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<main>
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<h1 id="数据布局和构建"><a class="header" href="#数据布局和构建">数据布局和构建</a></h1>
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<p>聪明的读者应该已经想到了：让 <code>Rc</code> 可变，就需要使用 <code>RefCell</code> 的配合。关于 <code>RefCell</code> 的一切，在之前的章节都有介绍，还不熟悉的同学请移步<a href="https://course.rs/advance/smart-pointer/cell-refcell.html">这里</a>。</p>
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<p>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。</p>
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<h2 id="数据布局"><a class="header" href="#数据布局">数据布局</a></h2>
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<p>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：</p>
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<pre><pre class="playground"><code class="language-rust edition2021">
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<span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>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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<span class="boring">}
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</span></code></pre></pre>
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<p>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。</p>
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<h2 id="构建"><a class="header" href="#构建">构建</a></h2>
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<p>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:</p>
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<pre><pre class="playground"><code class="language-rust edition2021">
|
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<span class="boring">#![allow(unused)]
|
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</span><span class="boring">fn main() {
|
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|
</span>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 }
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>> cargo build
|
|
|
|
|
|
|
|
|
|
|
|
**一大堆 DEAD CODE 警告，但是好歹可以成功编译**
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<h2 id="push"><a class="header" href="#push">Push</a></h2>
|
|
|
|
|
|
<p>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 <code>push</code> 操作也要复杂的多。</p>
|
|
|
|
|
|
<p>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 <code>head</code> 或 <code>tail</code> 指针，但是对于空链表，则需要同时使用它们。</p>
|
|
|
|
|
|
<p>一个验证方法 <code>methods</code> 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，尾部的节点亦是如此。</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>pub fn push_front(&mut self, elem: T) {
|
|
|
|
|
|
let new_head = Node::new(elem);
|
|
|
|
|
|
match self.head.take() {
|
|
|
|
|
|
Some(old_head) => {
|
|
|
|
|
|
// 非空链表，将新的节点跟老的头部相链接
|
|
|
|
|
|
old_head.prev = Some(new_head.clone());
|
|
|
|
|
|
new_head.next = Some(old_head);
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
None => {
|
|
|
|
|
|
// 空链表，需要设置 tail 和 head
|
|
|
|
|
|
self.tail = Some(new_head.clone());
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>> cargo build
|
|
|
|
|
|
|
|
|
|
|
|
error[E0609]: no field `prev` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
|
|
|
|
|
|
--> src/fourth.rs:39:26
|
|
|
|
|
|
|
|
|
|
|
|
|
39 | old_head.prev = Some(new_head.clone()); // +1 new_head
|
|
|
|
|
|
| ^^^^ unknown field
|
|
|
|
|
|
|
|
|
|
|
|
error[E0609]: no field `next` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
|
|
|
|
|
|
--> src/fourth.rs:40:26
|
|
|
|
|
|
|
|
|
|
|
|
|
40 | new_head.next = Some(old_head); // +1 old_head
|
|
|
|
|
|
| ^^^^ unknown field
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<p>虽然有报错，但是一切尽在掌握，今天真是万事顺利啊！</p>
|
|
|
|
|
|
<p>从报错来看，我们无法直接去访问 <code>prev</code> 和 <code>next</code>，回想一下 <code>RefCell</code> 的使用方式，修改代码如下:</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>pub fn push_front(&mut self, elem: T) {
