boost python: how to call a C++ virtual function

问题

I have python embedded in a C++ application. The C++ calls python and passes it as an argument a C++ object. that object has some virtual functions and can be a base class for some derived class. How do I make boost::python understand that it's a virtual function?

consider the following:
in C++:

class Base {
public:
  virtual void func();
}

class Derived {
public:
  virtual void func();
}

BOOST_PYTHON_MODULE(module_api) {
  class_<Base>("Base")
    .def("func", &Base::func);  // ?? what should I put here?
}

int main() {
  //... initialization
  Derived derived;
  main_namespace["pyentry"](&derived);
}

in python:

def pyentry(baseref):
  baseref.func()    # here I want Derived::func() to be called

What am I doing wrong here?

回答1:

The problem here is that Boost.Python is deep-copying your derived-class object and slicing it into a base object when it converts it to Python; there's no need to tell Boost.Python about a function being virtual at all unless you need to override it in Python (and it sounds like you don't).

It's doing that copy to be safe: it's making certain that the object Python is given won't be deleted by C++ while Python still has a reference to it. And it's slicing it to Base - I think - because it doesn't know anything about Derived.

I can think of two ways to fix this:

• Provide a trivial wrapper for Derived. Boost.Python will still copy it when converting it to Python, but it won't slice it into a Base anymore when it does so.

• Register a shared_ptr conversion for Base (via register_ptr_to_python< boost::shared_ptr<Base> >()), create your Derived instance in a shared_ptr, and pass that as the argument to your Python function. Now Boost.Python knows the C++ object can't get deleted while the Python object exists, because the Python wrapper holds a shared_ptr reference to it.

来源：https://stackoverflow.com/questions/10539373/boost-python-how-to-call-a-c-virtual-function