Hi, I have an internally generated templated functor that naturally takes fusion sequences. since this function can also be called by the user I want to make it also "unfused" For example, the struct is struct F{ typedef result_of::make_map< p , q , double, double >::type mapped_arguments; template<class Args> struct result{ typedef double type; }; double operator()(mapped_arguments const& args) const{ return at_key<p>(args) + at_key<q>(args); } }; and can be called as F f; f(make_map(one, two)); (I need this because of internal conventions of the library) but I want it to be also called as f(one, two); // somehow (I need this because the user can call it) I tried this struct F : unfused{ typedef result_of::make_map< p , q , double, double >::type mapped_arguments; template<class Args> struct result{ typedef double type; }; double operator()(mapped_arguments const& args) const{ return at_key<p>(args) + at_key<q>(args); } }; but I get: usr/include/boost/utility/result_of.hpp:80: error: invalid use of incomplete type ‘const struct F’ I then tried this struct F_impl{ typedef result_of::make_map< p , q , double, double >::type mapped_arguments; template<class Args> struct result{ typedef double type; }; double operator()(mapped_arguments const& args) const{ return at_key<p>(args) + at_key<q>(args); } }; struct F : F_impl, unfused{ F() : unfused((F_impl const&)*this){} F(F_impl const& f) : F_impl(f), unfused((F_impl const&)*this){} template<class Args> struct result{ typedef typename F_impl::result<Args>::type type; }; using F_impl::operator(); }; which seems to work but looks and feels awkward. It works in the sense that F f; can be called clog << f(make_map(1.,2.)) << endl; clog << fu(uno, dos) << endl; Is there any simpler way to get this. Thank you, Alfredo