on Wed Dec 17 2008, "Roman Yakovenko" <roman.yakovenko-AT-gmail.com> wrote:
> 2008/12/18 lin yun <[email protected]>: >> Hi, folks: >> >> I am trying to wrap a c++ function that returns a boost::shared_ptr<some >> class> type using boost.python, is that possible? > > Yes. I believe Py++ deals with boost::shared_ptr without invoking the > user. Just use default call policy. Boost.Python itself also deals with shared_ptr without involving the user. -- Dave Abrahams BoostPro Computing http://www.boostpro.com _______________________________________________ Cplusplus-sig mailing list [email protected] http://mail.python.org/mailman/listinfo/cplusplus-sig
