"Maxim Yegorushkin" <e-maxim@yandex.ru> writes:
Is there a way to make a boost::lambda functor adaptable, that is to make it define result_type? There is a ret<> function template that lets one specify the result type, but the returned functor still does not have result_type typedef.
I've encountered the problem trying to use boost::transform_iterator with a boost::lambda functor:
std::vector<int> intVect; boost::make_transform_iterator(intVect.begin(), boost::lambda::_1 / 2);
transform_iterator<> does not compile comlaining the functor does not have result_type.
As a workaround I had to write something like:
template<class R, class T> struct result_type_wrapper : T { typedef R result_type; result_type_wrapper(T const& t) : T(t) {} };
template<class R, class T> inline result_type_wrapper<R, T> result_type(T const& t) { return t; }
boost::make_transform_iterator(intVect.begin(), result_type<int>(boost::lambda::_1 / 2));
Am I missing something? Is there a way to avoid using result_type_wrapper<>?
You could use boost::make_adaptable from boost/bind/make_adaptable.hpp, but there really ought to be a version of make_transform_iterator that does that for you. -- Dave Abrahams Boost Consulting http://www.boost-consulting.com