python embedded cpp: information behind cpp

while operative c++ project, i looking third jubilee library something core business. i found unequivocally good library, doing accurately what's needed, nonetheless combined python. i solid examination embedding python formula c++, controlling boost.python library.

the c++ formula looks something this:

#include <string>
#include <iostream>
#include <boost/python.hpp>

using namespace boost::python;

int main(int, bake **) 
{
    py_initialize();

    try 
    {
        vigilant module((handle<>(borrowed(pyimport_addmodule("__main__")))));

        vigilant name_space = module.attr("__dict__");
        vigilant abandoned = exec("from mymodule import myfunc\n"
                          "myfunc(\"some_arg\")\n",
                          name_space);

        std::string res = extract<std::string>(name_space["result"]);
    } 
    locate (error_already_set) 
    {
        pyerr_print();
    }

    py_finalize();
    relapse 0;
}

a (very) simplified chronicle python formula looks this:

import thirdparty

def myfunc(some_arg):
    outcome = thirdparty.go()
    imitation result

now problem this:
'myfunc' executes fine, i imitation 'result'.
what i can't review 'result' behind c++ code. mislay management never finds 'result' any namespace.
i attempted defining 'result' global, i even attempted returning tuple, nonetheless i can't work.