Noah, I am aware of such constructs and possibilities. The problem arises if you have the following class: template<class T> class ptr_like_type { public: T* operator->(); //... }; In this case you can not generically know thre result type of the operator-> (because it depends on specialization of type T) and therefore you can't say if this function is defined or not. The only thing would be the use of decltype or result_of operator. That is what I ment. Best Regards, Ovanes -----Original Message----- From: Noah Roberts [mailto:roberts.noah@gmail.com] Sent: Monday, January 22, 2007 7:06 PM To: boost-users@lists.boost.org Subject: Re: [Boost-users] 'is_pointer_like' or 'is_smart_pointer' orsimilar,again Ovanes Markarian wrote:
Just a small addition. You can also specialize the template in another file, that is seen in the translation unit. But anyway you need to specialize it. I hope with a new C++09 standard (it is still pretty far away), it will be possible to inspect if a class has a specific member function (using typeof or decltype operators). Currently it is only possible with has_xxx macros.
http://unitlib.svn.sourceforge.net/viewvc/unitlib/trunk/include/unitlib/meta func/is_quantity_operand.hpp?revision=6&view=markup A meta function that is composed of two meta functions that check for the existence of a function with a particular signature in the passed in type and for the existence of a typedef of a particular name. Uses SFINAE for testing mechanics. I don't know that this would help with the OP's problem except to make sure there is a smart pointer that responds to a particular interface. If that is what is needed then something like that would help. _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users