|
|
|
|
|
|
let new_head = Node::new(elem);
|
|
|
|
|
|
match self.head.take() {
|
|
|
|
|
|
Some(old_head) => {
|
|
|
|
|
|
old_head.borrow_mut().prev = Some(new_head.clone());
|
|
|
|
|
|
new_head.borrow_mut().next = Some(old_head);
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
None => {
|
|
|
|
|
|
self.tail = Some(new_head.clone());
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
|
|
|
|
|
|
warning: field is never used: `elem`
|
|
|
|
|
|
--> src/fourth.rs:12:5
|
|
|
|
|
|
|
|
|
|
|
|
|
12 | elem: T,
|
|
|
|
|
|
| ^^^^^^^
|
|
|
|
|
|
|
|
|
|
|
|
|
= note: #[warn(dead_code)] on by default
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>嘿，我又可以了！既然状态神勇，那就趁热打铁，再来看看 <code>pop</code>。</p>
|
|
|
|
|
|
<h2 id="pop"><a class="header" href="#pop">Pop</a></h2>
|
|
|
|
|
|
<p>如果说 <code>new</code> 和 <code>push</code> 是在构建链表，那 <code>pop</code> 显然就是一个破坏者。</p>
|
|
|
|
|
|
<p>何为完美的破坏？按照构建的过程逆着来一遍就是完美的！</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>pub fn pop_front(&mut self) -> Option<T> {
|
|
|
|
|
|
self.head.take().map(|old_head| {
|
|
|
|
|
|
match old_head.borrow_mut().next.take() {
|
|
|
|
|
|
Some(new_head) => {
|
|
|
|
|
|
// 非空链表
|
|
|
|
|
|
new_head.borrow_mut().prev.take();
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
None => {
|
|
|
|
|
|
// 空链表
|
|
|
|
|
|
self.tail.take();
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
old_head.elem
|
|
|
|
|
|
})
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
|
|
|
|
|
|
error[E0609]: no field `elem` on type `std::rc::Rc<std::cell::RefCell<fourth::Node<T>>>`
|
|
|
|
|
|
--> src/fourth.rs:64:22
|
|
|
|
|
|
|
|
|
|
|
|
|
64 | old_head.elem
|
|
|
|
|
|
| ^^^^ unknown field
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>哎，怎么就不长记性呢，又是 <code>RefCell</code> 惹的祸:</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>pub fn pop_front(&mut self) -> Option<T> {
|
|
|
|
|
|
self.head.take().map(|old_head| {
|
|
|
|
|
|
match old_head.borrow_mut().next.take() {
|
|
|
|
|
|
Some(new_head) => {
|
|
|
|
|
|
new_head.borrow_mut().prev.take();
|
|
|
|
|
|
self.head = Some(new_head);
|
|
|
|
|
|
}
|
|
|
|
|
|
None => {
|
|
|
|
|
|
self.tail.take();
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
old_head.borrow_mut().elem
|
|
|
|
|
|
})
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
|
|
|
|
|
|
error[E0507]: cannot move out of borrowed content
|
|
|
|
|
|
--> src/fourth.rs:64:13
|
|
|
|
|
|
|
|
|
|
|
|
|
64 | old_head.borrow_mut().elem
|
|
|
|
|
|
| ^^^^^^^^^^^^^^^^^^^^^^^^^^ cannot move out of borrowed content
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>额... 我凌乱了，看上去 <code>Box</code> 是罪魁祸首，<code>borrow_mut</code> 只能返回一个 <code>&mut Node<T></code>，因此无法拿走其所有权。</p>
|
|
|
|
|
|
<p>我们需要一个方法来拿走 <code>RefCell<T></code> 的所有权，然后返回给我们一个 <code>T</code>， 翻一翻<a href="https://doc.rust-lang.org/std/cell/struct.RefCell.html">文档</a>，可以发现下面这段内容:</p>
|
|
|
|
|
|
<blockquote>
|
|
|
|
|
|
<p><code>fn into_inner(self) -> T</code></p>
|
|
|
|
|
|
</blockquote>
|
|
|
|
|
|
<blockquote>
|
|
|
|
|
|
<p>消费掉 RefCell 并返回内部的值</p>
|
|
|
|
|
|
</blockquote>
|
|
|
|
|
|
<p>喔，看上去好有安全感的方法:</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>old_head.into_inner().elem
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
|
|
|
|
|
|
error[E0507]: cannot move out of an `Rc`
|
|
|
|
|
|
--> src/fourth.rs:64:13
|
|
|
|
|
|
|
|
|
|
|
|
|
64 | old_head.into_inner().elem
|
|
|
|
|
|
| ^^^^^^^^ cannot move out of an `Rc`
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>...看走眼了，没想到你浓眉大眼也会耍花枪。 <code>into_inner</code> 想要拿走 <code>RecCell</code> 的所有权，但是还有一个 <code>Rc</code> 不愿意，因为 <code>Rc<T></code> 只能让我们获取内部值的不可变引用。</p>
|
|
|
|
|
|
<p>大家还记得我们之前实现 <code>Drop</code> 时用过的方法吗？在这里一样适用：</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>Rc::try_unwrap(old_head).unwrap().into_inner().elem
