The obvious thing is to annotate your enum:
// generic code
#include <algorithm>
template <typename T>
struct enum_traits {};
template<typename T, size_t N>
T *endof(T (&ra)[N]) {
return ra + N;
}
template<typename T, typename ValType>
T check(ValType v) {
typedef enum_traits<T> traits;
const T *first = traits::enumerators;
const T *last = endof(traits::enumerators);
if (traits::sorted) { // probably premature optimization
if (std::binary_search(first, last, v)) return T(v);
} else if (std::find(first, last, v) != last) {
return T(v);
}
throw "exception";
}
// "enhanced" definition of enum
enum e {
x = 1,
y = 4,
z = 10,
};
template<>
struct enum_traits<e> {
static const e enumerators[];
static const bool sorted = true;
};
// must appear in only one TU,
// so if the above is in a header then it will need the array size
const e enum_traits<e>::enumerators[] = {x, y, z};
// usage
int main() {
e good = check<e>(1);
e bad = check<e>(2);
}
You need the array to be kept up to date with e
, which is a nuisance if you're not the author of e
. As Sjoerd says, it can probably be automated with any decent build system.
In any case, you're up against 7.2/6:
For an enumeration where emin is the
smallest enumerator and emax is the
largest, the values of the enumeration
are the values of the underlying type
in the range bmin to bmax, where bmin
and bmax are, respectively, the
smallest and largest values of the
smallest bit-field that can store emin
and emax. It is possible to define an
enumeration that has values not
defined by any of its enumerators.
So if you aren't the author of e
, you may or may not have a guarantee that valid values of e
actually appear in its definition.