这是我的代码:
struct Node<T> {
data: T,
next: Option<Box<Node<T>>>,
}
impl<T> Node<T> {
fn new(data : T) -> Node<T> {
Node { data: data, next: None }
}
fn new_with_next(data: T, next: Option<Box<Node<T>>>) -> Node<T> {
Node { data: data, next: next }
}
}
struct LinkedList<T> {
head: Box<Node<T>>,
size: u8,
}
impl<T> LinkedList<T> {
fn new(data: T) -> LinkedList<T> {
let new_node = Node::new(data);
let head = Box::new(new_node);
LinkedList { head: head, size: 1 }
}
fn insert(&mut self, data: T) {
let mut next = Some(self.head); // <-- error here
let new_node = Node::new_with_next(data, next);
self.head = Box::new(new_node);
self.size += 1;
}
}
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我收到此错误:
src\linked_list.rs:28:29: 28:33 error: cannot move out of borrowed content [E0507]
src\linked_list.rs:28 let next = Some(self.head);
^~~~
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我不明白错误,也不了解如何修复错误.我试图给一个参考self.head来Some,但我改变了数据类型这里面Some这种方式.
问题是你self通过引用,因此你不能移出它head(它将变得无效),这let mut next = Some(self.head);将做.
std::mem::replace 这是一个很好的功能:
fn insert(&mut self, data: T) {
let mut next = std::mem::replace(&mut self.head, Box::new(Node::new(data)));
self.head.next = Some(next);
self.size += 1;
}
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它可以让你用一个新节点替换头,并得到旧的同时,这锈是高兴,因为head保持有效,一直以来现.