Base class of non exposed derived class in Python side?

示されている Base、Derived、DerivedTwo、factory、factoryTwo の定義を使用して、例を再現します。まず Boost.Python の class_ 登録、bases 宣言、shared_ptr の変換動作を読み、DerivedTwo を登録した場合と登録しない場合で、返される Python 型を比較します。型の選択動作が説明され、fix の想定される扱いまたは対象範囲が文書化されていれば完了です。

Hello, I have a use case where a polymorphism factory can instantiate object of a non exposed class. I have a multi-level inheritance scheme and the last level is not exposed to Python. What I try to understand is why the Python object of the last level takes the type of the top-most class in the inheritance hierarchy and not the one just above?

Here is a minimal example to reproduce the behavior:

C++ side

class Base
{
public:
    Base() {}
    virtual ~Base() = default;
};

class Derived: public Base
{
public:
    Derived(int value) { m_value = value; }
    ~Derived() = default;

    int getValue() { return m_value; }

private:
    int m_value;
};

class DerivedTwo: public Derived
{
public:
    DerivedTwo(int value): Derived(value) {}
    ~DerivedTwo() = default;
};

std::shared_ptr<Base> factory()
{
    return std::make_shared<Derived>(9);
}

std::shared_ptr<Base> factoryTwo()
{
    return std::make_shared<DerivedTwo>(99);
}

BOOST_PYTHON_MODULE(mymodule)
{
    using namespace boost::python;

    class_<Base, std::shared_ptr<Base>, boost::noncopyable>("Base", init<>());

    class_<Derived, bases<Base>, std::shared_ptr<Derived>, boost::noncopyable>("Derived", init<int>())
        .def("get_value", &Derived::getValue)
    ;

    def("factory", &factory);
    def("factory_two", &factoryTwo);
}

Python side

import mymodule

obj = mymodule.factory()
print(type(obj))

obj2 = mymodule.factory_two()
print(type(obj2))

Output

<class 'mymodule.Derived'>
<class 'mymodule.Base'> <-- WHY? 

Obviously, if I expose DerivedTwo class, all is OK.
Thanks for your help.