I see from the source code that the transform range is calling make_transform_iterator which takes two arguments, and that "If Reference is use_default then the reference member of transform_iterator is result_of<UnaryFunction(iterator_traits<Iterator>::reference)>::type." The documentation for result_of explains the need, and shows how to use a result template instead of a result_type simple type in this case. My code follow from the example: struct ASCII_lower { // tell details to result_of template, since VS8 doesn't support decltype keyword. template <typename> struct result; template<typename F, typename T> struct result<F(T)> { typedef T type; }; template< typename CharT > CharT operator() (const CharT& ch) const { if (ch >= 'A' && ch <= 'Z') return ch+0x20; return ch; // unchanged } }; But, I still get an error that result_type is not found! B:\boost/iterator/transform_iterator.hpp(42) : error C2039: 'result_type' : is not a member of 'vst::internal::ASCII_lower' B:\boost/mpl/eval_if.hpp(41) : see reference to class template instantiation 'boost::detail::function_object_result<UnaryFunc>' being compiled with [ UnaryFunc=vst::internal::ASCII_lower ] B:\boost/iterator/iterator_adaptor.hpp(172) : see reference to class template instantiation 'boost::mpl::eval_if<C,F1,F2>' being compiled with [ C=boost::is_same<boost::use_default,boost::use_default>, F1=boost::detail::function_object_result<vst::internal::ASCII_lower>, F2=boost::mpl::identity<boost::use_default> ] It appears to be using function_object_result, not result_of as documented. And that is: template <class UnaryFunc> struct function_object_result { typedef typename UnaryFunc::result_type type; }; which simply turns result_type into type to make eval_if happy, and doesn't do anything fancier. So, it doesn't work as documented, and neither provides another way of passing the parameter in (exposing more arguments from the underlying transform_iterator). Is that a bug, or what? —John