what is wrong with c++ streams when using boost.python?

Question

Update 2: I'm not sure why this is still being upvoted (March 2014). This appears to be fixed since I asked this question many years ago. Make sure you're using a recent version of boost.

UPDATE: Perhaps C++ streams need to be initialized in order to format numbers, and the initialization is not happening when the shared library is loaded in Python?

I am calling

cout << 1 << "!" << endl;

in a method that is exported to a shared library via boost.python. It doesn't print anything, but if I do

cout << "%" << "!" << endl;

it works.

This is important because I want to do this:

ostream& operator <<(ostream &os, const Bernoulli& b) {
    ostringstream oss;
    oss << b.p() * 100.0 << "%";
    return os << oss.str();
}

I exposed that by doing this:

BOOST_PYTHON_MODULE(libdistributions)
{
    class_<Bernoulli>("Bernoulli")
        .def(init<>())
        .def(init<double>())

        .def("p", &Bernoulli::p)
        .def("set_p", &Bernoulli::set_p)
        .def("not_p", &Bernoulli::not_p)

        .def("Entropy", &Bernoulli::Entropy)
        .def("KL", &Bernoulli::KL)
        .def(self_ns::str(self))
    ;
}

but when I call the str method in python on a Bernoulli object, I get nothing. I suspect the simpler cout problem is related.

I did not have problem with iostreams and boost.python... maybe the problem comes from a more subtile bug? However, the technique in the docs (1dl.us/dAD) didn't work for me, I had to write .def("__str__", &print_wrapper<Bernouilli>). What is self_ns?. Also, in your method, why not just {return os << b.p() * 100.0 << "%"; }? — rafak
– rafak, Commented Dec 6, 2010 at 19:59
@rafak the reason for doing it via ostringstream is explained in this question: stackoverflow.com/questions/2249018/… — Neil G
– Neil G, Commented Dec 6, 2010 at 20:31
@rafak where is print_wrapper defined? I can't find it in boost. — Neil G
– Neil G, Commented Dec 6, 2010 at 20:35
I should have said that it is mine: template <class C> inline std::string print_wrapper(const C& obj) { std::ostringstream os; os << obj; return os.str(); } — rafak
– rafak, Commented Dec 6, 2010 at 21:03
Possible duplicate of: stackoverflow.com/questions/2828903/… — vz0
– vz0, Commented Jan 25, 2011 at 4:17

Community · Accepted Answer · 2017-05-23 11:43:09Z

3

I also run into this problem a while ago, using self_ns as outlined in the answers here Build problems when adding `__str__` method to Boost Python C++ class

The reason for using self_ns is explained by Dave himself here http://mail.python.org/pipermail/cplusplus-sig/2004-February/006496.html

Just for the sake of debugging, please try

inline std::string toString(const Bernoulli& b) 
{
   std::ostringstream s;
   s << b;
   return s.str(); 
}

And replace .def(self_ns::str(self)) with

class_<Bernoulli>("Bernoulli")
[...]
.def("__str__", &toString)

edited May 23, 2017 at 11:43

CommunityBot

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answered Nov 12, 2012 at 10:56

H. Brandsmeier

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6 Comments

Neil G Over a year ago

I used self_ns as shown in the question, and it didn't fix this problem.

H. Brandsmeier Over a year ago

Can you try .def(self_ns::str(self_ns::self))?

H. Brandsmeier Over a year ago

Also you could try to add a `std::cout << "hello world << std::endl" inside your operator<< and report if your function gets called at all.

Neil G Over a year ago

Hi, I did try that code. I'm pretty sure that Boost.Python is not properly initializing the integer-to-string code for streams. Using a printf works, but using ostream doesn't.

H. Brandsmeier Over a year ago

@Neil G: Did you try what I wrote under "Just for the sake of debugging", did that work?

|

UncleZeiv · Accepted Answer · 2011-01-07 14:21:12Z

2

Have you tried using boost::format instead? Since you are already using boost, it shouldn't be too much of a hassle.

boost::format( "%d%%" ) % ( b.p() * 100.0 )

Another thing, try passing std::endl explicitly to the output os.

answered Jan 7, 2011 at 14:21

UncleZeiv

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6 Comments

Neil G Over a year ago

Thanks. The number is still not being shown, but the "%" sign is there. It appears that the buffer in which the number is stringified is not being created.

UncleZeiv Over a year ago

Hm. I'll try to reproduce the problem then.

Neil G Over a year ago

Thank you. Please let me know if you can't reproduce it and if you ever find a solution.

UncleZeiv Over a year ago

Ok, I wrote an empty Bernoulli class, with p defined as double p() {return 2.5;}; in Python, str(b) prints 250% as expected. I used boost 1.42 and Python 2.5. Which versions are you using?

Neil G Over a year ago

I'm using boost 1.48 and Python 3.1 now.

|

Vukasin Toroman · Accepted Answer · 2011-01-11 19:55:22Z

0

have you tried flushing the stream prior to using the .str() method?

oss << b.p() * 100.0 << "%" << std::flush;

answered Jan 11, 2011 at 19:55

Vukasin Toroman

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