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<p><code>Rc::try_unwrap</code> 返回一个 <code>Result</code>，由于我们不关心 <code>Err</code> 的情况( 如果代码合理，这里不会是 <code>Err</code> )，直接使用 <code>unwrap</code> 即可。</p>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
|
|
|
|
|
|
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
|
|
|
|
|
|
--> src/fourth.rs:64:38
|
|
|
|
|
|
|
|
|
|
|
|
|
64 | Rc::try_unwrap(old_head).unwrap().into_inner().elem
|
|
|
|
|
|
| ^^^^^^
|
|
|
|
|
|
|
|
|
|
|
|
|
= note: the method `unwrap` exists but the following trait bounds were not satisfied:
|
|
|
|
|
|
`std::rc::Rc<std::cell::RefCell<fourth::Node<T>>> : std::fmt::Debug`
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>额，<code>unwrap</code> 要求目标类型是实现了 <code>Debug</code> 的，这样才能在报错时提供 <code>debug</code> 输出，而 <code>RefCell<T></code> 要实现 <code>Debug</code> 需要它内部的 <code>T</code> 实现 <code>Debug</code>，而我们的 <code>Node</code> 并没有实现。</p>
|
|
|
|
|
|
<p>当然，我们可以选择为 <code>Node</code> 实现，也可以这么做：</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>Rc::try_unwrap(old_head).ok().unwrap().into_inner().elem
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
</code></pre>
|
|
|
|
|
|
<p>终于成功的运行了，下面依然是惯例 - 写几个测试用例 :</p>
|
|
|
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
|
|
|
|
|
<span class="boring">#![allow(unused)]
|
|
|
|
|
|
</span><span class="boring">fn main() {
|
|
|
|
|
|
</span>#[cfg(test)]
|
|
|
|
|
|
mod test {
|
|
|
|
|
|
use super::List;
|
|
|
|
|
|
|
|
|
|
|
|
#[test]
|
|
|
|
|
|
fn basics() {
|
|
|
|
|
|
let mut list = List::new();
|
|
|
|
|
|
|
|
|
|
|
|
// Check empty list behaves right
|
|
|
|
|
|
assert_eq!(list.pop_front(), None);
|
|
|
|
|
|
|
|
|
|
|
|
// Populate list
|
|
|
|
|
|
list.push_front(1);
|
|
|
|
|
|
list.push_front(2);
|
|
|
|
|
|
list.push_front(3);
|
|
|
|
|
|
|
|
|
|
|
|
// Check normal removal
|
|
|
|
|
|
assert_eq!(list.pop_front(), Some(3));
|
|
|
|
|
|
assert_eq!(list.pop_front(), Some(2));
|
|
|
|
|
|
|
|
|
|
|
|
// Push some more just to make sure nothing's corrupted
|
|
|
|
|
|
list.push_front(4);
|
|
|
|
|
|
list.push_front(5);
|
|
|
|
|
|
|
|
|
|
|
|
// Check normal removal
|
|
|
|
|
|
assert_eq!(list.pop_front(), Some(5));
|
|
|
|
|
|
assert_eq!(list.pop_front(), Some(4));
|
|
|
|
|
|
|
|
|
|
|
|
// Check exhaustion
|
|
|
|
|
|
assert_eq!(list.pop_front(), Some(1));
|
|
|
|
|
|
assert_eq!(list.pop_front(), None);
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
<span class="boring">}
|
|
|
|
|
|
</span></code></pre></pre>
|
|
|
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<pre><code class="language-shell">$ 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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</code></pre>
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<h2 id="drop"><a class="header" href="#drop">Drop</a></h2>
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<p>在<a href="">循环引用章节</a>，我们介绍过 <code>Rc</code> 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 <code>drop</code> 我们的链表时，两个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。</p>
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<p>所以，这里最好的实现就是将每个节点 <code>pop</code> 出去，直到获得 <code>None</code>:</p>
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<pre><pre class="playground"><code class="language-rust edition2021">
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<span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>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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<span class="boring">}
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</span></code></pre></pre>
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<p>细心的读者可能已经注意到，我们还未实现在链表尾部 <code>push</code> 和 <code>pop</code> 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。</p>
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