How do I pass Rc>> to a function accepting Rc>>?

Question

I have originally asked this question here, but it was marked as duplicate, although it duplicates only a part of it in my opinion, so I have created a more specific one:

Answer 1

(An older revision of this answer essentially advised to clone the underlying struct and put it in a new Rc> object; this was necessary at the time on stable Rust, but since not long after that time, Rc> will coerce to Rc> without trouble.)

Drop the Box<> wrapping, and you can coerce Rc> to Rc> freely and easily. Recalling that cloning an Rc just produces another Rc, increasing the refcount by one, you can do something like this:

use std::rc::Rc;
use std::cell::RefCell;

trait MyTrait {
    fn trait_func(&self);
}

#[derive(Clone)]
struct MyStruct1;
impl MyStruct1 {
    fn my_fn(&self) {
        // do something
    }
}

impl MyTrait for MyStruct1 {
    fn trait_func(&self) {
        // do something
    }
}

fn my_trait_fn(t: Rc>) {
    t.borrow_mut().trait_func();
}

fn main() {
    // (The type annotation is not necessary here, but helps explain it.
    // If the `my_str.borrow().my_fn()` line was missing, it would actually
    // be of type Rc> instead of Rc>,
    // essentially doing the coercion one step earlier.)
    let my_str: Rc> = Rc::new(RefCell::new(MyStruct1));
    my_trait_fn(my_str.clone());
    my_str.borrow().my_fn();
}

As a general rule, see if you can make things take the contained value by reference, ideally even generically—fn my_trait_fn(t: &T) and similar, which can typically be called as my_str.borrow() with automatic referencing and dereferencing taking care of the rest—rather than the whole Rc> thing.