conversion of boost::shared_ptr in boost::python function call

Question

Consider the following example:

#include "Python.h"
#include <boost/python.hpp>
#include <boost/shared_ptr.hpp>

class A {};

class B : public A{};

void foo(boost::shared_ptr<A>& aptr) { }

BOOST_PYTHON_MODULE(mypy)
{
  using namespace boost::python;   
  class_<A, boost::shared_ptr<A> >("A", init<>());
  class_<B, boost::shared_ptr<B>, bases<A> >("B", init<>());
  def("foo", foo);
}

if I call the python code

import mypy
b = mypy.B()
mypy.foo(b)

I get

ArgumentError: Python argument types in
    mypy.foo(B)
did not match C++ signature:
    foo(boost::shared_ptr<A> {lvalue})

I have googled around quite alot, but I can't find a good explanation / fix / workaround for this. Any help is quite welcome!

What happens if you have foo take a boost::python::object argument, and then extract the shared_ptr from that? — Useless
– Useless, Commented May 18, 2012 at 17:55
Sure, that would work, but I was hoping there would be a way without writing any additional wrappers for foo (there are quite a lot of those that I would need ) — Hans
– Hans, Commented May 18, 2012 at 23:27

jbosch · Accepted Answer · 2012-05-19 01:03:14Z

4

The problem is that you're asking for a non-const reference to a shared_ptr<A>, and your b instance in Python simply doesn't contain one; it contains a shared_ptr<B>. While shared_ptr<B> can be implicitly converted to shared_ptr<A>, shared_ptr<B>& cannot be implicitly converted to shared_ptr<A>&.

If you can modify foo to take a shared_ptr<A>, or shared_ptr<A> const &, that will solve your problem.

If not, you'll need to also wrap a version that accepts shared_ptr<B>&.

answered May 19, 2012 at 1:03

jbosch

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