Mailman 3 August 2005 - Cplusplus-sig

Embedded Python: how to enumerate function names?
by Bob Eaton Aug. 21, 2005

Aug. 21, 2005

In my embedded Python C++ code, I have a PyObject* to the module and its dictionary. From either of these, is there a way to enumerate the methods/functions defined in that file? I know I can at least get the function names with (((PyFunctionObject *)func) -> func_name), but I'm not clear on how to get the PyFunctionObject pointer to begin with. Is it the Keys or Values or Items of the dictionary? Also, once I have the function pointer, can I enumerate its parameters and return type? And finally, what's the difference between a Function and a Method? Any help will be greatly appreciated. Bob

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small error within implicitly_convertible example
by Roman Yakovenko Aug. 21, 2005

Aug. 21, 2005

Hi. There is an error, while compiling example: http://boost.org/libs/python/doc/v2/implicit.html. I've got errors on compilers: gcc 3.3 ( from debian 3.1 ) and on msvc 7.1 ( win XP ). The error caused by the fact that X::operator int() defined as non const. Adding const to the definition fix the problem. Roman

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Strange behavior when split modules
by Eric Jardim Aug. 20, 2005

Aug. 20, 2005

Hi all, While wrapping the Qt4 code, I decided to split my single "Qt.so" module in many modules (that by now are: Namespace, Core and Gui), and I am using a package called Qt, that do some essential Python initializations on "__init__.py". Everything was Ok while spliting the modules, and it compiled Ok, and most of my tests ran Ok. But a strange thing is happening! There is a method called "parent" defined on the "QObject" class that returns the object's parent (or None if not it exists) that is another "QObject". As this is just a reference to a known object, I am using the "return_value_policy<reference_existing_object>" policy. The "QObject" class is defined on the "Core" module. On the other hand. I have a lot of classes defined on the "Gui" module, that are children (direct or indirect) of QObject. One of them, for example, is the "QWidget" class. When I call "parent" from a "QWidget" object it return an object with a different Python object, and I loose it's reference because the refcount of that especific new Python object is lost. I also noted that this behavior happend with other methods that return "non-new" objects. All I have to say is that when modules were together in one this problem didn't occur. To be sure of that, I just moved the QWidget class (and QPaintDevice, that is another base class of QWidget) to the "Core" module, and everything worked again. So, am I not allowed to defined derived classes in different modules, or am I doing something wrong? By now, it is not a problem defining everything in a single module, but if someone may want to write extension of the python-qt4 library, that would be a problem. By the way, the project can be downloaded from: http://developer.berlios.de/projects/python-qt4/ Thanks, [Eric Jardim]

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exec test failure
by Stefan Seefeld Aug. 19, 2005

Aug. 19, 2005

I submitted a new test that runs the new 'exec' and 'exec_file' functions. I tested it locally (obviously), but now I see it fail (http://engineering.meta-comm.com/boost-regression/CVS-HEAD/developer/python…). The error is always the same 'RuntimeError: exec.py : no such file', which I don't understand. The file is checked into the libs/python/test/ directory. Could somebody with more insight into the boost.build logic and Jamfile syntax please help ? Thanks, Stefan

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Converting from Python integer to C++ by reference
by Saikat Chakrabarti Aug. 18, 2005

Aug. 18, 2005

Hello, I am attempting to expose a class in C++ with a method that takes a reference to an integer. Thus, a simple version of what I have is as follows: class C { public: void SetMe(int& i); } and SetMe is implemented as follows: void C::SetMe(int& i) { i = 5; } and my goal is to be able to call this method from Python with an integer and have the value of that integer in Python set. Thus, in Python, I would like to do: c = C() i = 0 c.SetMe(i) and have i's value be 5. However, when I do this, I get an exception on line 292 of arg_from_python.hpp (at the end of the constructor for reference_arg_from_python) that simply says I have an access violation. If I get rid of the reference and made SetMe simply declared as "void SetMe(int i)", the code works fine. Also, if I change int to long (as I believe Python integers are actually C++ long types), but keep the reference (thus, having void SetMe(long& i), it gives me the same error. It seems like I am doing something fundamentally wrong. How should I fix this problem? Thanks, Saikat Chakrabarti

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Re: [C++-sig] boost.python iterators
by David Abrahams Aug. 18, 2005

