casting - How do I clone a Rc trait object and cast it to another trait object?

Question

This is a follow up question from Rust dynamic cast trait object between different taits. The solution provided there works really well when we use references for trait objects, but this time I am trying to do the same with Rc pointers. For example

• I have a super trait named TraitAB and 2 traits named TraitA and TraitB
• So when I first create a trait object of type TraitAB instead of using a Box, now I use an Rc pointer.
• I need a variable of type TraitA to be a reference of ab

Here I made a very minimal example:

use std::rc::Rc;

trait TraitAB: TraitA + TraitB {
    fn as_a(&self) -> Rc<dyn TraitA>;
    fn as_b(&self) -> Rc<dyn TraitB>;
}

trait TraitA {}
trait TraitB {}

struct MyType {}

impl TraitAB for MyType {
    fn as_a(&self) -> Rc<dyn TraitA> {
        Rc::clone(self)
    }
    fn as_b(&self) -> Rc<dyn TraitB> {
        Rc::clone(self)
    }
}

impl TraitA for MyType {}
impl TraitB for MyType {}

fn main() {
    let a: Rc<dyn TraitA>;
    let b: Rc<dyn TraitB>;
    {
        let mut ab: Rc<dyn TraitAB> = Rc::new(MyType {});
        a = ab.as_a();
        b = ab.as_b();
    }
}

This doesn't work though. According to the error messages:

error[E0308]: mismatched types
  --> src/main.rs:15:19
   |
15 |         Rc::clone(self)
   |                   ^^^^ expected struct `std::rc::Rc`, found struct `MyType`
   |
   = note: expected reference `&std::rc::Rc<dyn TraitA>`
              found reference `&MyType`

error[E0308]: mismatched types
  --> src/main.rs:18:19
   |
18 |         Rc::clone(self)
   |                   ^^^^ expected struct `std::rc::Rc`, found struct `MyType`
   |
   = note: expected reference `&std::rc::Rc<dyn TraitB>`
              found reference `&MyType`

as_a and as_b can't know self is actually an Rc pointer. Is there a way to do the cast of a cloned shared pointer?

Answer 1

methods as_a and as_b can't know self is actually an Rc pointer.

Actually, that's not true! There's a rarely used feature that allows self to be taken as various standard kinds of references (Rc<Self>, Box<Self>, etc.).

That means that you can rewrite your TraitAB as

trait TraitAB : TraitA + TraitB {
    fn as_a(self: Rc<Self>) -> Rc<dyn TraitA>;
    fn as_b(self: Rc<Self>) -> Rc<dyn TraitB>;
}

Unfortunately, as written, as_a and as_b move self: Rc<Self>, since Rc<T> doesn't implement Copy (only Clone). One way to fix this is to simply clone ab before passing it into these methods. This also means that you don't need to clone the self inside the method. (playground link)

let ab: Rc<dyn TraitAB> = Rc::new(MyType{});
let _a: Rc<dyn TraitA> = ab.clone().as_a();
let _b: Rc<dyn TraitB> = ab.clone().as_b();

Using the nightly-only feature arbitrary_self_types, it's possible to make as_a and as_b take self as &Rc<Self> (which looks weird to me since it's a reference to a reference). This allows ab.as_a() to be called without moving ab. The only problem with this approach is that TraitAB is no longer object-safe¹, so Rc<dyn TraitAB> no longer works. (playground link).

---

1. According to the tracking issue for arbitrary self types, the object safety question is still open. I'm not really sure what the rules are right now.