rust - 如何获取 rc::Ref 对 rc::Weak> 指向的节点的引用？

Question

我正在编写一个链表，其中存在链接前一个节点的弱引用。

use std::cell::{Ref, RefCell, RefMut};
use std::rc::{Rc, Weak};

pub struct List<T> {
    head: NextLink<T>,
    tail: PrevLink<T>,
}

type NextLink<T> = Option<Rc<RefCell<Node<T>>>>;
type PrevLink<T> = Weak<RefCell<Node<T>>>;

struct Node<T> {
    elem: T,
    prev: PrevLink<T>,
    next: NextLink<T>,
}

impl<T> Node<T> {
    fn new(elem: T) -> Rc<RefCell<Self>> {
        Rc::new(RefCell::new(Self {
            elem,
            prev: Weak::new(),
            next: None,
        }))
    }
}

impl<T> List<T> {
    pub fn new() -> Self {
        Self {
            head: None,
            tail: Weak::new(),
        }
    }

    // add to tail
    pub fn push(&mut self, elem: T) {
        let new_node = Node::new(elem);
        match self.tail.upgrade().take() {
            Some(old_tail) => {
                new_node.borrow_mut().prev = Rc::downgrade(&old_tail);
                old_tail.borrow_mut().next = Some(new_node);
            }
            None => {
                self.tail = Rc::downgrade(&new_node);
                self.head = Some(new_node);
            }
        }
    }

    pub fn pop(&mut self) -> Option<T> {
        self.tail.upgrade().map(|old_tail| {
            match old_tail.borrow_mut().prev.upgrade() {
                Some(new_tail) => {
                    new_tail.borrow_mut().next = None;
                    self.tail = Rc::downgrade(&new_tail);
                }
                None => {
                    self.head.take();
                    self.tail = Weak::new();
                }
            };
            Rc::try_unwrap(old_tail).ok().unwrap().into_inner().elem
        })
    }

    // add to head
    pub fn unshift(&mut self, elem: T) {
        let new_node = Node::new(elem);
        match self.head.take() {
            Some(old_head) => {
                old_head.borrow_mut().prev = Rc::downgrade(&new_node);
                new_node.borrow_mut().next = Some(old_head);
            }
            None => {
                self.tail = Rc::downgrade(&new_node);
                self.head = Some(new_node);
            }
        }
    }

    pub fn shift(&mut self) -> Option<T> {
        self.head.take().map(|old_head| {
            match old_head.borrow_mut().next.take() {
                Some(new_head) => {
                    new_head.borrow_mut().prev = Weak::new();
                    self.head = Some(new_head);
                }
                None => {
                    self.tail = Weak::new();
                }
            };
            Rc::try_unwrap(old_head).ok().unwrap().into_inner().elem
        })
    }

    pub fn peek_head(&self) -> Option<Ref<T>> {
        self.head
            .as_ref()
            .map(|node| Ref::map(node.borrow(), |node| &node.elem))
    }

    pub fn peek_tail(&self) -> Option<Ref<T>> {
        self.tail
            .upgrade()
            .map(|node| Ref::map(node.borrow(), |node| &node.elem))
    }

    pub fn peek_head_mut(&mut self) -> Option<RefMut<T>> {
        self.head
            .as_ref()
            .map(|node| RefMut::map(node.borrow_mut(), |node| &mut node.elem))
    }

    pub fn peek_tail_mut(&mut self) -> Option<RefMut<T>> {
        unimplemented!()
    }
}

impl<T> Drop for List<T> {
    fn drop(&mut self) {
        while let Some(old_head) = self.head.take() {
            self.head = old_head.borrow_mut().next.take();
        }
    }
}

存在编译错误:

error[E0515]: cannot return value referencing function parameter `node`
   --> src/lib.rs:106:25
    |
106 |             .map(|node| Ref::map(node.borrow(), |node| &node.elem))
    |                         ^^^^^^^^^----^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    |                         |        |
    |                         |        `node` is borrowed here
    |                         returns a value referencing data owned by the current function

好像是当尾节点的弱引用升级为Rc时，这个Rc是这个函数拥有的，最后会被消费掉。

我该如何解决？

Answer 1

你的问题

使用 Weak::upgrade 和 RefCell::borrow :

fn example<T>(foo: Weak<RefCell<T>>) {
    let strong: Rc<RefCell<T>> = foo.upgrade().unwrap();
    let the_ref: Ref<T> = strong.borrow();
}

你的问题

self.head.as_ref()返回 Option<&Rc<_>> . self.tail.upgrade()返回 Option<Rc<_>> .当您调用 Option::map ，内部值(value)的所有权转移到闭包，你得到你发布的错误。您可以调用Option::as_ref ，但是 Option由函数拥有，因此不允许返回引用，因为一旦 Rc 就不再有效被摧毁。

您唯一可以做的就是返回 Rc来自函数。

另见:

• How do I borrow a reference to what is inside an Option<T>?
• Is there any way to return a reference to a variable created in a function?
• How do I return a reference to something inside a RefCell without breaking encapsulation?