Reputation: 16188
I currently have the following:
namespace py=boost::python;
//C++
void f() {
std::cout << "hello world\n";
}
//I am not precious about this, if it can be done without a module that would be great
BOOST_PYTHON_MODULE(test)
{
py::def("f", f);
}
int main() {
auto main_module =py::import("__main__");
auto main_namespace =main_module.attr("__dict__");
//???????
auto result=py::exec_file("t.py", main_namespace);
}
//t.py
f()
I am trying to call f, but I am not sure of the glue required to get it to work. With classes I can do
int main() {
//...
py::obejct p_my_type=py::class_<my_type>("my_type").def("f", &my_type::f);
main_namespace["my_type"]=p_my_type;
//...
however boost::python::def
doesn't appear to return a boost::python::object
like class_
does
My questions are, how do I get the first test case to work as expected? And secondly is the way in which I am exposing my types in the second code snippet "correct"?
Upvotes: 3
Views: 969
Reputation: 16188
The fix was simple but wasn't mentioned in the doc on this page:
http://www.boost.org/doc/libs/1_55_0/libs/python/doc/tutorial/doc/html/python/embedding.html
I needed to do this:
auto main_module =py::import("__main__");
auto main_namespace =main_module.attr("__dict__");
inittest();
auto result=py::exec_file("t.py", main_namespace);
from test import f
f()
Upvotes: 1