更新时间:2023-01-13 14:20:57
是否可以指定一个自定义函数来评估布尔上下文中的类枚举?
Is it possible to specify a custom function to evaluate a class enum in a boolean context?
是的,但不是自动的。手动调用一个函数仍然比提供的其他替代方法更优雅。
Yes, but not automatically. Manually calling a function is still more elegant than the other alternatives presented.
只需选择一个不错的函数名称,例如 any
,并实施它。重载解析将确保您的函数与所有其他函数配合良好。
Simply pick a nice function name, such as any
, and implement it. Overload resolution will make sure your function plays well with all others.
bool any( E arg )
{ return arg != E::none; }
...
if ( any( flags ) ) {
...
对我来说看起来不错。
更新::如果您希望将其应用于多种枚举类型,则可以将其用作模板:
Update: if you want this to apply to several enumeration types, it can be templated:
template< typename enum_type > // Declare traits type
struct enum_traits {}; // Don't need to declare all possible traits
template<>
struct enum_traits< E > { // Specify traits for "E"
static constexpr bool has_any = true; // Only need to specify true traits
};
template< typename enum_type > // SFINAE makes function contingent on trait
typename std::enable_if< enum_traits< enum_type >::has_any,
bool >::type
any( enum_type e )
{ return e != enum_type::none; }
我在其他事情上一直使用这种机制,从未遇到任何副作用或问题:v)。
I've been using this sort of mechanism for other things and never encountered any side effects or issues :v) .
您可以跳过特征并将SFINAE条件设置为 enum_type :: none == enum_type :: none
,仅检查 none
和相等运算符的存在,但这将不太明确和安全。
You could skip the trait and set the SFINAE condition to something like enum_type::none == enum_type::none
, to merely check for the presence of none
and the equality operator, but that would be less explicit and safe.