Aug. 18, 2005

Dirk, you're in the wrong mailing list/newsgroup. Try the Python C++ sig (where followups are directed). Dirk Steenpass <steenpaz(a)sdf-eu.org> writes: > Dear all, > > I am looking for a way to create a python savvy iterator by wrapping > a container that provides begin and end member functions taking > arguments. Taking arguments, hmm. > The reference manual entry on boost::python::range left me unsure whether > the appended code would compile (it does not). > > The sentence: > > "The first form creates a Python callable object which, when invoked, converts > its argument to a Target object x, and creates a Python iterator which > traverses [bind(start,_1)(x), bind(finish,_1)(x)), applying NextPolicies to the > iterator's next() function." > > suggested to me that a single argument, i.e., the target reference is passed to > the accessor functions. Correct. > This would imply that the accessor takes exactly one argument. (???) Correct. > However, the subsequent explanations about the deduction of Target left me with > the impression that accessor functions and accessor member function pointers > may take several arguments. I understand. I think the text is misleading. I should delete "arguments..." > Any help is appreciated, Your problem is that you're forgetting the implicit "this" argument. In Python that corresponds to the initial explicit "self." Probably your simplest solution would be something like: python::object Elements(A& a, A::SizeType n) { return python::range<python::default_call_policies, A>( boost::bind(&A::begin, _1, n) , boost::bind(&A::end, _1, n) )(boost::ref(a)); } Then wrap that. We create a new Python function object on the fly each time and invoke it. The bind expressions just bind n into the accessors so that they take only a single argument. HTH, > dirk > > //--- A [n] example ----------------------------------------------------------- > // Does not compile with gcc 3.3.6 and gcc 4.0.2 > > #include <vector> > > #include <boost/python.hpp> > > using namespace std; > using namespace boost::python; > > class A > { > public: > typedef std::vector<int> VecInt; > typedef VecInt::iterator Iter; > typedef VecInt::size_type SizeType; > > public: > A(): vec_(100, 23) {} > > Iter Begin(SizeType ignored) { return vec_.begin(); } > Iter End(SizeType ignored) { return vec_.end(); } > > private: > VecInt vec_; > }; > > BOOST_PYTHON_MODULE(A) > { > class_<A>("A") > .def("Elements", range(&A::Begin, &A::End)) > ; > } > > // Just allowing standalone compilation. > int main(void) > { > return 0; > } -- Dave Abrahams Boost Consulting www.boost-consulting.com

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Re: [C++-sig] Boost.Python.function.__signatures__
by Ult Mundane Aug. 18, 2005

Aug. 18, 2005

Ralf, I'm using your function signatures patch, thank you very much, it is excellent and a huge timesaver for documenting exported C++ interfaces. I wonder if you've made any updates since you posted it last month. Is your patch checked into CVS somewhere? I'm looking into making two changes: 1. At the moment, parameters which are classes exported to Python are reported with their C++ names, not their Python names (i.e., if you export the class with a different name, the docstring doesn't match the name you use in Python). Glancing over the Boost.Python code, this is over my head at the moment; is this something that seems reasonable/doable to you? 2. Parameters which are built-in C++ types are reported as single-letter names (e.g., i = int, v = void, b = boolean, x = long long, j = size_t, etc.). In your example below, it shows the real C++ type names. I wonder what's different about my configuration. Any comments or suggestions are welcome. Thanks, Ult Ralf W. Grosse-Kunstleve wrote: > --- Nick Rasmussen <nick(a)ilm.com> wrote: >> > - I guess currently the docstrings are static. They'd have to become >> dynamic. >> > This seems to be a bigger project. I think we'd need a volunteer. >> >> I'm not sure what you mean by dynamic, > > It turns out my guess was wrong. Sorry. > > I looked some more and here is the result: > > http://cci.lbl.gov/~rwgk/boost_python/doc_signatures/function.hpp_2005_07_2… > http://cci.lbl.gov/~rwgk/boost_python/doc_signatures/function.hpp_2005_07_2… > > http://cci.lbl.gov/~rwgk/boost_python/doc_signatures/function.cpp_2005_07_2… > http://cci.lbl.gov/~rwgk/boost_python/doc_signatures/function.cpp_2005_07_2… > > Summary: > > - I've removed my __signatures__ method. > > - I've patched the getter for __doc__ to *always* append the list of > signatures. E.g.: > >>>> help(rational.lcm) > > Help on built-in function lcm: > > lcm(...) > C++ signature: > lcm(int, int) -> int > >>>> help(rational.int().numerator) > > Help on method numerator: > > numerator(...) method of boost_rational_ext.int instance > C++ signature: > numerator(boost::rational<int> {lvalue}) -> int > > > This looks like *the* solution to me. Thanks a lot for your brilliant idea, > Nick! > > Personally I don't think we need the facilities to disable appending the > signatures. Are there other opinions? > > Dave, could you please briefly review my patch? It is still very simple: > forward class list in function.hpp + list signatures(bool > show_return_type=false) const; declaration; in function.cpp I've move some code > from the body of argument_error() to the new signatures() function and I've > patched the body of function_get_doc() to append the result of signatures(). > > Cheers, > Ralf > > > __________________________________________________ > Do You Yahoo!? > Tired of spam? Yahoo! Mail has the best spam protection around > http://mail.yahoo.com > _______________________________________________ > C++-sig mailing list > C++-sig(a)python.org > http://mail.python.org/mailman/listinfo/c++-sig > > >

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Boost.Python at runtime
by Allen Bierbaum Aug. 17, 2005

