Rust，如何从 Rc> 复制内部值并返回它

Brief:我是Rust的新手，所以我决定通过实现一个双向链表来练习.出于调试目的，我实现了get()方法，但未能将值从Rc<RefCell<_>>中复制出来.(很抱歉问了这么愚蠢的问题)

Problem:我try 在.get()中返回Result<T, &'static str>，其中T是 node 中存储的数据类型，&str是错误消息字符串.borrow 判断器告诉我不能返回对方法内变量的引用，所以我try 将内部值复制出来并返回它，但失败了.

Source code:

use std::{rc::Rc, cell::RefCell};

struct Node<T> {
    data: Option<T>,
    prev: Option<Rc<RefCell<Node<T>>>>,
    next: Option<Rc<RefCell<Node<T>>>>,
}

impl<T> Node<T> {
    /// Instantiate a new dummy node.
    /// This node is used to mark the start and end of the list.
    /// It is not counted in the size of the list.
    fn new() -> Self {
        Node {
            data: None,
            prev: None,
            next: None,
        }
    }
    /// Instantiate a new content node.
    /// This node is used to store data.
    /// It is counted in the size of the list.
    fn from(data: T) -> Self {
        Node {
            data: Some(data),
            prev: None,
            next: None,
        }
    }
}

struct List<T> {
    head: Rc<RefCell<Node<T>>>,
    tail: Rc<RefCell<Node<T>>>,
    size: usize,
}

impl<T> List<T> {
    pub fn new() -> Self {
        let head = Rc::new(RefCell::new(Node::new()));
        let tail = Rc::new(RefCell::new(Node::new()));
        head.borrow_mut().next = Some(Rc::clone(&tail));
        tail.borrow_mut().prev = Some(Rc::clone(&head));
        List { head, tail, size: 0 }
    }
    pub fn prepend(&self, data: T) {
        let node = Rc::new(RefCell::new(Node::from(data)));
        let mut head = self.head.borrow_mut();
        
        node.borrow_mut().next = Some(head.next.take().unwrap());
        node.borrow_mut().prev = Some(Rc::clone(&self.head));
        head.next = Some(Rc::clone(&node));
        if let Some(next) = node.borrow().next.as_ref() {
            next.borrow_mut().prev = Some(Rc::clone(&node));
        };
    }
    pub fn append(&self, data: T) {
        let node = Rc::new(RefCell::new(Node::from(data)));
        let mut tail = self.tail.borrow_mut();
        
        node.borrow_mut().prev = Some(Rc::clone(&tail.prev.take().unwrap()));
        node.borrow_mut().next = Some(Rc::clone(&self.tail));
        tail.prev = Some(Rc::clone(&node));
        if let Some(prev) = node.borrow().prev.as_ref() {
            prev.borrow_mut().next = Some(Rc::clone(&node));
        };
    }
    pub fn get(&self, index: isize) -> Result<T, &'static str> {
        let mut current: Rc<RefCell<Node<T>>> = Rc::clone(self.head.borrow().next.as_ref().unwrap());
        for _ in 0..index {
            let tmp = Rc::clone(current.borrow().next.as_ref().ok_or("Index out of range")?);
            current = tmp;
        }
        let result = current.borrow().data.as_ref().ok_or("Index out of range")?;  // error[E0716]
        Ok(*result)  // error[E0507]
    }
}
/*
error[E0716]: temporary value dropped while borrowed
  --> src\linked.rs:74:22
   |
74 |         let result = current.borrow().data.as_ref().ok_or("Index out of range")?;
   |                      ^^^^^^^^^^^^^^^^                                           - temporary value is freed at the end of this statement
   |                      |
   |                      creates a temporary value which is freed while still in use
75 |         Ok(*result)
   |            ------- borrow later used here
   |
help: consider using a `let` binding to create a longer lived value
   |
74 ~         let binding = current.borrow();
75 ~         let result = binding.data.as_ref().ok_or("Index out of range")?;
   |

error[E0507]: cannot move out of `*result` which is behind a shared reference
  --> src\linked.rs:75:12
   |
75 |         Ok(*result)
   |            ^^^^^^^ move occurs because `*result` has type `T`, which does not implement the `Copy` trait
*/

I've Tried:

• This post about making a reference into value美元，但这就是我想要做的，但失败了.
• This post about modifying a RefCell，但这无济于事.我是想把它还回go ，而不是变异它.
• This post about how to borrow a RefCell，但我不能返回借来的值，因为借来的Rc是短期的(但内在价值不是).
• This post about how to return something inside a RefCell和this post about how to return it with .map()，但当我try 使用.into()时，编译器会说"特征界限不满意"，如果我删除.into()，borrow 判断器会抱怨"无法移出".
• This post about using Rc::try_unwarp()，但它不会起作用，因为内部数据有多个所有者.

Also:我可能做错了这些，如果其中一个帖子解决了我的问题，但我没有正确地执行，请原谅我，并请教我如何正确地执行它.非常感谢.