Re: implementation: specialization of member of template class

Specialization of member of template class is impossible according to C
++ standard. The following code:
template<typename T> struct enbl_if {};
template<> struct enbl_if<int> { typedef int type; };
template<typename T> struct disble_if {typedef T type; };
template<> struct disble_if<int> {};
template<typename T>
struct test
{
   template<typename T2>
   typename enbl_if<T2>::type get() const
   {
      std::cout << "enbl_if" << std::endl;
      return T2();
   }
   template<typename T2>
   typename disble_if<T2>::type get() const
   {
      std::cout << "disble_if" << std::endl;
      return T2();
   }
};
int main()
{
   int i = test<int>().get<int>();
   long l = test<int>().get<long>();
   std::string s = test<int>().get<std::string>();
   return 0;
}
prints:
enbl_if
disble_if
disble_if
and so, it works as full specialization for "int". I was able to
compile it under gcc 4.2, VC2003-2005, comeua online. Is it correct
code?