Mailman 3 June 2007 - Cplusplus-sig

C++ static function as __init__
by Piotr Jaroszyński June 14, 2007

June 14, 2007

Hello, I want to do possibly strange thing: struct Foo { Foo() {} static void init_me(PyObject * self) { ??? (I want to initialize self with a new Foo object using its constructor) } }; (...) bp::class_<Foo> foo("Foo", bp::no_init); foo.def("__init__", &Foo::init_me); The above "works" as in the signature matches, but I have no idea how to initialize the object properly given the PyObject self pointer. I was trying to dig in the code how is bp::init really working but I failed miserably :/ Any clues? ;] -- Best Regards, Piotr Jaroszyński

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__init__() should return None, not 'NoneType' with Boost.Python?
by Phil June 14, 2007

June 14, 2007

Hi, I'm a newbie having difficulty getting a very simple C++ class to expose in Python. I've narrowed this down to the simplest case I can to reproduce the error (code below - all in one cpp). I'm using Mac OS X 10.4.9, GCC 4.01, Python 2.5 and Boost 1.33. I'm using XCode to compile the below as a BSD Dynamic Library. It builds fine. I'm then renaming the resulting library from .dylib to .so (necessary extra step on Mac OS). On importing into Python I see get this: $ python Python 2.5.1 (r251:54863, Jun 10 2007, 17:57:56) [GCC 4.0.1 (Apple Computer, Inc. build 5250)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> import libphil_boost >>> testing = libphil_boost.MyTest() Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: __init__() should return None, not 'NoneType' >>> I can 'see' it like this tho: >>> libphil_boost <module 'libphil_boost' from 'libphil_boost.so'> >>> libphil_boost.MyTest <class 'libphil_boost.MyTest'> >>> Using exactly the same method I can expose more basic C++ types. For example a basic enum or function works fine (for example the Hello World example in Boost docs). But I see the same error for any type of class I choose to try it with. I've tried googling this, and I've had no luck. Has anyone seen this before, or have any ideas what I may be doing wrong? Thanks for any help, Phil ______ #include <boost/python.hpp> using namespace boost::python; class MyTest { }; BOOST_PYTHON_MODULE(libphil_boost) { class_<MyTest>("MyTest"); }

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len function?
by Ron Brown, Jr. June 13, 2007

June 13, 2007

This is probably a ridiculous question, but how do I find the length of a boost::python::list object? I can't seem to find a straightforward answer in any of my searches. Thanks, Ron

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Py++: auto_ptr held_type
by Anand, Kumar June 13, 2007

June 13, 2007

I am trying to expose a "MyTestClass" to python using py++ and with held_type as std::auto_ptr. Since Py++ creates a wrapper class for every C++ class that it exposes, should held_type be "std::auto_ptr<MyTestClass>" or "std::auto_ptr<MyTestClass_Wrapper>"? Can you point as to why one would choose one against the other? Thanks Kumar

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Py++ Error for function operator *()
by Haridev, Meghana June 13, 2007

