[C++-sig] Getting a b::p::object for a PyObject and vice-versa.
Murray Cumming
murrayc at murrayc.com
Thu Feb 4 21:16:41 CET 2010
Can someone confirm that this is correct:
1.
To get a boost::python::object that wraps an existing PyObject* (when
you get a PyObject from a C function not wrapped in boost::python), can
someone confirm that this is correct:
Either:
a)
PyObject* c_object = get_the_c_object(); //returns a reference.
boost::python::handle<> handle(c_object);
boost::python::object cpp_object(handle);
or
b)
PyObject* c_object = get_the_c_object(); //returns no extra reference.
boost::python::handle<> handle(boost::python::borrowed(cObject));
boost::python::object cpp_object(handle);
The code for b) can be simplified by doing
boost::python::object cpp_object(boost::python::borrowed(cObject))
but I don't see a way to simplify a)
I also see allow_null here, but like borrowed, it's not documented:
http://www.boost.org/doc/libs/1_41_0/libs/python/doc/v2/handle.html#allow_null-spec
so I can only guess at its purpose.
In general I wish this was much simpler and stated clearly somewhere.
2.
How can I get the underlying PyObject* from a boost::python::object
(when I need it to call a C function not wrapped in boost::python)?
I guessed at this, but it doesn't seem to be working for me:
boost::python::extract<PyObject*> extractor(cpp_object);
if(extractor.check())
{
PyObject* c_object = extractor;
}
--
murrayc at murrayc.com
www.murrayc.com
www.openismus.com
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