[C++-sig] Can't use make_function to wrap base class method? (original) (raw)

David Abrahams dave at boost-consulting.com
Tue Dec 13 23:17:09 CET 2005

Alex Mohr <amohr at pixar.com> writes:

Seems I can't wrap a base class method of a derived class if I use makefunction. Is this expected? Here's an example.

#include <boost/python.hpp> using namespace boost::python; class Base { public: int method() { return 1; } }; class Derived : public Base {}; BOOSTPYTHONMODULE(Foo) { class("Derived", "", init<>()) .def("method", makefunction(&Derived::method)) .def("method2", &Derived::method) ; } >>> from Foo import * >>> Derived().method() Traceback (most recent call last): File "", line 1, in ? Boost.Python.ArgumentError: Python argument types in Derived.method(Derived) did not match C++ signature: method(Base {lvalue}) >>> Derived().method2() 1 Seems like this should work. Wrapping a method in makefunction seems to make it not. This becomes cumbersome since the only way I know of to supply returnvaluepolicies in .addproperty is to use makefunction.

You either need to wrap Base or you need to 

    .def(
         "method", 
          make_function(
            (int (Derived::*)()) &Derived::method
          )
     )

class_ has the information about the derived class (it's a template argument), so it can do the cast automagically. However, make_function is just a function, so it can't do anything fancy with Derived. The function pointer's type is

int (Base::*)()

even when you get it via &Derived::method.

HTH,

-- Dave Abrahams Boost Consulting www.boost-consulting.com

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