SFINAE compiler troubles

Question

The following code of mine should detect whether T has begin and end methods:

template 
struct is_contai         


        

Answer 1

So, here's how I go about debugging these things.

First, comment out the negative alternative so you get an error instead of just a mismatch. Next, try to instantiate the type you're putting in the function with one of the items that do not work.

At this step, I was able to instantiate your sfinae object but it still wasn't working. "This lets me know it IS a VS bug, so the question then is how to fix it." -- OBS

VS seems to have troubles with SFINAE when done the way you are. Of course it does! It works better when you wrap up your sfinae object. I did that like so:

template 
struct sfinae 
{
  // typedef typename U::const_iterator it_t; - fails to compile with non-cont types.  Not sfinae
  template < typename U, typename IT, IT (U::*)() const, IT (U::*)() const >
  struct type_ {};

  typedef type_ type;
};

Still wasn't working, but at least I got a useful error message:

error C2440: 'specialization' : cannot convert from 'overloaded-function' to 'std::_Tree_const_iterator<_Mytree> (__thiscall std::set<_Kty>::* )(void) const'

This lets me know that &U::end is not sufficient for VS (ANY compiler) to be able to tell which end() I want. A static_cast fixes that:

  typedef type_(&U::begin),static_cast(&U::end)> type;

Put it all back together and run your test program on it...success with VS2010. You might find that a static_cast is actually all you need, but I left that to you to find out.

I suppose the real question now is, which compiler is right? My bet is on the one that was consistent: g++. Point to the wise: NEVER assume what I did back then.

Edit: Jeesh... You are wrong!

Corrected version:

template 
struct is_container
{
    template  
    struct sfinae 
    {
      //typedef typename U::const_iterator it_t;
      template < typename U, typename IT, IT (U::*)() const, IT (U::*)() const >
      struct type_ {};

      typedef type_(&U::begin),static_cast(&U::end)> type;
    };

    template  static char test(typename sfinae::type*);
    template  static long test(...);

    enum { value = (1 == sizeof test(0)) };
};



#include 
#include 
#include 
#include 
#include 

-- The debugging above is sensible, but the assumption about the compiler was wrong headed. G++ should have failed for the reason I emphasized above.