Thorsten Ottosen wrote:
Hi,
What is the reason that function_traits is not defined for function objects? Is it impossible?
Hi Thorsten, Some weeks ago I uploaded a first proposal for an extension to the Type Traits library capable of handling (all kinds of) function pointers. These could be applied to a (pointer to) the operator() member of a functor type, if it's not a member template (as Doug mentioned earlier). So it does work for for types of the STL functor templates or alike. Here is an example: The headers required to compile this, a preliminary documentation and a more complete example can be found in the yahoo files section (function_pointer_traits.zip). [code] #include <iostream> #include <typeinfo> #include <functional> #include <boost/type_traits/function_pointer_arity.hpp> #include <boost/type_traits/function_pointer_result.hpp> // This is used to feed a type traits class with the type deduced // from its function argument template < template <typename> class Trait, typename T > typename Trait<T>::type trait_func (T) { return typename Trait<T>::type (); } // Create some functor type typedef std::less<int> my_functor_type; int main() { using namespace std; using namespace boost; // Extract some of operator()'s properties // and make some (not very formatted) output cout << trait_func <function_pointer_arity> ( & my_functor_type::operator() ) << endl << typeid ( trait_func <function_pointer_result> ( & my_functor_type::operator() ) ).name() << endl; return 0; } [/code] In case you are using it, playing with it or just looking at it, any comments are welcome ! Regards, Tobias