June 13, 2007

Hi Roman, We've been using Py++ 0.8.5 for a while now and it has been working great. I'm in the process of upgrading to Py++ 0.9.0. I currently have the following installed: 1. Gccxml (cvs-snapshot June 5, 2007): There is some problem with the extraction of the tarball at https://sourceforge.net/project/showfiles.php?group_id=118209&package_id =146545, so I'm just using the latest from cvs. And I am positive that this is not the source of the problem I'm running into. 2. Pygccxml (0.9.0) 3. Py++ (0.9.0) ----------------------------------------------------------------------- When I moved from Py++ 0.8.5 to Py++ 0.9.0, Py++ fails for the following function declaration during the module_builder's build_code_creator() phase: operator * (const ACE_Time_Value &tv, double d) { //..... } Here's the Error message: self.moduleBuilder.build_code_creator(self.moduleName, create_casting_constructor=flag_casting_constr) File "/opt/data/BTBLD_rhine_1080/thirdParty/lib/python2.4/site-packages/pyplu splus/module_builder/builder.py", line 247, in build_code_creator self.__code_creator = creator.create() File "/opt/or-tree/mharidev/thirdParty/lib/python2.4/site-packages/pyplusplus /module_creator/creator.py", line 315, in create self._adopt_free_operator( operator ) File "/opt/or-tree/mharidev/thirdParty/lib/python2.4/site-packages/pyplusplus /module_creator/creator.py", line 185, in _adopt_free_operator assert isinstance( arg_type, declarations.declarated_t ) AssertionError ----------------------------------------------------------------------- I narrowed down the Error to this part of the code (Py++ 0.9.0) in function _adopt_free_operator() of pyplusplus/module_creator/creator.py(line 178): arg_type = declarations.base_type( operator.arguments[0].type ) if isinstance( arg_type, declarations.fundamental_t ): arg_type = declarations.base_type( operator.arguments[1].type ) elif isinstance( arg_type, declarations.declarated_t ) and arg_type.declaration.ignore: arg_type = declarations.base_type( operator.arguments[1].type ) else: pass assert isinstance( arg_type, declarations.declarated_t ) ----------------------------------------------------------------------- Note that I don't have this problem with the earlier version (Py++ 0.8.5) when the code looked like this and didn't have the extra check: arg_type = declarations.base_type( operator.arguments[0].type ) if isinstance( arg_type, declarations.fundamental_t ): arg_type = declarations.base_type( operator.arguments[1].type ) assert isinstance( arg_type, declarations.declarated_t ) ----------------------------------------------------------------------- Any ideas to how this can be fixed? Thanks for your help, -Meghana.

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Re: [C++-sig] Problems instantiating class Boost Python class
by David Abrahams June 13, 2007

June 13, 2007

on Mon Jun 11 2007, Phil <phil.beadling-AT-homechoice.co.uk> wrote: > Hi, > > I'm a newbie having some problems instantiating a class written in C++ > in Python using Boost.Python. > > I can get a very simple example working when I expose a function outside > of a class, that returns a string, but the below fails as shown, despite > building. The enum in the same library however works fine. > > I've googled the error returned from Python and I get zero hits - has > anyone get any idea what I'm doing wrong? > > I'm using Python 2.5, GCC 4.0.1 and Boost 1.33 on Mac OS X 10.4.9. I > don't think it matters but I'm building the solution as a dynamic C++ > bsd library in XCode and the changing the Mac OS X .dylib extention to > .so as is required by Python. This all works for simpler examples and > the enum in this example. > > Error: TypeError: __init__() should return None, not 'NoneType' Phil, First, I suggest you take your question to a more subject-specific mailing list: http://www.boost.org/more/mailing_lists.htm#cplussig I have no clue what could be going wrong here; I've never seen that problem. My only suggestion to you is to further reduce your test case to the absolute minimal example that reproduces the behavior. That almost always makes the problem obvious, or at least easily diagnosed. Your example is small, but surely not minimal yet. -- Dave Abrahams Boost Consulting http://www.boost-consulting.com The Astoria Seminar ==> http://www.astoriaseminar.com

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inheritance and polyphormism
by Luca Sbardella June 12, 2007

June 12, 2007

hi, How do I expose properly a class tree so that classes can be derived from python in a polymorphic way? I've approached the problem is the following way. class base { public: virtual void method1(){} }; class derived: public base { public: virtual void method2(){} }; I then wrote two class templates to wrap the classes in the following way template<class T> class base_wrap: public T, public boost::python::wrapper<T> { void method1() {....} }; template<class T> class derived_wrap: public base_wrap<T> { void method2() {....} }; And finally class_<base_wrap<base>, boost::noncopyable>("base") .def("method1", &base::method1) ; class_<derived_wrap<derived>, boost::noncopyable>("derived") .def("method1", &derived::method1) <======== this seems redundant .def("method2", &derived::method2) ; This works fine, however I does't seem to be the optimal solution (see the comment on class_<derived_wrap<derived>>). Is there a better way? Thanks Luca

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[c++-sig] Is it a BUG? boost.python makes doctest impossible to recursive test.
by 甜瓜 June 11, 2007

