Boost.Python: Unable to wrap C++ "operator<<" to expose print functionality in Python

Question

I am trying to wrap a C++ class that handles binary values using Boost.Python. For this class, the "<<" operator has been defined as

std::ostream &operator<<(std::ostream &output, const bin &inbin);

I tried wrapping it with

class_<bin>("bin", init<>())
    .def(str(self));

However, the compilation throws this error:

boost/python/def_visitor.hpp:31:9: error: no matching function for call to
‘boost::python::api::object::visit(boost::python::class_<itpp::bin>&) const’

I am not sure how to resolve this error, anyone have an idea?

Reference: http://www.boost.org/doc/libs/1_31_0/libs/python/doc/tutorial/doc/class_operators_special_functions.html

Why is this tagged SWIG and ctypes when it's about boost.python? — Flexo, Jan 30 '17 at 18:27
@flexo The issue could be something basic related to wrapping operator<< for python — vidit, Jan 30 '17 at 19:14
If you write a boost.python interface it has nothing to do with a swig one or a ctypes one. — Flexo, Jan 30 '17 at 19:34
I agree with @Flexo, the swig and ctypes tags should go away... — fedepad, Jan 30 '17 at 23:33
Which boost version are you using? And what about trying `.def(self_ns::str(self));` instead of `.def(str(self));`. Or `.def(self_ns::str(self_ns::self))` if the previous doesn't work? Does it help a bit? — fedepad, Jan 30 '17 at 23:38
And might be related to this post: http://stackoverflow.com/questions/4362402/what-is-wrong-with-c-streams-when-using-boost-python — fedepad, Jan 30 '17 at 23:39
Yes, that helps a lot @fedepad. The build works fine, but the output of "print obj" is still not perfect (prints the object type and address instead of stream output). Thanks. — vidit, Jan 31 '17 at 09:17
Maybe at this point would be nice if you could share your implementation of `std::ostream &operator<<(std::ostream &output, const bin &inbin);` if you can. — fedepad, Jan 31 '17 at 10:16
And did you try this `.def(self_ns::str(self_ns::self))` or `.def(self_ns::str(self));`? — fedepad, Jan 31 '17 at 10:21
@fedepad Yes, that is what solved the build issue. I'm trying to wrap a third-party library and the "operator<<" implementation can be found at http://itpp.sourceforge.net/4.3.1/vec_8h_source.html. I wrapped with .def(self_ns::str(self)), but trying "print obj" results in "<[classname] object at [memory address]>" — vidit, Jan 31 '17 at 11:04
yes, clear, but please try `.def(self_ns::str(self_ns::self));` also instead of `.def(self_ns::str(self));` before moving forward thinking about other stuff... — fedepad, Jan 31 '17 at 11:17
I did, both .def(self_ns::str(self_ns::self)) and .def(self_ns::str(self)); give same result — vidit, Jan 31 '17 at 11:21
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/134473/discussion-between-fedepad-and-vid). — fedepad, Jan 31 '17 at 11:44

score 6 · Accepted Answer · edited May 23 '17 at 12:33

This somehow seems to be covered in the following SO posts:
what is wrong with c++ streams when using boost.python?
Build problems when adding `__str__` method to Boost Python C++ class

So, according to the previous posts, you should try to substitute

.def(str(self));

with either

.def(self_ns::str(self));

or

.def(self_ns::str(self_ns::self))

It seems to have resolved the same problem in those cases.

If the above doesn't work, try writing a custom wrapper. For example, you may define a print_wrap function as:

#include <sstream>
std::string print_wrap(const classname &c)
{
  std::ostringstream oss;
  oss << "HelloWorld " << c.var1 << " " << c.var2;
  return oss.str();
}

And then in the class_<classname>("py_classname") definition use,

.def("__str__", &print_wrap)

Then in Python you should be able to get

>>> obj = py_classname(1,2)  #var1=1, var2=2
>>> print obj
>>> HelloWorld 1 2

Boost.Python: Unable to wrap C++ "operator<<" to expose print functionality in Python

1 Answers1