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@ -0,0 +1,316 @@
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# 数据布局和构建
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聪明的读者应该已经想到了:让 `Rc` 可变,就需要使用 `RefCell` 的配合。关于 `RefCell` 的一切,在之前的章节都有介绍,还不熟悉的同学请移步[这里](https://course.rs/advance/smart-pointer/cell-refcell.html)。
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好了,绝世神兵在手,接下来...我们将见识一个绝世啰嗦的数据结构...如果你来自 GC 语言,那很可能就没有见识过这种阵仗。
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## 数据布局
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双向链表意味着每一个节点将同时指向前一个和下一个节点,因此我们的数据结构可能会变成这样:
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```rust
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use std::rc::Rc;
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use std::cell::RefCell;
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pub struct List<T> {
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head: Link<T>,
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tail: Link<T>,
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}
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type Link<T> = Option<Rc<RefCell<Node<T>>>>;
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struct Node<T> {
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elem: T,
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next: Link<T>,
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prev: Link<T>,
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}
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```
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耳听忐忑,心怀忐忑,尝试编译下,竟然顺利通过了,thanks god! 接下来再来看看该如何使用它。
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## 构建
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如果按照之前的构建方式来构建新的数据结构,会有点笨拙,因此我们先尝试将其拆分:
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```rust
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impl<T> Node<T> {
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fn new(elem: T) -> Rc<RefCell<Self>> {
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Rc::new(RefCell::new(Node {
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elem: elem,
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prev: None,
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next: None,
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}))
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}
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}
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impl<T> List<T> {
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pub fn new() -> Self {
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List { head: None, tail: None }
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}
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}
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```
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```rust
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> cargo build
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**一大堆 DEAD CODE 警告,但是好歹可以成功编译**
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```
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## Push
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很好,再来向链表的头部推入一个元素。由于双向链表的数据结构和操作逻辑明显更加复杂,因此相比单向链表的单行实现,双向链表的 `push` 操作也要复杂的多。
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除此之外,我们还需要处理一些关于空链表的边界问题:对于绝大部分操作而言,可能只需要使用 `head` 或 `tail` 指针,但是对于空链表,则需要同时使用它们。
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一个验证方法 `methods` 是否有效的办法就是看它是否能保持不变性, 每个节点都应该有两个指针指向它: 中间的节点被它前后的节点所指向,而头部的端节点除了被它后面的节点所指向外,还会被链表本身所指向,尾部的端节点亦是如此。
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```rust
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pub fn push_front(&mut self, elem: T) {
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let new_head = Node::new(elem);
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match self.head.take() {
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Some(old_head) => {
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// 非空链表,将新的节点跟老的头部相链接
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old_head.prev = Some(new_head.clone());
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new_head.next = Some(old_head);
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self.head = Some(new_head);
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}
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None => {
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// 空链表,需要设置 tail 和 head
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self.tail = Some(new_head.clone());
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self.head = Some(new_head);
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}
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}
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}
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```
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```rust
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cargo build
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error[E0609]: no field `prev` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
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--> src/fourth.rs:39:26
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39 | old_head.prev = Some(new_head.clone()); // +1 new_head
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| ^^^^ unknown field
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error[E0609]: no field `next` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
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--> src/fourth.rs:40:26
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40 | new_head.next = Some(old_head); // +1 old_head
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| ^^^^ unknown field
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```
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虽然有报错,但是一切尽在掌握,今天真是万事顺利啊!
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从报错来看,我们无法直接去访问 `prev` 和 `next`,回想一下 `RefCell` 的使用方式,修改代码如下:
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```rust
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pub fn push_front(&mut self, elem: T) {
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let new_head = Node::new(elem);
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match self.head.take() {
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Some(old_head) => {
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old_head.borrow_mut().prev = Some(new_head.clone());
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new_head.borrow_mut().next = Some(old_head);
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self.head = Some(new_head);
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}
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None => {
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self.tail = Some(new_head.clone());
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self.head = Some(new_head);
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}
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}
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}
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```
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```shell
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> cargo build
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warning: field is never used: `elem`
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--> src/fourth.rs:12:5
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12 | elem: T,
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| ^^^^^^^
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= note: #[warn(dead_code)] on by default
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```
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嘿,我又可以了!既然状态神勇,那就趁热打铁,再来看看 `pop`。
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## Pop
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如果说 `new` 和 `push` 是在构建链表,那 `pop` 显然就是一个破坏者。
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何为完美的破坏?按照构建的过程逆着来一遍就是完美的!