June 11, 2007

Today, I encounter a tough problem about using doctest in class exposed by boost.python. For traditional python class which is define in python: class P: """ >>> .... # this line will be tested by doctest """ def func(self): """ >>>... # this line will be tested by doctest by recursively discovering __doc__ """ pass However, if I expose the same class by boost.python, class C { public: void func() {} }; char* doc_str[] = { // __doc__ for class ">>>...", // this line will be tested by doctest // __doc__ for func ">>>..." // this line will NOT be tested by doctest }; BOOST_PYTHON_MODULE(...) { class_<C>("C", doc_str[0], init<>()) .def("func", &C::func, doc_str[1]); } In above case, doctest can only discover the __doc__ for class, but can NOT go further to discover __doc__ for method. I throughly studied the reason, and found the problem is: In doctest.py class DocTestFinder: def _find(self, tests, obj, name, module, source_lines, globs, seen): ..... # Look for tests in a class's contained objects. if inspect.isclass(obj) and self._recurse: for valname, val in obj.__dict__.items(): ..... # Recurse to methods, properties, and nested classes. BUT IT FAILS. if ((inspect.isfunction(val) or inspect.isclass(val) or isinstance(val, property)) and self._from_module(module, val)): .... inspect.isfunction(val) ALWAYS returns False, because the type of 'val' corresponding to 'func' in class C is: <Boost.Python.function object at 0x....> while for class P, the type of 'val' is <function func at 0x....>, which will get True. Therefore, the strange type of the class' method exposed by boost.python suppresses the doctest to recursively discover __doc__. What should I do? Is is a bug for boost.python? Thank you --- ShenLei

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Call policy for static property?
by Albert Strasheim June 11, 2007

June 11, 2007

Hello all I'm trying to add a static property to a class, where this property contains an instance of another class. #include <boost/python/class.hpp> namespace py = boost::python; struct Foo{}; struct Bar{}; static Bar& get_bar() { static Bar bar; return bar; } void export_foo() { py::class_<Bar, boost::noncopyable>("Bar", py::no_init); py::class_<Foo, boost::noncopyable>("Foo", py::no_init) .add_static_property("theonlybar", &get_bar); } so that I can write mymodule.Foo.theonlybar in Python and get this single instance of the Bar class. Unfortunately, this code doesn't compile as-is. I get: error C2027: use of undefined type 'boost::python::detail::specify_a_return_value_policy_to_wrap_functions_returning<T>' c:\program files\boost\boost_1_34_0\boost\python\detail\caller.hpp which makes sense, since get_bar probably needs a return_internal_reference call policy. I tried to add it with make_function: make_function(&get_bar, py::return_internal_reference<>()) but it seems make_function is only for wrapping member functions, because I get: IndexError: boost::python::with_custodian_and_ward_postcall: argument index out of range when I try to access the property. Is there a way to specify a call policy for the function called by a static property getter? Alternatively, is there another way to accomplish what I'm trying to do? Thanks for your time. Regards, Albert

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boost::shared_ptr getting back from Python with a PyObject inside
by Gabriel Becedillas June 11, 2007

June 11, 2007

I have the following problem: 1_) A boost::shared_ptr<T> is converted to python. 2_) The PyObject holding the boost::shared_ptr<T> is converted back from python, (apparently) yielding the same boost::shared_ptr<T>. 3_) The boost::shared_ptr<T> hits refcount 0 somewhere else in C++Land without holding the Python GIL. 4_) Python's API gets called without holding the GIL because a sp_counted_impl_pd with a shared_ptr_deleter modifies the PyObject's refcount and Py_Dealloc gets called (check shared_ptr_deleter::operator()). This is wrong because you shouldn't use Python's API without holding the GIL. In my particular case I don't know when or where a boost::shared_ptr<T> will hit refcount 0 (thats the funny part of using smart pointers). I took a look around and found that: 1_) shared_ptr_from_python is putting the shared_ptr_deleter. 2_) shared_ptr_to_python is checking if the boost::shared_ptr<T> holds a shared_ptr_deleter to get the PyObject inside. Is there a way to make shared_ptr_from_python return the original boost::shared_ptr<T> without changing the sp_counted_impl_pd/shared_ptr_deleter ? Thanks.

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