如何将 C++11 枚举类用于标志

问题描述

假设我有这样一个类:

enum class Flags : char
{
    FLAG_1 = 1;
    FLAG_2 = 2;
    FLAG_3 = 4;
    FLAG_4 = 8;
};

现在我可以有一个具有类型标志的变量并分配一个值 7 例如吗?我可以这样做吗:

Now can I have a variable that has type flags and assign a value 7 for example? Can I do this:

Flags f = Flags::FLAG_1 | Flags::FLAG_2 | Flags::FLAG_3;

或

Flags f = 7;

出现这个问题是因为在枚举中我没有为 7 定义值.

This question arises because in the enum I have not defined value for 7.

推荐答案

您需要编写自己的重载 operator|(大概还有 operator& 等).

You need to write your own overloaded operator| (and presumably operator& etc.).

Flags operator|(Flags lhs, Flags rhs)
{
    return static_cast<Flags>(static_cast<char>(lhs) | static_cast<char>(rhs));
}

只要值在枚举值的范围内(否则为 UB；[expr.static.cast]/p10)，整数到枚举类型(范围或非范围)的转换是明确定义的.对于具有固定基础类型的枚举(这包括所有作用域枚举；[dcl.enum]/p5)，枚举值的范围与基础类型([dcl.enum]/p8)的值范围相同.如果底层类型不固定，规则会更棘手 - 所以不要这样做:)

Conversion of an integer to an enumeration type (scoped or not) is well-defined as long as the value is within the range of enumeration values (and UB otherwise; [expr.static.cast]/p10). For enums with fixed underlying types (this includes all scoped enums; [dcl.enum]/p5), the range of enumeration values is the same as the range of values of the underlying type ([dcl.enum]/p8). The rules are trickier if the underlying type is not fixed - so don't do it :)

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