Re: [boost] [function types] cv-qualification on function types

From: Tobias Schwinger (tschwinger_at_[hidden])
Date: 2005-07-08 15:44:19

Peter Dimov wrote:
> Tobias Schwinger wrote:
> 
> 
>>>It is not too clear whether it applies to partial template
>>>specialization, since we don't really "form" a type - we just
>>>consider a possibitity: 
>>>
>>>    typedef int my_const_function() const;
>>>
>>>    template<typename T> struct remove_const
>>>       { typedef T  type; };
>>>
>>>    template<typename T> struct remove_const< T const >
>>>       { typedef T type; };
>>>
>>>    // somwhere else
>>>    remove_const<my_const_function>::type // ill-formed?
> 
> 
> Well formed, returns my_const_function. 

Wasn't anywhere near obvious to me before reading #295, though...

> The specialization doesn't match.
> 

Where do I find it in the standard and/or what should happen in the following two 
cases?

   template<typename T>             struct remove_member_pointer
   { typedef T type; };
   template<typename T, typename C> struct remove_member_pointer< T C::* >
   { typedef T type; };

   remove_member_pointer< my_const_function >::type // <-- case 1

   template<typename T, typename C> struct add_member_pointer
   { typedef T C::*type; };

   class X;

   add_member_pointer< my_const_function, X >::type // <-- case 2

Thanks,

Tobias