我在这段代码中做错了什么来实现队列的"取消排队"功能?

Question

我正在尝试为队列实现dequeue函数，但我对借用检查器的工作原理感到困惑。我在这段代码中做错了什么？

use std::cell::RefCell;
use std::rc::Rc;
use std::mem::replace;
type Link<T> = Option<Rc<RefCell<Node<T>>>>;
struct Node<T>{
item: T,
next: Link<T>
}
pub struct Queue<T>{
first: Link<T>,
last: Link<T>,
length: usize
}
impl<T> Queue<T>{
pub fn new() -> Queue<T> {
Queue {
first: None,
last: None,
length: 0
}
}
pub fn is_empty(&self) -> bool {
self.length == 0
}
pub fn size(&self) -> usize {
self.length
}
pub fn enqueue(&mut self, item: T) {
let temp = self.last.take();
self.last = Some(Rc::new(RefCell::new(Node{
item,
next: None
})));
if self.is_empty() {
self.first = self.last.clone();
} else {
let temp = temp.unwrap();
temp.borrow_mut().next = self.last.clone();
}
self.length += 1;
}
pub fn dequeue(&mut self) -> Result<T, String>{
if let Some(ref mut value) = self.first.take() {
let mut temp = *(value.borrow_mut());
let next = *(temp.next.unwrap().borrow_mut());
let old_value = replace(&mut temp, next);
return Ok(old_value.item);
}
Err("Queue is empty".to_owned())
}
}

在获得了对Some内部值的可变引用后，我想替换为节点的next字段正在引用的节点。我需要拥有Some内价值的所有权吗？我能做到吗？

Answer 1

这是dequeue的实现：

pub fn dequeue(&mut self) -> Result<T, String> {
// First, disconnect `self.last` from the element it is pointing,
// since it will have to be updated anyway. If there is no elemen in the
// queue, we're done.
let first = try!(self.first.take().ok_or("Queue is empty".to_owned()));
// if there are two Rc's pointing to this node, then this must be the
// only node, so `self.last` has to go
if Rc::strong_count(&first) == 2 {
self.last = None;
}
let first_node = Rc::try_unwrap(first).ok().expect(
"This should be the only owner of the node"
).into_inner();
self.first = first_node.next;
self.length -= 1;
Ok(first_node.item)
}

这是完整的代码。我还添加了一个dequeue_back，使这几乎是一个双端队列和一些测试。