Hi being an enthusiastic user of boost python, I often missed the possibility to use function objects, see http://www.boost.org/doc/libs/1_55_0/libs/python/todo.html#wrapping-function.... Now I succeeded in a workaround to do so. The code below allows to use function objects, lambdas and std::function<> with arbitrary signature. It should work for c++11 compliant compilers (declspec must be implemented), I tested it with Visual Studio 2012 and gcc4.8.1. For the latter the code must be included prior to #include <boost/python.hpp>. Up to now it didn't break my existing code and the definitions with lambdas work as expected. Regards Martin ------------------------------------------------------------------------------------------ #ifndef BOOST_NO_CXX11_DECLTYPE # include <boost/mpl/erase.hpp> # include <boost/mpl/next.hpp> # include <boost/utility.hpp> # include <boost/function_types/is_callable_builtin.hpp> # include <boost/mpl/size.hpp> # include <boost/mpl/greater.hpp> # include <boost/mpl/size_t.hpp> namespace boost { namespace python { namespace detail { template<typename T, typename Enable = void> struct fun_sign { typedef T type; }; template<typename T> struct fun_sign< T, typename enable_if< mpl::greater< mpl::size<T>, mpl::size_t<1> >
::type
{ typedef typename mpl::erase< T, typename mpl::next< typename mpl::begin< T >::type >::type
::type type; };
template<typename T> typename disable_if<typename function_types::is_callable_builtin<T>, typename fun_sign<typename function_types::components<decltype(&T::operator())>::types>::type>::type get_signature(T, void* = 0) { return typename fun_sign<typename function_types::components<decltype(&T::operator())>::types>::type(); } }}} //#endif