Can I overload operators on enum types in C++?

Question

For example, if I have:

typedef enum { year, month, day } field_type;

inline foo operator *(field_type t,int x)
{
   return foo(f,x);
}
inline foo operator -(field_type t)
{
   return t*-1;
}
int operator /(distance const &d,field_type v)
{
  return d.in(v);
}

Because if I do not define such operators it is actually legal to write day*3 and it would be translated into 6?

So is it legal?

At least gcc and intel compiler accept this without a warning.

Clearification:

I do not want default arithmetic operations, I want my own operations that return non-integer type.

Answer 1

Yes, operator overloading can be done on enum and class types. The way you do it is fine, but you should use + to promote the enumeration, instead of *-1 or something (the purpose ultimately is to avoid infinite recursion because -t):

inline foo operator -(field_type t) {
   return -+t;
}

This will scale well to other operations. + will promote the enumeration to an integer type that can represent its value, and then you can apply - without causing infinite recursion.

---

Notice that your operator* does only allow you to do enum_type * integer, but not the other way around. It may be worth considering the other direction too.

Also notice that it's always a bit dangerous to overload operators for operands that builtin-operators already accept (even if only by implicit conversions). Imagine that distance has a converting constructor taking int (as in distance(int)), then given your operator/ the following is ambiguous

// ambiguous: operator/(int, int) (built-in) or
//            operator/(distance const&, field_type) ?
31 / month;

For this, maybe it's better to make field_type a real class with the appropriate operators, so that you can exclude any of such implicit conversions from begin on. Another good solution is provided by C++0x's enum class, which provides strong enumerations.