generics - 我如何施放 Rc> 到 Rc>?

Question

我正在尝试转换Rc<RefCell<Data>>为Rc<RefCell<dyn Interface>>（Dataimplements Interface），但在通用方法中是不可能的：

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

trait Interface {
    fn pouet(&self);
}

struct Data {}

impl Interface for Data {
    fn pouet(&self) {
        println!("pouet");
    }
}

fn helper<T>(o: &Rc<RefCell<T>>)
where
    T: Interface,
{
    let t = o as &Rc<RefCell<dyn Interface>>;
    work(t);
}

fn work(o: &Rc<RefCell<dyn Interface>>) {
    o.borrow().pouet();
}

fn main() {
    // work
    {
        let o = Rc::new(RefCell::new(Data {}));
        work(&(o as Rc<RefCell<dyn Interface>>));
    }
    // raise an compile error
    {
        let o = Rc::new(RefCell::new(Data {}));
        helper(&o);
    }
}

我对非原始强制转换有一个编译错误：

error[E0605]: non-primitive cast: `&Rc<RefCell<T>>` as `&Rc<RefCell<dyn Interface>>`
  --> src/main.rs:20:13
   |
20 |     let t = o as &Rc<RefCell<dyn Interface>>;
   |             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ an `as` expression can only be used to convert between primitive types or to coerce to a specific trait object

操场

Answer 1

非常感谢，我明白了

解决方案是

fn helper<T>(o: &Rc<RefCell<T>>)
where
    T: Interface + 'static,
{
    let t = o.clone() as Rc<RefCell<dyn Interface>>;
    work(&t);
}

或者

fn helper<T>(o: Rc<RefCell<T>>)
where
    T: Interface + 'static,
{
    let t = o as Rc<RefCell<dyn Interface>>;
    work(&t);
}

谢谢