[C++-sig] returning a list of extension types

[David Abrahams]

on Fri Jul 11 2008, ""Alkis Evlogimenos ('Αλκης Ευλογημένος)"" <alkis-AT-evlogimenos.com> wrote:

> I have a class I want to expose to python, which has a method that returns a list of instances
> of itself. What I did was to define a function:
>
> list MakeListOfFoo(Foo* foo) {
>   list foos;
>   foos.append(new Foo());
>   foos.append(new Foo());
>   return foos;
> }
>
> and in the module definition:
>
> BOOST_PYTHON_MODULE(Bar) {
>   class_<Foo, boost::noncopyable>("Foo")
>       .def("MakeListOfFoo", &MakeListOfFoo);
> }
>
> The above while it returns a list, it doesn't convert Foo* to a python
> Foo object. How do I go about doing that?

What type is "list," really?  Is there a reason not to do something like

  boost::python::list MakeListOfFoo(Foo* foo) {
    boost::python::list foos;
    foos.append(Foo());  // No "new"
    foos.append(Foo());
    return foos;
  }

?

If you really need reference semantics, use boost::shared_ptr:

  boost::python::list MakeListOfFoo(Foo* foo) {
    boost::python::list foos;

    foos.append(boost::shared_ptr<Foo>(new Foo));
    foos.append(boost::shared_ptr<Foo>(new Foo));

    return foos;
  }

If your list is boost::python::list, what you've done above would leak
the things you are trying to append, and store a copy in the list.

-- 
Dave Abrahams
BoostPro Computing
http://www.boostpro.com