C ++-将类型映射到枚举

Question

Is is possible to make compilation-time Type -> Enum Series mapping?

Illustrating with an example:

Let's say, I have some Type and a enumerated value:

typedef int Type;

enum Enumerated { Enum1, Enum2, Enum3, Enum4 };

and now I somehow state the following: "let's associate Enum1 and Enum4 with type Type (don't know how to implement this yet).

---

Now I want to be able to check the following (better to be done using mpl in compilation time):

If some arbitrary type and enum are actually being mapped to each other:

template <typename ArbitraryType, Enumerated E>
struct check_at_compile_time {
   // Somehow tricky evaluate this
   static const bool value;
};

so that the results are the following:

check_at_compile_time<Type, Enum1>::value evaluates to TRUE

check_at_compile_time<Type, Enum2>::value evaluates to FALSE

check_at_compile_time<Type, Enum4>::value evaluates to TRUE

check_at_compile_time<int, Enum3>::value evaluates to FALSE

If someone knows a nice way to implement this, please help me. Maybe something using boost::mpl , I'm not sure.

Thanks.

Answer 1

You can do it this way even without boost.mpl:

template< Enumerated Value > struct Enumerated2Type { typedef void type; enum { value = false }; };
#define DEFINE_ENUMERATED_TYPE(TYPE, ENUM) template<> struct Enumerated2Type<ENUM> { typedef TYPE type; enum { value = true }; }
DEFINE_ENUMERATED_TYPE(int, Enum1);
DEFINE_ENUMERATED_TYPE(bool, Enum2);
DEFINE_ENUMERATED_TYPE(double, Enum3);
DEFINE_ENUMERATED_TYPE(std::string, Enum4);

And you check you can do this way:

template <typename ArbitraryType, Enumerated E>
struct check_at_compile_time {
    static bool const value = Enumerated2Type< E >::value && boost::is_same< ArbitraryType, typename Enumerated2Type< E >::type >::value;
};

Untested but should work like this.