Convert a Box<T> into a PyClass

[JackAshwell11]

Say if I have this struct:

#[pyclass]
#[derive(Clone)]
pub struct Leaf {
    left: Option<Box<Leaf>>,
    right: Option<Box<Leaf>>,
    parent: Option<Box<Leaf>>,
}

How can I access the left, right, and parent fields from Python since Leaf is already a PyClass, but it is inside of a Box?

[adamreichold]

If you want to hand out references to those Leaf objects to Python code, store them in the Py smart pointer (i.e. on Python's heap with shared mutability using reference counting) instead of using Box.

Note that since using Py implies shared ownership, you might need to use interior mutability of your Leaf structure is not immutable, e.g. using Option<Py<PyCell<Leaf>>> which would the Python-compatible equivalent of Option<Rc<RefCell<Leaf>>.

[adamreichold]

Just wasn't sure if it was the best solution.

If it is only for tests, it should be the simplest solution. Note that you can use Option::cloned.

[JackAshwell11]

When using Option::cloned, I get the error expected Option<Leaf>, found Cloned<&mut Option<Box<Leaf>>>. The code:

    #[getter(left)]
    fn get_left(&self) -> Option<Leaf> {
        self.left.cloned()
    }

[adamreichold]

The call Option::as_ref is still required, i.e. self.leaf.as_ref().cloned().

[JackAshwell11]

That still returns a Option<Box<Leaf>>

[adamreichold]

Of course, the hint on Option::cloned was just code golfing to avoid the manual closure. If references are to be shared with Python, Py<_> is still the way to go.