C++ recursive type traits

Question

I'm trying to implement a template class (named Get<> here) that, given a structure H, the type Get<H>::type is the H itself if the qualified-id H::der doesn't exist, and is Get<H::der>::type otherwise. I can't understand what's wrong with the following code:

#include <iostream>
#include <typeinfo>
using namespace std;

template<class U, class V = void>
struct Get
{
  static const char id = 'A';
  typedef U type;
};

template<class U>
struct Get<U,typename U::der>
{
  static const char id = 'B';
  typedef typename Get<typename U::der>::type type;
};

struct H1
{ };
struct H2
{ typedef double der; };
struct H3
{ typedef void der; };
struct H4
{ typedef H2 der; };

void print(char id, const char* name)
{
  cout << id << ", " << name << endl;
}
int main(int , char *[])
{
  print(Get<H1>::id, typeid(Get<H1>::type).name()); // prints "A, 2H1", OK
  print(Get<H2>::id, typeid(Get<H2>::type).name()); // prints "A, 2H2", why?
  print(Get<H3>::id, typeid(Get<H3>::type).name()); // prints "B, v"  , OK
  print(Get<H4>::id, typeid(Get<H4>::type).name()); // prints "A, 2H4", why?
}

I'd like some help to make this code behave as expected. More specifically, I wish that Get< H2 >::type was equal to double, and the same for Get< H4 >::type.

Answer 1

The template Get<> has a default template parameter -- this is very dangerous. Depending on whether V is equal int, void or double you get different results. This is what happens:

Get<H2>::type is Get<H2, void> on the first place (with id='A'). Now comes the check, whether there is an specialization of it. Your B is Get<U,typename U::der> which becomes Get<U, double>. But this doesn't match whith Get<H2, void>, so A is chosen. Things get interesting with Get<H2>::type. Then the variant B is also Get<U, void> and provides a better match. However, the approach can't work for all types.

This is how I would have implemented Get:

template<class U>
class Get
{
  template <typename T, typename = typename T::der>
  static typename Get<typename T::der>::type test(int);
  template <typename T>
  static T test(...);
public:
  typedef decltype(test<U>(0)) type;
};