Boost.Python: how to output a factory method in python

I have a C ++ API that is built into a dll, now I want an api using python, so I decided to rebuild the dll with boost.python, but I'm new to boost.python.

The API is in the CMyApi.h file , it has a factory way like:

class __declspec(dllexport) CMyApi
{
public:
    static CMyApi* CreateUserApi(const char *path = "");

    virtual void Init() = 0;

protected:
    ~CMyApi(){};
};

      

        
        
        
      

    

CMyApi is a virtual class and CApiImpl inherits from CMyApi, it is defined in CApiImpl.h :

class CApiImpl : public CMyApi
{
public:
    CApiImpi();
    ~CApiImpi();
    virtual void Init();
};

      

        
        
        
      

    

CApiImpl.cpp contains the CApiImpl implementation and the CMyApi :: CreateUserApi implementation:

static CMyApi* CMyApi::CreateUserApi(const char *path)
{
    return new CApiImpl();
}

      

        
        
        
      

    

I don't want to expose CApiImpl for the api user, I would like the API to be called by CMyApi, so the user has CMyApi.h and a DLL file.

CMyApi wraps like this:

class CMyApiWrapper : public CMyApi, public wrapper<CMApi>
{
    void Init()
    {
        this->get_override("Init")();
    }
};

BOOST_PYTHON_MODULE(CMyApiWrapper)
{
    class_<CMyApiWrapper>("CMyApiWrapper")
        .def("CreateUserApi", &CMyApi::CreateUserApi)
        .staticmethod("CreateUserApi")
        .def("Init", pure_virtual(&CMyApi::Init));
}

      

        
        
        
      

    

The above BOOST_PYTHON_MODULE cannot pass the assembly, what's wrong in my program?