Re: to_python with c++ class??

From: rwgk_at_[hidden]
Date: 2001-08-31 00:58:41

Ideas:
    
namespace boost { namespace python {
  template class 
  class_builder<CMyClass>; // explicitly instantiate
}} // namespace boost::python

int main(...)
    ...
    // now I need a PyObject Pointer for my object a
    PyObject* pA = BOOST_PYTHON_CONVERSION::to_python(a); 

Caveat: I have no experience with embedding Python.

Ralf

--- In boost_at_y..., yahoo_at_p... wrote:
> Hi,
> just tried to figure out this by myself and by reading the docs and 
> archives, but I failed:-(
> 
> Here is my problem.
> 
> I have an app which should 
> a) export objects to python and
> b) embedds python.
> 
> I managed to build a extension module using boost, which builds 
> wrapper classes for my c++ objects.
> 
> In my App I include the extension dll to register my apps c++ 
objects
>  
> In the embedded python I load this extension module and get the 
> objects using the wrapped registry.
> 
> This works really fine. I can get access the apps objects from 
python 
> and change their attributes from there. 
> 
> What I don't know is how I can create an PyObject* from one of my 
> classes to call a Python function with it? (embedded Python)
> 
> Code should like that:
> 
> #include "ext_mod.h"
> int main(int, char**)
> {
>     Py_Initialize();
>     CMyClass a;  // create one of my apps class objects
> 
>     PyObject *pMod = PyImport_ImportModule("ext_mod"); // load the 
> extension module 
> 
>     // now I need a PyObject Pointer for my object a
>     PyObject* pA = ?????(a);    
> 
>     // after that I would like to add it to the modules dict
>     PyModule_AddObject(pMod,"a",pA);
> 
>     Py_Finalize();
> } 
>     
> How do I accomplish this?
> 
> In the definition for my ext_mod module (boost) I use 
> class_builder<CMyClass> myclass(mymod,"a"); 
> to build an extension class. But how do I use this to call to_python
> ()?
> 
> How can I export this from my DLL if I have to use it to get my 
> PyObject* from CMyClass?
> 
> Thanks a lot 
> Joerg