I think you can… You’ll have to add operators for bitmasky things. I didn’t do it here but you could easily overload any relational operator. /** * */ // NOTE: I changed to a more descriptive and consistent name // This needs to be a real bitmask type. enum class file_permissions : int { no_perms … Read more
The no-cast solution is to use switch. However, you can generate a pseudo-switch using templates. The principle is to recursively process all values of the enum using a template list (or a parameter pack). So, here are 3 methods I found. Test enum: enum class Fruit { apple, banana, orange, pineapple, lemon }; The vanilla … Read more
A type defined with enum class or enum struct is not a a class but a scoped enumeration and can not have a default constructor defined. The C++11 standard defines that your PinID pid = PinID(); statement will give a zero-initialization. Where PinID was defined as a enum class. It also allows enum types in … Read more
Check the type you are deriving the enum class from exists. In this case, there was no typedef specified for int8.
No, it’s not. Actually, an enum class is no class at all. The class keyword is only used because suddenly changing the unscoped enum to a scoped enum would have mean reworking all enums codes. So the committee decided that to distinguish between new-style and old-style enums, the new ones would be tagged with class, … Read more
I use a functor object to calculate hash of enum class: struct EnumClassHash { template <typename T> std::size_t operator()(T t) const { return static_cast<std::size_t>(t); } }; Now you can use it as 3rd template-parameter of std::unordered_map: enum class MyEnum {}; std::unordered_map<MyEnum, int, EnumClassHash> myMap; So you don’t need to provide a specialization of std::hash, the … Read more
Not directly, but you could use the following trick: enum class Example { A, B, C, D, E, Count }; Then the cardinality is available as static_cast<int>(Example::Count). Of course, this only works nicely if you let values of the enum be automatically assigned, starting from 0. If that’s not the case, you can manually assign … Read more
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 … Read more