Aug. 17, 2005

Has anyone every used or even tried to use Boost.Python to dynamically create python types at runtime (instead of the standard compile time method)? I am working on an very large extensible C++ class library that has a reflective interface that I can extend as much as I need. So instead of trying to create an enormous wrapper at compile time (that would be quickly out of date) I have been investigating the possibility of dynamically creating the python class/type wrappers on demand. So for example if I discovered at run-time that I wanted a wrapper for a class Hello I would: - Ask Hello's reflective interface for information about the class and it's bases - Create an associated boost.python class object - For each method Hello reports - add method to the wrapper class using signature and method info from Hello I would also need to handle registering associated smart pointers, etc. The details are definitely complex, but before I get to far into it I was wondering if anyone had done anything similar or had tried and could tell me how it went or if it is impossible with the current system. Thanks, Allen

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Re: [C++-sig] how do I instantiate templates in pyste with types defined in different files?
by Alexis H. Rivera-Rios Aug. 17, 2005

Aug. 17, 2005

That doesn't do it. My template declaration is in one file. The actual types are in another. There is no way for me to tell pyste where the actual types are when I try to instantiate the class. For example, Class("A","a.h") Class("B","b.h") Test = Template("Test","test.h") Test("A","testA") Test("B","testB") <-- these didn't work because when the pyste code generator tries to instantiate the class, it doesn't include the a.h or b.h because there is no mechanism to tell it how to do so. This didn't work either: Include("a.h") Test("A","testA") I tried declaring the class as a typedef: ---- somefile.h ---- #include "test.h" #include "a.h" typedef Test<A> TestA Then in pyste Class("TestA","somefile.h") Pyste doesn't complain, but doesn't generate any functions or constructors or anything. It would generate something like this: class_<Test<A>, boost::noncopyable >("TestA", no_init) ; Any ideas? Or ways around this? Thanks, Alexis > template <class A, class B> > class Test > { > A a; > B b; > }; > > Now I want to export an instanciate this template > class using the types Astruct1, Astruct2 defined in > atruct1.h and astruct2.h respectively. > > In pyste, I write this: > > Test = Template("Test","test.h") > Test(["Astruct1","Astruct2","Test1") Programming Tutorial: In Python: To do this, do this In Perl: To do this, do this or this or this or this... In C: To do this, do this, but be careful In C++: To do this, do this, but don't do this, be careful of this, watch out for this, and whatever you do, don't do this __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [C++-sig] problem chaning object state using a python class that inherits from c++ class
by Alexis H. Rivera-Rios Aug. 17, 2005

Aug. 17, 2005

David, Here's a simpler version: struct Data { int a; }; class B { public: virtual void Update(Data& a) = 0; ~B() {} }; class D : public B { virtual void Update(Data& a) { a.a = 10; } }; class C { public: void SetBptr(B* newB) { Bptr = newB; } void Update() { // this doesn't seem to work correctly if Bptr is a python object Bptr->Update(d); } int GetData_a() const { return d.a; } private: Data d; B* Bptr; }; here is the generated code from pyste: // Boost Includes ============================================================== #include <boost/python.hpp> #include <boost/cstdint.hpp> // Includes ==================================================================== #include "test.h" // Using ======================================================================= using namespace boost::python; // Declarations ================================================================ namespace { struct B_Wrapper: B { B_Wrapper(PyObject* py_self_, const B& p0): B(p0), py_self(py_self_) {} B_Wrapper(PyObject* py_self_): B(), py_self(py_self_) {} void Update(Data& p0) { call_method< void >(py_self, "Update", p0); } PyObject* py_self; }; }// namespace // Module ====================================================================== BOOST_PYTHON_MODULE(PysteError) { class_< Data >("Data", init< >()) .def(init< const Data& >()) .def_readwrite("a", &Data::a) ; scope* B_scope = new scope( class_< B, boost::noncopyable, B_Wrapper >("B", init< >()) .def("Update", pure_virtual(&B::Update)) ); register_ptr_to_python< boost::shared_ptr< B > >(); delete B_scope; class_< C >("C", init< >()) .def(init< const C& >()) .def("SetBptr", &C::SetBptr, with_custodian_and_ward< 1, 2 >()) .def("Update", &C::Update) .def("GetData_a", &C::GetData_a) ; class_< D, bases< B > >("D", init< >()) .def(init< const D& >()) ; } Here is the test code: Expected 10 (good) >>> from PysteError import * >>> c = C() >>> d = D() >>> c.SetBptr(d) >>> c.Update() >>> print c.GetData_a() 10 Expected -1 (bad) >>> class E(B): ... def Update(self,data): ... data.a = -1 ... >>> e = E() >>> cc = C() >>> cc.SetBptr(e) >>> cc.Update() >>> print cc.GetData_a() 0 Expected -1 (good) >>> data = Data() >>> e.Update(data) >>> print data.a -1 Hope this exposes the problem better. I honestly don't have any idea why this doesn't work. Not even a clue where to start looking for a bug. Alexis Programming Tutorial: In Python: To do this, do this In Perl: To do this, do this or this or this or this... In C: To do this, do this, but be careful In C++: To do this, do this, but don't do this, be careful of this, watch out for this, and whatever you do, don't do this __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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