use std::mem;
pub struct List<T> {
head: Link<T>,
tail: Link<T>, // NEW!
}
type Link<T> = Option<Box<Node<T>>>;
struct Node<T> {
elem: T,
next: Link<T>,
}
impl<T> List<T> {
pub fn new() -> Self {
List { head: None, tail: None }
}
pub fn push(&mut self, elem: T) {
let new_tail = Box::new(Node {
elem: elem,
// When you push onto the tail, your next is always None
next: None,
});
// swap the old tail to point to the new tail
let old_tail = mem::replace(&mut self.tail, Some(new_tail));
match old_tail {
Some(mut old_tail) => {
// If the old tail existed, update it to point to the new tail
old_tail.next = Some(new_tail);
}
None => {
// Otherwise, update the head to point to it
self.head = Some(new_tail);
}
}
}
}
> cargo build
error[E0382]: use of moved value: `new_tail`
--> src/fifth.rs:38:38
|
26 | let new_tail = Box::new(Node {
| -------- move occurs because `new_tail` has type `std::boxed::Box<fifth::Node<T>>`, which does not implement the `Copy` trait
...
33 | let old_tail = mem::replace(&mut self.tail, Some(new_tail));
| -------- value moved here
...
38 | old_tail.next = Some(new_tail);
| ^^^^^^^^ value used here after move
使用 move value: new tail
没有实现 Copy，所以我们不能把它分配到两个位置。更重要的是，Box 拥有它所指向的东西，并且会在它 drop 时释放它。如果我们编译了 push 实现，我们就可以双倍释放列表的尾部了！实际上，正如我们所写的那样，我们的代码会在每次推送时释放 old_tail。哎呀！
好吧，我们知道如何制作一个非所有权指针。 那只是参考！
pub struct List<T> {
head: Link<T>,
tail: Option<&mut Node<T>>, // NEW!
}
type Link<T> = Option<Box<Node<T>>>;
struct Node<T> {
elem: T,
next: Link<T>,
}
impl<T> List<T> {
pub fn new() -> Self {
List { head: None, tail: None }
}
pub fn push(&mut self, elem: T) {
let new_tail = Box::new(Node {
elem: elem,
// When you push onto the tail, your next is always None
next: None,
});
// Put the box in the right place, and then grab a reference to its Node
let new_tail = match self.tail.take() {
Some(old_tail) => {
// If the old tail existed, update it to point to the new tail
old_tail.next = Some(new_tail);
old_tail.next.as_mut().map(|node| &mut **node)
}
None => {
// Otherwise, update the head to point to it
self.head = Some(new_tail);
self.head.as_mut().map(|node| &mut **node)
}
};
self.tail = new_tail;
}
}
pub struct List<'a, T> {
head: Link<T>,
tail: Option<&'a mut Node<T>>, // NEW!
}
type Link<T> = Option<Box<Node<T>>>;
struct Node<T> {
elem: T,
next: Link<T>,
}
impl<'a, T> List<'a, T> {
pub fn new() -> Self {
List { head: None, tail: None }
}
pub fn push(&mut self, elem: T) {
let new_tail = Box::new(Node {
elem: elem,
// When you push onto the tail, your next is always None
next: None,
});
// Put the box in the right place, and then grab a reference to its Node
let new_tail = match self.tail.take() {
Some(old_tail) => {
// If the old tail existed, update it to point to the new tail
old_tail.next = Some(new_tail);
old_tail.next.as_mut().map(|node| &mut **node)
}
None => {
// Otherwise, update the head to point to it
self.head = Some(new_tail);
self.head.as_mut().map(|node| &mut **node)
}
};
self.tail = new_tail;
}
}
cargo build
error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
--> src/fifth.rs:35:27
|
35 | self.head.as_mut().map(|node| &mut **node)
| ^^^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 18:5...
--> src/fifth.rs:18:5
|
18 | / pub fn push(&mut self, elem: T) {
19 | | let new_tail = Box::new(Node {
20 | | elem: elem,
21 | | // When you push onto the tail, your next is always None
... |
39 | | self.tail = new_tail;
40 | | }
| |_____^
note: ...so that reference does not outlive borrowed content
--> src/fifth.rs:35:17
|
35 | self.head.as_mut().map(|node| &mut **node)
| ^^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 13:6...
--> src/fifth.rs:13:6
|
13 | impl<'a, T> List<'a, T> {
| ^^
= note: ...so that the expression is assignable:
expected std::option::Option<&'a mut fifth::Node<T>>
found std::option::Option<&mut fifth::Node<T>>
cargo build
warning: field is never used: `elem`
--> src/fifth.rs:9:5
|
9 | elem: T,
| ^^^^^^^
|
= note: #[warn(dead_code)] on by default
哦，嘿，成功了! 太好了！
让我们也做一下 pop 吧:
pub fn pop(&'a mut self) -> Option<T> {
// Grab the list's current head
self.head.take().map(|head| {
let head = *head;
self.head = head.next;
// If we're out of `head`, make sure to set the tail to `None`.
if self.head.is_none() {
self.tail = None;
}
head.elem
})
}
并为此写一个快速测试:
mod test {
use super::List;
#[test]
fn basics() {
let mut list = List::new();
// Check empty list behaves right
assert_eq!(list.pop(), None);
// Populate list
list.push(1);
list.push(2);
list.push(3);
// Check normal removal
assert_eq!(list.pop(), Some(1));
assert_eq!(list.pop(), Some(2));
// Push some more just to make sure nothing's corrupted
list.push(4);
list.push(5);
// Check normal removal
assert_eq!(list.pop(), Some(3));
assert_eq!(list.pop(), Some(4));
// Check exhaustion
assert_eq!(list.pop(), Some(5));
assert_eq!(list.pop(), None);
}
}
cargo test
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:68:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
68 | list.push(1);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:69:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
69 | list.push(2);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:70:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
70 | list.push(3);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
....
** WAY MORE LINES OF ERRORS **
....
error: aborting due to 11 previous errors
🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀🙀
我的天啊。
编译器对我们所有人的呕吐没有错。我们刚犯了一个主要的Rust罪过：我们在自己的内部存储了对自己的引用。 不知何故，我们设法说服了Rust，这在我们的push和pop实施中是完全合理的（我们确实为之震惊）。 我相信原因是Rust不能仅仅通过 push 和 pop 来告诉引用是对我们自己的引用，或者更确切地说，Rust根本没有这个概念。 引用失败的行为只是一种新出现的行为。