and then use cast:

#include <type_traits> //for std::underlying_type

typedef std::underlying_type<my_fields>::type utype;

utype a = static_cast<utype>(my_fields::field);

这样，您不必或者你不必在枚举类的定义中提及它 enum class my_fields：int {....}

With this, you don't have to assume the underlying type, or you don't have to mention it in the definition of the enum class like enum class my_fields : int { .... } or so.

您甚至可以编写一个泛型转换函数，该函数应该能够转换任何 > 枚举类到其底层的整数类型：

You can even write a generic convert function that should be able to convert any enum class to its underlying integral type:

template<typename E>
constexpr auto to_integral(E e) -> typename std::underlying_type<E>::type 
{
   return static_cast<typename std::underlying_type<E>::type>(e);
}

然后使用它：

auto value = to_integral(my_fields::field);

auto redValue = to_integral(Color::Red);//where Color is an enum class!

由于函数声明为 constexpr ，您可以在需要常量表达式的地方使用它：

And since the function is declared to be constexpr, you can use it where constant expression is required:

int a[to_integral(my_fields::field)]; //declaring an array

std::array<int, to_integral(my_fields::field)> b; //better!

希望有帮助。

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