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```rust
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pub fn pop_front(&mut self) -> Option<T> {
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self.head.take().map(|old_head| {
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match old_head.borrow_mut().next.take() {
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Some(new_head) => {
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// 非空链表
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new_head.borrow_mut().prev.take();
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self.head = Some(new_head);
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}
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None => {
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// 空链表
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self.tail.take();
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}
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}
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old_head.elem
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})
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}
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```
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```shell
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> cargo build
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error[E0609]: no field `elem` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
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--> src/fourth.rs:64:22
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64 | old_head.elem
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| ^^^^ unknown field
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```
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哎,怎么就不长记性呢,又是 `RefCell` 惹的祸:
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```rust
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pub fn pop_front(&mut self) -> Option<T> {
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self.head.take().map(|old_head| {
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match old_head.borrow_mut().next.take() {
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Some(new_head) => {
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new_head.borrow_mut().prev.take();
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self.head = Some(new_head);
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}
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None => {
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self.tail.take();
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}
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}
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old_head.borrow_mut().elem
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})
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}
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```
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```shell
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cargo build
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error[E0507]: cannot move out of borrowed content
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--> src/fourth.rs:64:13
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64 | old_head.borrow_mut().elem
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| ^^^^^^^^^^^^^^^^^^^^^^^^^^ cannot move out of borrowed content
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```
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额... 我凌乱了,看上去 `Box` 是罪魁祸首,`borrow_mut` 只能返回一个 `&mut Node<T>`,因此无法拿走其所有权。
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我们需要一个方法来拿走 `RefCell<T>` 的所有权,然后返回给我们一个 `T`, 翻一翻[文档](https://doc.rust-lang.org/std/cell/struct.RefCell.html),可以发现下面这段内容:
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> `fn into_inner(self) -> T`
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> 消费掉 RefCell 并返回内部的值
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喔,看上去好有安全感的方法:
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```rust
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old_head.into_inner().elem
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```
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```shell
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> cargo build
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error[E0507]: cannot move out of an `Rc`
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--> src/fourth.rs:64:13
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64 | old_head.into_inner().elem
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| ^^^^^^^^ cannot move out of an `Rc`
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```
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...看走眼了,没想到你浓眉大眼也会耍花枪。 `into_inner` 想要拿走 `RecCell` 的所有权,但是还有一个 `Rc` 不愿意,因为 `Rc<T>` 只能让我们获取内部值的不可变引用。
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大家还记得我们之前实现 `Drop` 时用过的方法吗?在这里一样适用:
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```rust
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Rc::try_unwrap(old_head).unwrap().into_inner().elem
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```
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`Rc::try_unwrap` 返回一个 `Result`,由于我们不关心 `Err` 的情况( 如果代码合理,这里不会是 `Err` ),直接使用 `unwrap` 即可。
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```shell
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> cargo build
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error[E0599]: no method named `unwrap` found for type `std::result::Result<std::cell::RefCell<fourth::Node<T>>, std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>>` in the current scope
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--> src/fourth.rs:64:38
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64 | Rc::try_unwrap(old_head).unwrap().into_inner().elem
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| ^^^^^^
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= note: the method `unwrap` exists but the following trait bounds were not satisfied:
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`std::rc::Rc<std::cell::RefCell<fourth::Node<T>>> : std::fmt::Debug`
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```
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额,`unwrap` 要求目标类型是实现了 `Debug` 的,这样才能在报错时提供 `debug` 输出,而 `RefCell<T>` 要实现 `Debug` 需要它内部的 `T` 实现 `Debug`,而我们的 `Node` 并没有实现。
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当然,我们可以选择为 `Node` 实现,也可以这么做:
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```rust
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Rc::try_unwrap(old_head).ok().unwrap().into_inner().elem
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```
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```shell
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cargo build
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```
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终于成功的运行了,下面依然是惯例 - 写几个测试用例 :
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```rust
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#[cfg(test)]
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mod test {
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use super::List;
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#[test]
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fn basics() {
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let mut list = List::new();
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// Check empty list behaves right
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assert_eq!(list.pop_front(), None);
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// Populate list
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list.push_front(1);
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list.push_front(2);
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list.push_front(3);
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// Check normal removal
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assert_eq!(list.pop_front(), Some(3));
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assert_eq!(list.pop_front(), Some(2));
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// Push some more just to make sure nothing's corrupted
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list.push_front(4);
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list.push_front(5);
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// Check normal removal
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assert_eq!(list.pop_front(), Some(5));
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assert_eq!(list.pop_front(), Some(4));
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// Check exhaustion
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assert_eq!(list.pop_front(), Some(1));
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assert_eq!(list.pop_front(), None);
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}
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}
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```
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```shell
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cargo test
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Running target/debug/lists-5c71138492ad4b4a
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running 9 tests
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test first::test::basics ... ok
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test fourth::test::basics ... ok
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test second::test::iter_mut ... ok
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test second::test::basics ... ok
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test fifth::test::iter_mut ... ok
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test third::test::basics ... ok
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test second::test::iter ... ok
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test third::test::iter ... ok
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test second::test::into_iter ... ok
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test result: ok. 9 passed; 0 failed; 0 ignored; 0 measured
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```
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## Drop
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在[循环引用章节](),我们介绍过 `Rc` 最怕的就是引用形成循环,而双向链表恰恰如此。因此,当使用默认的实现来 `drop` 我们的链表时,两个端节点会将各自的引用计数减少到 1, 然后就不会继续减少,最终造成内存泄漏。
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所以,这里最好的实现就是将每个节点 `pop` 出去,直到获得 `None`:
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```rust
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impl<T> Drop for List<T> {
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fn drop(&mut self) {
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while self.pop_front().is_some() {}
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}
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}
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```
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细心的读者可能已经注意到,我们还未实现在链表尾部 `push` 和 `pop` 的操作,但由于所需的实现跟之前差别不大,因此我们会在后面直接给出,下面先来看看更有趣的。
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