Just return boost::python::object by value; it's actually a smart
pointer that carries a PyObject* and uses Python's reference counting,
and you can implicitly create one from boost::python::list (since list
derives from object).
Just returning boost::python::list by value might also work, but that
copy constructor might be overloaded to do deep copies (I'm not sure),
and you seem to be trying to avoid that.
Jim
On 08/09/2011 02:21 PM, fjanoos wrote:
Hello,
I'm new to boost::python and was having trouble wrapping a C++ function of
the form
void FindVCs(int vId, vector<int>& vcs);
Here vcs is allocated in the caller and populated by FindVCs.
Initially, I considered wrapping it something like this:
boost::python::list* FindVCs_wrap(int vid)
{
vector<int> vcs;
FindVCs(vid,vcs);
//wrap and return
boost::python::list* out_list = new boost::python::list;
unsigned int num_elems = vcs.size();
for( unsigned int idx = 0; idx< num_elems; idx++){
out_list->append(vcs[idx]);
}
return out_list;
}
however, this gives me compile time errors
Error 5 error C2027: use of undefined type
'boost::python::detail::specify_a_return_value_policy_to_wrap_functions_returning<T>'
Then I change the return type to
boost::shared_ptr<boost::python::list> FindVCs_wrap(int vid)
{
...
return boost::shared_ptr<boost::python::list>(out_list);
}
This compiles fine but then at runtime Python raises:
TypeError: No to_python (by-value) converter found for C++ type: class
boost::shared_ptr<class boost
::python::list>
Any ideas of what I am doing wrong ?
thanks,
-fj
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