我试图为队列实现dequeue
函数,但我很困惑借阅检查器是如何工作的。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())
}
}
获得Some
内部值的可变引用后,我想替换节点的next
字段所引用的节点。我是否需要获取Some
内的值的所有权?我甚至可以这样做吗?
答案 0 :(得分:1)
以下是dequeue
:
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)
}
Here is the complete code。我还添加了dequeue_back
来使这几乎成为双端队列和一些测试。