Murray Cumming wrote:

On Fri, 2009-02-27 at 07:35 -0800, Ralf W. Grosse-Kunstleve wrote:

...

The trick is to use boost::python::handle:

http://www.boost.org/doc/libs/1_38_0/libs/python/doc/v2/handle.html

For example:

object python_file((handle<>(borrowed(obj_ptr))));

Thanks, but like much of the documentation, I find that page very unclear. Is there anything that states more obviously what should be done in the two cases I mentioned.

The best documentation I could find was this page on the Python wiki, taken from a email to this list: http://wiki.python.org/moin/boost.python/handle The only error is that references to "null_ok" should be "allow_null". At least that's the way I got it to work. -- Anthony Foglia Princeton Consultants (609) 987-8787 x233

On Fri, 2009-02-27 at 08:34 -0800, Alex Mohr wrote:

Murray Cumming wrote:

...

I can't find any definitive documentation that tells me how I should get a boost::python::object to wrap an existing PyObject*. I guess that there's a way to do it that uses an existing reference, and a way that takes an extra reference.

You can construct a boost::python::object with a boost::python::handle<>. boost::python::handle<> is sort of a "smart" PyObject *. It manages the python object reference count automatically.

Oh, I assumed that boost::python::object did this already. If not, is there any reason not to use boost::python::handle<boost::python::object> everywhere instead of a boost::python::object? One of my main reasons for using boost::python instead of the C API is to avoid worrying so much about the reference counting. For instance, I don't want to explicitly increment the ref when returning a PyObject, or remember whether I should take a ref when given a PyObject from various functions.

When you create a boost::python::handle<> you can tell it whether to bump the reference count or not. Given a PyObject *p:

// Use with a "new reference" -- *doesn't* bump ref count. handle<>(p);

// Use with a "borrowed reference" -- *does* bump ref count. handle<>(borrowed(p));

Alex _______________________________________________ Cplusplus-sig mailing list Cplusplus-sig@python.org http://mail.python.org/mailman/listinfo/cplusplus-sig -- Murray Cumming murrayc@murrayc.com www.murrayc.com www.openismus.com

Oh, I assumed that boost::python::object did this already.

It does. Think of bp::handle<> as a low-level, bare-bones, smart PyObject*. Think of bp::object as that, *plus* all the fancy c++ API that maps to python API (like obj.attr("foo"), []-operator, ()-operator, etc). I believe bp::object holds a handle<> internally.

If not, is there any reason not to use boost::python::handle<boost::python::object> everywhere instead of a boost::python::object?

Don't do that. You can't do that. It won't compile. Use bp::object everywhere unless you're convinced you have a good reason to do otherwise. If you have a PyObject *p and you want a bp::object, construct it via: object(handle<>(p)) // when p's a new reference object(handle<>(borrowed(p))) // when p's a borrowed reference

One of my main reasons for using boost::python instead of the C API is to avoid worrying so much about the reference counting. For instance, I don't want to explicitly increment the ref when returning a PyObject, or remember whether I should take a ref when given a PyObject from various functions.

Yup. Just use bp::object everywhere. I recommend you read through the reference manual: http://www.boost.org/doc/libs/1_38_0/libs/python/doc/v2/object.html http://www.boost.org/doc/libs/1_38_0/libs/python/doc/v2/handle.html Also look at the tutorials and examples and tests in the boost.python distribution. Alex

On 02/11/2010 07:18 AM, Murray Cumming wrote:

On Fri, 2009-02-27 at 15:18 -0800, Alex Mohr wrote: [snip]

...

If you have a PyObject *p and you want a bp::object, construct it via:

object(handle<>(p)) // when p's a new reference

I find that I have to split the handle and object over two lines. If I do this boost::python::object cppobject(boost::python::handle<>(cObject)); and then something like this: if(!cppobject.ptr()) doSomething()

then I get this weird compiler error on the if() line:

glom/python_embed/glom_python.cc:229: error: request for member ‘ptr’ in ‘cppobject’, which is of non-class type ‘boost::python::api::object(boost::python::handle<_object>)’

It's as if doing it on one line has changed the type.

It does ! In C++, if something may be interpreted as a declaration, it is a declaration. And in the above case, the first line may be interpreted as a declaration of a function "cppobject" returning a bp::object, taking a handle<>. If you split things into three statements, by first instantiating a (named) handle<>, then pass that to the cppobject constructor, the error will go away, since that line then is no longer ambiguous. HTH, Stefan -- ...ich hab' noch einen Koffer in Berlin...