Question

Using boost python I need create nested namespace.

Assume I have following cpp class structure:

namespace a
{
    class A{...}
    namespace b
    {
         class B{...}
    }
}

Obvious solution not work:

BOOST_PYTHON_MODULE( a ) {
    boost::python::class_<a::A>("A")
     ...
    ;
    BOOST_PYTHON_MODULE(b){
        boost::python::class_<a::b::B>("B")
        ...
    ;
    }
}

It causes compile-time error: linkage specification must be at global scope

Is there any way to declare class B that would be accessed from Python as a.b.B?

Answer 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