[C++-sig] boost::python::object referencing c++ object

[David Abrahams]

"Bjorn Pettersen" <BPettersen at NAREX.com> writes:

> I've got the following code to assign a wrapped object to a Python
> variable:
>
>   template <class T>
>   void set(const std::string& name, const T& value) {
>     PyObject* key = PyString_FromString(name.c_str());
>     object val(value);
>     PyDict_SetItem(interpreter()->mainmodule(), key, val.ptr());
>     Py_DECREF(key);
>   }

Why make life hard for yourself?

  template <class T>
  void set(const std::string& name, const T& value) {
    interpreter()->mainmodule()[name] = value;
  }

Of course, this relies on interpeter()->mainmodule() returning an
object (or dict, or...)

> and I'd like to be able to set() a variable to a c++ object, then later
> mutate the object and have the changes visible to Python. I haven't
> traced through the code completely, but it looks like the object ctor
> makes a copy(?) 

More precisely, it converts its argument into a Python object.
If your C++ object is a wrapped class type, it will be copied.

> Is there any way to accomplish what I'd like to do?

I'm not sure exactly what you want.

Perhaps you'd like the resulting Python object to contain a raw
pointer to the argument?  In that case, the caveat is that if the
lifetime of the C++ object ends before that of the Python object, that
pointer will dangle and using the Python object may cause a crash.

There is a way to do that, but it's more convoluted than it should be:

  template <class T>
  T& identity(T& x)
  {
      return x;
  }

  template <class T>
  object get_object_reference(T& x)
  {
      // build a function object around identity
      object f
          = make_function(
                  &identity<T>, return_value_policy<reference_existing_object>());

      // and call
      return f(x);
  }

HTH,
-- 
                       David Abrahams
   dave at boost-consulting.com * http://www.boost-consulting.com
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