Boost::Python raw_function returning void

Question

Using Boost::Python, the normal mechanism for wrapping functions works correctly with C++ functions returning void. Unfortunately, the normal mechanism also has limitations, specifically with regards to the function arity it supports. So I need to use boost::python::raw_function to wrap my function, but it doesn't compile when my function returns void. Here's a simple test case:

#include <boost/python.hpp>
#include <boost/python/raw_function.hpp>

void entry_point(boost::python::tuple args, boost::python::dict kwargs) {  }

BOOST_PYTHON_MODULE(module)
{
  boost::python::def("entry_point", boost::python::raw_function(&entry_point));
}

Which gives the error:

/usr/local/include/boost/python/raw_function.hpp: In member function ‘PyObject* boost::python::detail::raw_dispatcher::operator()(PyObject*, PyObject*) [with F = void (*)(boost::python::tuple, boost::python::dict)]’:

/usr/local/include/boost/python/object/py_function.hpp:94: instantiated from ‘PyObject* boost::python::objects::full_py_function_impl::operator()(PyObject*, PyObject*) [with Caller = boost::python::detail::raw_dispatcher, Sig = boost::mpl::vector1]’

void.cpp:8: instantiated from here

/usr/local/include/boost/python/raw_function.hpp:36: error: invalid use of void expression

For the moment, I can work around this by having my function return a dummy value, but that's somewhat unsatisfying. Have other people run into this problem?

You can get around the arity issues with the normal method by #define-ing the BOOST_PYTHON_MAX_ARITY macro. — Matthew Scouten
– Matthew Scouten, Commented Nov 8, 2010 at 19:32
Thanks for pointing this out. I'm doing this now instead of raw_function, it turns out to be faster on my machine. — Bryan Catanzaro
– Bryan Catanzaro, Commented Nov 29, 2010 at 5:57

Arlaharen · Accepted Answer · 2010-11-06 12:56:10Z

2

I think this is the way that raw_function() works. It expects your function to return a Python object.

In Python the closest thing you will get to a function returning void is a function returning None. I think that approach would be best (and not even that ugly) in your case:

#include <boost/python.hpp>
#include <boost/python/raw_function.hpp>

using namespace boost::python;

namespace
{
  object entry_point(tuple args, dict kwargs) 
  {  
    return object();
  } 
}

BOOST_PYTHON_MODULE(foo)
{
  def("entry_point", raw_function(&entry_point));
}

answered Nov 6, 2010 at 12:56

Arlaharen

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1 Comment

Bryan Catanzaro Over a year ago

Thanks, this is exactly what I was hoping to see.

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