Re: work round for std::distance's long arm.

Without the nested typedefs the following fails to compile
under vc7 and gcc 3.2.

/* distance.cpp */
#include <iterator>
#include <list>

namespace user
{
     template<typename Iter,typename Dist> struct Cursor
     {
#if 0
         typedef void iterator_category;
         typedef void difference_type;
         typedef void value_type;
         typedef void reference;
         typedef void pointer;
#endif
     };

     template<typename Iter,typename Dist>
         Dist distance(
         Cursor<Iter,Dist> const&,
         Cursor<Iter,Dist> const&)
     {
         return 0;
     }
}

int main()
{
     user::Cursor<std::list<int>::iterator,int> cur;

     distance(cur,cur); // XXX

     return 0;
}
/* distance.cpp end */

Because Cursor's instantiated with an iterator from std,
I guess what happens (at XXX) is the compiler looks in
namespace std and sees something like...

template<class Iter>
     typename iterator_traits<Iter>::difference_type
     distance<Iter,Iter) {...}

...and so needs to instantiate iterator_traits<Cursor> to get
the return type. But the default iterator_traits requires its
parameter to have nested typedefs for iterator_category etc.
Have I got that right?

This being the case, what's the best work round?

I know I can use void typedefs as above but that adds noise.

Is there a better way, other than not using the name distance?

Thanks, Louis.