暴露多态性与boost python

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我开始变得非常沮丧,尝试用boost :: python将简单的C多态性暴露给python.

我在C中有以下结构：

struct Base {
    int typeID;
};

struct Derived : public Base {
    int derivedProperty;
}

//and some more from base derived types....    

Base *returnSomethingDerivedFromBase(...) {
    Derived *ret = new Derived;
    ret->derivedProperty = 1234;
    return ret;
}

BOOST_PYTHON_MODULE(foo) 
{
    class_<Base>("Base")
        .add_property("baseProperty", &Base::baseProperty);

    class_<Derived, bases<Base> >("Derived")
        .add_property("derivedProperty", &Derived::derivedProperty);

    def("returnSomethingDerivedFromBase", returnSomethingDerivedFromBase);    
}

在Python中我只想拥有以下内容：

object = returnSomethingFromDerived() #object is of type Base
if object.typeID = 1:
    #here i want to cast to Derived and access "derivedProperty"
    #but this is not working :-( :
    object.__class__ = Derived

有没有办法实现这一点？或者这不可能像在C中一样吗？

非常感谢你的帮助！！

解决方法:

好吧,我错过了Base类中的虚拟析构函数.所以这是它的工作原理：

struct Base {
    virtual ~Base() {}
    int typeID;
};

struct Derived : public Base {
    int derivedProperty;
}

//and some more from base derived types....    

Base *returnSomethingDerivedFromBase(...) {
    Derived *ret = new Derived;
    ret->derivedProperty = 1234;
    return ret;
}

BOOST_PYTHON_MODULE(foo) 
{
    class_<Base>("Base")
        .add_property("baseProperty", &Base::baseProperty);

    class_<Derived, bases<Base> >("Derived")
        .add_property("derivedProperty", &Derived::derivedProperty);

    def("returnSomethingDerivedFromBase", returnSomethingDerivedFromBase, return_value_policy<manage_new_object>());    
}

但现在我有一个不同的问题.当我尝试在元组中返回此类型时,我再次丢失类型信息：

tuple returnSomethingDerivedFromBase(...) {
    Derived *ret = new Derived;
    ret->derivedProperty = 1234;
    return make_tuple(ret);
}

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