Exposing C++ interface in boost python

Question

Sample code to illustrate:

struct Base
{
  virtual int foo() = 0;
};

struct Derived : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

Base* get         


        

Answer 1

Abstract C++ classes cannot be exposed in this manner to Boost.Python. The Boost.Python tutorial gives examples as to how to expose pure virtual functions. In short, when decorating methods with boost::python::pure_virtual, a wrapper type needs to be created to allow C++ to polymorphic resolve the virtual function, and the virtual function implementation will delegate resolving the function polymorphically in the Python object's hierarchy.

struct BaseWrap : Base, boost::python::wrapper<Base>
{
  int foo()
  {
    return this->get_override("foo")();
  }
};

...

boost::python::class_<BaseWrap>("Base", ...)
  .def("foo", boost::python::pure_virtual(&Base::foo))
  ;

For details, when a type is exposed via boost::python::class_, HeldType defaults to the type being exposed, and the HeldType is constructed within a Python object. The class_ documentation states:

Template Parameter:

• T: The class being wrapped
• HeldType: Specifies the type that is actually embedded in a Python object wrapping a T instance [...]. Defaults to T.

Hence, the boost::python::class_<Base> will fail, because T = Base and HeldType = Base, and Boost.Python will try to instantiate an object of HeldType into a Python object that represents an instance of Base. This instantiation will fail as Base is an abstract class.

---

Here is a complete example showing the use of a BaseWrap class.

#include <boost/python.hpp>

struct Base
{
  virtual int foo() = 0;
  virtual ~Base() {}
};

struct Derived : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

Base* get_base()
{
  return new Derived;
}

namespace python = boost::python;

/// @brief Wrapper that will provide a non-abstract type for Base.
struct BaseWrap : Base, python::wrapper<Base>
{
  BaseWrap() {}

  BaseWrap(const Base& rhs)
    : Base(rhs)
  {}

  int foo()
  {
    return this->get_override("foo")();
  }
};

BOOST_PYTHON_MODULE(example)
{
  python::class_<BaseWrap>("Base")
    .def("foo", python::pure_virtual(&Base::foo));
    ;

  python::def("get_base", &get_base,
              python::return_value_policy<python::manage_new_object>());
}

and its usage:

>>> import example
>>> class Spam(example.Base):
...     pass
... 
>>> Spam().foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
RuntimeError: Pure virtual function called
>>> class Egg(example.Base):
...     def foo(self):
...         return 100
... 
>>> e = Egg()
>>> e.foo()
100
>>> d = example.get_base()
>>> d.foo()
42

---

It is possible to expose an abstract class in Boost.Python by exposing it with no default initializer (boost::python::no_init) and non-copyable (boost::noncopyable). The lack of an initializer prevents Python types from deriving from it effectively preventing overriding. Additionally, the implementation detail that Base::foo() is implemented within C++ by Derived is inconsequential. If Python should not know about a foo() method at all, then omit exposing it via def().

#include <boost/python.hpp>

struct Base
{
  virtual int foo() = 0;
  virtual ~Base() {}
};

struct Derived
  : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

struct OtherDerived
  : public Base
{
  virtual int foo()
  {
    return 24;
  }
};

Base* get_base()
{
  return new Derived;
}

Base* get_other_base()
{
  return new OtherDerived;
}

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;
  python::class_<Base, boost::noncopyable>("Base", python::no_init)
    ;

  python::class_<Derived, python::bases<Base> >("Derived", python::no_init)
    .def("foo", &Base::foo)
    ;

  python::class_<OtherDerived, python::bases<Base> >(
      "OtherDerived", python::no_init)
    ;

  python::def("get_base", &get_base,
              python::return_value_policy<python::manage_new_object>());

  python::def("get_other_base", &get_other_base,
              python::return_value_policy<python::manage_new_object>());
}

Interactive usage:

>>> import example
>>> b = example.get_base()
>>> b.foo()
42
>>> b = example.get_other_base()
>>> b.foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'OtherDerived' object has no attribute 'foo'

Answer 2

Abstract classes actually CAN be exposed to Boost.Python without wrappers. The trick is defining your class with boost::noncopyable and avoiding pure_virtual method wrappers.

Here is the corrected code (tested with Boost.Python 1.47.0 and Python 2.7.6):

#include <boost/python/class.hpp>
#include <boost/python/def.hpp>
#include <boost/python/module.hpp>

struct Base
{
  virtual int foo() = 0;
};

struct Derived : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

Base* get_base()
{
  return new Derived;
}

BOOST_PYTHON_MODULE(libTestMod)
{
    namespace py = boost::python;

    py::class_<Base, boost::noncopyable>("Base", py::no_init)
        .def("foo", &Base::foo);

    py::def("get_base", get_base,
        py::return_value_policy<py::reference_existing_object>()); //ignore mem leak
}

Testing:

$ python
Python 2.7.6 (default, Mar 31 2014, 16:04:58) 
[GCC 4.7.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import libTestMod
>>> base = libTestMod.get_base()
>>> print base.foo()
42