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603 lines
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9 months ago
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<ol class="chapter"><li class="chapter-item affix "><a href="../../about-book.html">关于本书</a></li><li class="chapter-item affix "><a href="../../into-rust.html">进入 Rust 编程世界</a></li><li class="chapter-item affix "><a href="../../first-try/sth-you-should-not-do.html">避免从入门到放弃</a></li><li class="chapter-item affix "><a href="../../community.html">社区和锈书</a></li><li class="spacer"></li><li class="chapter-item affix "><a href="../../some-thoughts.html">Datav: 可编程的数据可视化平台和可观测性平台</a></li><li class="chapter-item affix "><li class="part-title">Rust 语言基础学习</li><li class="spacer"></li><li class="chapter-item "><a href="../../first-try/intro.html"><strong aria-hidden="true">1.</strong> 寻找牛刀,以便小试</a><a class="toggle"><div>❱</div></a></li><li><ol class="section"><li class="chapter-item "><a href="../../first-try/installation.html"><strong aria-hidden="true">1.1.</strong> 安装 Rust 环境</a></li><li class="chapter-item "><a href="../../first-try/editor.html"><strong aria-hidden="true">1.2.</strong> 墙推 VSCode!</a></li><li class="chapter-item "><a href="../../first-try/cargo.html"><strong aria-hidden="true">1.3.</strong> 认识 Cargo</a></li><li class="chapter-item "><a href="../../first-try/hello-world.html"><strong aria-hidden="true">1.4.</strong> 不仅仅是 Hello world</a></li><li class="chapter-item "><a href="../../first-try/slowly-downloading.html"><strong aria-hidden="true">1.5.</strong> 下载依赖太慢了?</a></li></ol></li><li class="chapter-item "><a href="../../basic/intro.html"><strong aria-hidden="true">2.</strong> Rust 基础入门</a><a class="toggle"><div>❱</div></a></li><li><ol class="section"><li class="chapter-item "><a href="../../basic/variable.html"><strong aria-hidden="true">2.1.</strong> 变量绑定与解构</a></li><li class="chapter-item "><a href="../../basic/base-type/index.html"><strong aria-hidden="true">2.2.</strong> 基本类型</a><a class="toggle"><div>❱</div></a></li><li><ol class="section"><li class="chapter-item "><a href="../../basic/base-type/numbers.html"><strong aria-hidden="true">2.2.1.</strong> 数值类型</a></li><li class="chapter-item "><a href="../../basic/base-type/char-bool.html"><strong aria-hidden="true">2.2.2.</strong> 字符、布尔、单元类型</a></li><li class="chapter-item "><a href="../../basic/base-type/statement-expression.html"><strong aria-hidden="true">2.2.3.</strong> 语句与表达式</a></li><li class="chapter-item "><a href="../../basic/base-type/function.html"><strong aria-hidden="true">2.2.4.</strong> 函数</a></li></ol></li><li class="chapter-item "><a href="../../basic/ownership/index.html"><strong aria-hidden="true">2.3.</strong> 所有权和借用</a><a class="toggle"><div>❱</div></a></li><li><ol class="section"><li class="chapter-item "><a href="../../basic/ownership/ownership.html"><strong aria-hidden="true">2.3.1.</strong> 所有权</a></li><li class="chapter-item "><a href="../../basic/ownership/borrowing.html"><strong aria-hidden="true">2.3.2.</strong> 引用与借用</a></li></ol></li><li class="chapter-item "><a href="../../basic/compound-type/intro.html"><strong aria-hidden="true">2.4.</strong> 复合类型</a><a class="toggle"><div>❱</div></a></li><li><ol class="section"><li class="chapter-item "><a href="../../basic/compound-type/string-slice.html"><strong aria-hidden="true">2.4.1.</strong> 字符串与切片</a></li><li class="chapter-item "><a href="../../basic/compound-type/tuple.html"><strong aria-hidden="true">2.4.2.</strong> 元组</a></li><li class="chapter-item "><a href="../../basic/compound-type/struct.html"><strong aria-hidden="true">2.4.3.</strong> 结构体</a></li><li class="chapter-item "><a href="../../basic/compound-type/enum.html"><strong aria-hidden="true">2.4.4.</strong> 枚举</a></li><li class="chapter-item "><a href="../../basic/compound-type/array.html"><strong aria-hidden="true">2.4.5.</strong> 数组</a></li></ol></li><li class="chapter-item "><a href="../../basic/flow-contro
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<h1 id="数据布局"><a class="header" href="#数据布局">数据布局</a></h1>
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<p>那么单向链表的队列长什么样?对于栈来说,我们向一端推入( push )元素,然后再从同一端弹出( pop )。对于栈和队列而言,唯一的区别在于队列从末端弹出。</p>
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<p>栈的实现类似于下图:</p>
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<pre><code class="language-shell">input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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stack push X:
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[Some(ptr)] -> (X, Some(ptr)) -> (A, Some(ptr)) -> (B, None)
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stack pop:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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</code></pre>
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<p>由于队列是首端进,末端出,因此我们需要决定将 <code>push</code> 和 <code>pop</code> 中的哪个放到末端去操作,如果将 <code>push</code> 放在末端操作:</p>
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<pre><code class="language-shell">input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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flipped push X:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
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</code></pre>
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<p>而如果将 <code>pop</code> 放在末端:</p>
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<pre><code class="language-shell">input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
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flipped pop:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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</code></pre>
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<p>但是这样实现有一个很糟糕的地方:两个操作都需要遍历整个链表后才能完成。队列要求 <code>push</code> 和 <code>pop</code> 操作需要高效,但是遍历整个链表才能完成的操作怎么看都谈不上高效!</p>
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<p>其中一个解决办法就是保存一个指针指向末端:</p>
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<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>use std::mem;
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pub struct List<T> {
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head: Link<T>,
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tail: Link<T>, // NEW!
