[C++-sig] Re: instantiating python objects within C++

David Abrahams dave at boost-consulting.com
Wed May 28 12:54:37 CEST 2003

Stefan Seefeld <seefeld at sympatico.ca> writes:

> David Abrahams wrote:
>> The Boost.Python shared library has a converter registry where all of
>> this information is stored.
> ok, this seems really to be the key. How does this work ? 

You can "use the source, Luke" if you want to know how it's
implemented ;-)

> I'm a bit confused, as conceptually the lines
> {
>    python::class_<Base> base_type("Base");
>    //...
> }
> create a type 'Sandbox.Base', i.e. the equivalent in C++ would be
> something like 'Sandbox::Base'. However, 'Sandbox' isn't a type,


> so I wonder how I can access the above type object from inside C++.

I don't see what Sandbox being a type or a module has to do with it.
A Python type is just a kind of Python object.  When the class_< ... >
instantiation creates it, it holds onto it as well.

> Your example at the bottom of
> http://www.boost.org/libs/python/doc/tutorial/doc/derived_object_types.html
> hints a bit at how to instantiate such type objects, but I fail
> to see how I could do that from within a different 'scope'.

I don't know what you mean by 'scope' here.  Can't you spell out
explicitly what it is you'd like to do?  Code helps.

> What I'm really pondering about now is how to compare types, i.e.
> if I'v obtained a type from a python script, and I want to test
> whether this type derives from 'Sandbox.Base':
> from Sandbox import *
> class Derived(Base):
>    def foobar(self): pass
> I should be able to take the python::object that is the
> (wrapper around the) python type, and compare it with the 'base_type'
> from the above snippet, i.e. doing something like
> python::handle<> result(PyRun_File(fp, const_cast<char *>(file.c_str()),
> 				   Py_file_input,
> 				   globals.ptr(), globals.ptr()));
> python::object derived_type = globals["Derived"];
> I want to compare 'derived_type' with 'base_type', for example using
> Py_IsSubclass(derived_type.ptr(), base_type.ptr())
> But how can I access 'base_type' ? It's a local variable in the
> scope of 'BOOST_PYTHON_MODULE(Sandbox), or is it ?

Of course it is, the way you've written it.  You can always keep a
python::object 'save_base_type' around and assign into it:

       save_base_type = base_type;

A more-elegant way to hang onto it is:

python::object base_type;
       = python::class_<Base>("Base")

Dave Abrahams
Boost Consulting

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