克隆RC <RefCell<MyType> 特征对象并进行投射

Question

This question is related to Rust: Clone and Cast Rc pointer

Let's say I have this piece of code which works fine:

use std::rc::Rc;

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

trait TraitA {}
trait TraitB {}

struct MyType {}

impl TraitAB for MyType {
    fn as_a(self: Rc<Self>) -> Rc<dyn TraitA> {self}
    fn as_b(self: Rc<Self>) -> Rc<dyn TraitB> {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.clone().as_a();
        b = ab.clone().as_b();
    }
    // Use a and b.
}

Explaining the code a bit:

• I have type called MyType which implements TraitA and TraitB .
• The goal is to have a trait object TraitA be able to get casted to TraitB and viceversa.
• So I uses a supertrait that holds the methods to do the conversions.
• This works great for std::Rc smart pointers.

So far so good. But now I need a mutable reference of both a and b , but since a and b are actually the same type instance, Rust won't let me have 2 mutable references of the same thing.

So, the common pattern for this kind of problems is std::cell::RefCell .

Note: I believe this pattern is correct in this particular case because it's a common interior mutability problem. I'm not willing to actually change the reference, but the internal state of the type only.

So following that idea I changed the following lines:

trait TraitAB : TraitA + TraitB {
    fn as_a(self: Rc<RefCell<Self>>) -> Rc<RefCell<dyn TraitA>>;
    fn as_b(self: Rc<RefCell<Self>>) -> Rc<RefCell<dyn TraitB>>;
}
//...
let mut ab: Rc<RefCell<dyn TraitAB>> = Rc::new(RefCell::new(MyType{}));

But this changes won't compile. After some reading, I found that self can only be:

• self: Self // self
• self: &Self // &self
• self: &mut Self // &mut self
• self: Box<Self> // No short form
• self: Rc<Self> // No short form / Recently supported

So this means I can't use

self: Rc<RefCell<Self>>

for the self param.

So, the main question is: Is there a way to cast an Rc<RefCell<TraitA>> to an Rc<RefCell<TraitB> ? Thanks

Answer 1

You can work around this issue by not using a receiver in TraitAB 's casting methods (ie by declaring them as associated functions ):

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

The trait can then be implemented as

impl TraitAB for MyType {
    fn as_a(it: Rc<RefCell<MyType>>) -> Rc<RefCell<dyn TraitA>> {it}
    fn as_b(it: Rc<RefCell<MyType>>) -> Rc<RefCell<dyn TraitB>> {it}
}

These functions can then be called using the fully qualified syntax.

a = TraitAB::as_a(ab.clone());
b = TraitAB::as_b(ab.clone());

The TraitAB implementation for all types will be the same. To make this implementation available for all types implementing TraitA and TraitB , you can use a generic impl :

impl<T: TraitA + TraitB + 'static> TraitAB for T {
    fn as_a(it: Rc<RefCell<T>>) -> Rc<RefCell<dyn TraitA>> {it}
    fn as_b(it: Rc<RefCell<T>>) -> Rc<RefCell<dyn TraitB>> {it}
}

Note that T: 'static because the trait objects in the function return types have an implicit 'static lifetime bound.

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