Interior mutability - cannot return value referencing temporary value

[svenschmidt75]

Hi,

For fun, I implemented a queue with two stacks. The queue's peek method should take &self , instead of &mut self as the mutation is not observable from the outside. However, when using RefCell's, I get error
message

cannot return value referencing temporary value

which I understand, but don;t know a way around. Does anybody know how to do this? Code here.

Thank you,

[RustyYato]

You need to either return Ref or return a smart pointer that wraps Ref. You can't return a normal reference.

[alice]

Note that in your &mut self methods, you should prefer to use RefCell::get_mut instead of borrow or borrow_mut as get_mut is guaranteed to succeed. To solve the returning reference issue, you can return a Ref instead.

fn peek(&self) -> Option<Ref<u64>> {
    if self.dequeue_stack.borrow().is_empty() {
        let mut deque_ref = self.dequeue_stack.borrow_mut();
        let mut enque_ref = self.enqueue_stack.borrow_mut();
        while !enque_ref.is_empty() {
            let item = enque_ref.pop();
            deque_ref.push(item);
        }
        if deque_ref.is_empty() {
            return None;
        }
    }
    Some(Ref::map(self.dequeue_stack.borrow(), |stack| stack.peek().unwrap()))
}

playground

Note that the existence of this Ref will cause a panic if borrow_mut is called on dequeue_stack before the Ref is dropped. Thus the code above is designed to only call borrow_mut if items need to be moved over, which means that if a Ref still exists, the Queue is borrowed immutably, so you cannot call the &mut self methods that modify it, meaning that the api above is panic-free.

[svenschmidt75]

Thank you both for your replies.
I was hoping I can avoid Ref is the method signature as that would leak
implementation details.

[alice]

You can't avoid it with RefCell, sorry.

[RustyYato]

You can wrap the Ref in your own type, but you can't get rid of it.

[svenschmidt75]

OK, thank you both very much. Good to know...