Hi,

I have code similar to the following snippet:

#include #include #include #include namespace fusion=boost::fusion;

template<typename T> class A { public: struct B {};

B make_b() const { return B(); } };

template<typename T> A<T> make_a( T const& ) { return A<T>(); }

struct make_b { template<typename Sig> struct result;

template <typename DS> struct result { typedef typename DS::B type; };

template <typename T> typename T::B operator()(T const& t) const { return t.make_b(); } };

int main() { namespace fusion=boost::fusion;

fusion::as_list( fusion::transform( fusion::make_list( make_a( 1. ), make_a( 1 ) ), make_b() ) );

return 0; }

it fails to compile with an error like this:

... rather long context here ..\..\../boost/utility/result_of.hpp:83:1: error: invalid use of incomplete type ‘struct make_b::result’ test.cpp:27:35: error: declaration of ‘struct make_b::result’

What am I doing wrong?

You need to add a specialization result_of, in addition to result_of: template <typename DS> struct result {     typedef typename DS::B type; }; You need this because your operator() can be called with either a argument of type T, or an argument of type T const&. You don't need two versions of your operator(), because T is implicitly convertible to T const&, but the result<> template is instantiated with the exact argument type without any conversions being performed. This is subtle issue, I explain it in more detail here [1]. Thank goodness in C++11 we will be able to avoid it by using decltype. Regards, Nate [1] http://lists.boost.org/boost-users/2011/07/69684.php