AMDG Simonson, Lucanus J wrote:
//would not compile //template <int orient, typename T> //typename point_traits<T>::coordinate_type get(const T& point) { // return point_traits<T>::get<orient>(point); //}
//will compile when written this way template <int orient, typename T> typename point_traits<T>::coordinate_type get(const T& point) { point_traits<T> traits_instance; return traits_instance.get<orient>(point); }
int main() { point_data pt; //this line compiles int value0 = point_traits<point_data>::get<0>(pt); //this line now compiles int value = get<0>(pt); return 0; }
But it still bugs me that I don't know why the compiler thought that point_traits<T>::get was a function pointer and tried to apply the operator< with orient, but does the right thing when I instantiate the traits struct and use the . operator to access the same function. The . operator could also produce a function pointer.
You need to use point_traits<T>::template get<0> inside a template. The compiler should not accept the "." either, BTW. In Christ, Steven Watanabe