Create Boost-python Nested Namespace

Using boost python I need create nested namespace. Assume I have following cpp class structure: namespace a { class A{...} namespace b { class B{...} } }

Solution 1:

What you want is a boost::python::scope.

Python has no concept of 'namespaces', but you can use a class very much like a namespace:

#include <boost/python/module.hpp>
#include <boost/python/class.hpp>
#include <boost/python/scope.hpp>
using namespace boost::python;

namespace a
{
    class A{};

    namespace b
    {
         class B{};
    }
}

class DummyA{};
class DummyB{};

BOOST_PYTHON_MODULE(mymodule)
{
    // Change the current scope 
    scope a
        = class_<DummyA>("a")
        ;

    // Define a class A in the current scope, a
    class_<a::A>("A")
        //.def("somemethod", &a::A::method)
        ;

    // Change the scope again, a.b:
    scope b
        = class_<DummyB>("b")
        ;

    class_<a::b::B>("B")
        //.def("somemethod", &a::b::B::method)
        ;
}

Then in python, you have:

#!/usr/bin/env python
import mylib

print mylib.a,
print mylib.a.A
print mylib.a.b
print mylib.a.b.B

All a, a.A, a.b and a.b.B are actually classes, but you can treat a and a.b just like namespaces - and never actually instantiate them

Solution 2:

The trick with dummy classes is quite fine, but doesn't allow:

import mylib.a
from mylib.a.b import B

So, instead, use PyImport_AddModule(). You may find full featured examples in the following article: Packages in Python extension modules, by Vadim Macagon.

In short:

namespace py = boost::python;
std::string nested_name = py::extract<std::string>(py::scope().attr("__name__") + ".nested");
py::object nested_module(py::handle<>(py::borrowed(PyImport_AddModule(nested_name.c_str()))));
py::scope().attr("nested") = nested_module;
py::scope parent = nested_module;
py::class_<a::A>("A")...