Hi Boost, I am using boost::python to expose a C++ API into Python. The class I would like to export is: struct Subscriber { virtual ~Subscriber() = 0; static std::unique_ptr<Subscriber> create(); }; It's an interface with a factory function returning a std::unique_ptr<>. I would like to be able to export it in the following manner: BOOST_PYTHON_MODULE(pub_sub) { boost::python::class_< Subscriber , std::unique_ptr<Subscriber> , boost::noncopyable >("Subscriber") .def(boost::python::init<>, Subscriber::create) ; ; } I.e., the python wrapper would hold a std::unique_ptr<Subscriber> and use Subscriber::create() factory function as a constructor for Python Subscriber class. The above code doesn't compile with boost-1.51.0 for two reasons: * std::unique_ptr<> is not supported by boost::python. On the other hand, std::auto_ptr<> is. * The factory function can't be used used as a Python object constructor. I thought there could be a way to specialize certain boost::python traits/metaclasses to make it work with std::unique_ptr<> and factory functions as constructors, but couldn't figure it out. The workaround I managed to get working is: std::auto_ptr<Subscriber> createSubscriber() { return std::auto_ptr<Subscriber>(Subscriber::create().release()); } BOOST_PYTHON_MODULE(pub_sub) { // Expose the class. boost::python::class_< Subscriber , std::auto_ptr<Subscriber> , boost::noncopyable >("Subscriber_", boost::python::no_init) ; ; // Expose its factory function. boost::python::def("Subscriber", createSubscriber); } I.e., std::unique_ptr<> gets replaced with std::auto_ptr<>, the Python class is renamed to Subscriber_ and the factory function gets exported as Subscriber allowing for Python syntax: import pub_sub s = pub_sub.Subscriber() # however assert isinstance(s, pub_sub.Subscriber) # fails assert isinstance(s, pub_sub.Subscriber_) # succeeds So, I was wondering if there are ways to make my original Python class export syntax work? -- Maxim