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}
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type Link<T> = Option<Box<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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}
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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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pub fn push(&mut self, elem: T) {
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let new_tail = Box::new(Node {
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elem: elem,
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// 在尾端推入一个新节点时,新节点的下一个节点永远是 None
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next: None,
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});
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// 让 tail 指向新的节点,并返回之前的 old tail
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let old_tail = mem::replace(&mut self.tail, Some(new_tail));
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match old_tail {
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Some(mut old_tail) => {
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// 若 old tail 存在,则让该节点指向新的节点
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old_tail.next = Some(new_tail);
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}
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None => {
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// 否则,将 head 指向新的节点
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self.head = Some(new_tail);
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}
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}
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}
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}
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<span class="boring">}</span></code></pre></pre>
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<p>在之前的各种链表锤炼下,我们对于相关代码应该相当熟悉了,因此可以适当提提速 - 在写的过程中,事实上我碰到了很多错误,这些错误就不再一一列举。</p>
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<p>但是如果你担心不再能看到错误,那就纯属多余了:</p>
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<pre><code class="language-shell">$ cargo build
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error[E0382]: use of moved value: `new_tail`
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--> src/fifth.rs:38:38
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26 | let new_tail = Box::new(Node {
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| -------- move occurs because `new_tail` has type `std::boxed::Box<fifth::Node<T>>`, which does not implement the `Copy` trait
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...
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33 | let old_tail = mem::replace(&mut self.tail, Some(new_tail));
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| -------- value moved here
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...
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38 | old_tail.next = Some(new_tail);
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| ^^^^^^^^ value used here after move
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</code></pre>
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<p>新鲜出炉的错误,接好!<code>Box</code> 并没有实现 <code>Copy</code> 特征,因此我们不能在两个地方进行赋值。好在,可以使用没有所有权的引用类型:</p>
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<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
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</span>pub struct List<T> {
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head: Link<T>,
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tail: Option<&mut Node<T>>, // NEW!
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}
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type Link<T> = Option<Box<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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}
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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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pub fn push(&mut self, elem: T) {
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let new_tail = Box::new(Node {
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elem: elem,
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next: None,
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});
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let new_tail = match self.tail.take() {
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||
|
Some(old_tail) => {
|
||
|
old_tail.next = Some(new_tail);
|
||
|
old_tail.next.as_deref_mut()
|
||
|
}
|
||
|
None => {
|
||
|
self.head = Some(new_tail);
|
||
|
self.head.as_deref_mut()
|
||
|
}
|
||
|
};
|
||
|
|
||
|
self.tail = new_tail;
|
||
|
}
|
||
|
}
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<pre><code class="language-shell">$ cargo build
|
||
|
|
||
|
error[E0106]: missing lifetime specifier
|
||
|
--> src/fifth.rs:3:18
|
||
|
|
|
||
|
3 | tail: Option<&mut Node<T>>, // NEW!
|
||
|
| ^ expected lifetime parameter
|
||
|
</code></pre>
|
||
|
<p>好吧,结构体中的引用类型需要显式的标注生命周期,先加一个 <code>'a</code> 吧:</p>
|
||
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
||
|
</span><span class="boring">fn main() {
|
||
|
</span>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,
|
||
|
next: None,
|
||
|
});
|
||
|
|
||
|
let new_tail = match self.tail.take() {
|
||
|
Some(old_tail) => {
|
||
|
old_tail.next = Some(new_tail);
|
||
|
old_tail.next.as_deref_mut()
|
||
|
}
|
||
|
None => {
|
||
|
self.head = Some(new_tail);
|
||
|
self.head.as_deref_mut()
|
||
|
}
|
||
|
};
|
||
|
|
||
|
self.tail = new_tail;
|
||
|
}
|
||
|
}
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<pre><code class="language-shell">$ cargo build
|
||
|
|
||
|
error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
|
||
|
--> src/fifth.rs:35:27
|
||
|
|
|
||
|
35 | self.head.as_deref_mut()
|
||
|
| ^^^^^^^^^^^^
|
||
|
|
|
||
|
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_deref_mut()
|
||
|
| ^^^^^^^^^
|
||
|
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>>
|
||
|
</code></pre>
|
||
|
<p>好长... Rust 为啥这么难... 但是,这里有一句重点:</p>
|
||
|
<blockquote>
|
||
|
<p>the lifetime must be valid for the lifetime 'a as defined on the impl</p>
|
||
|
</blockquote>
|
||
|
<p>意思是说生命周期至少要和 <code>'a</code> 一样长,是不是因为编译器为 <code>self</code> 推导的生命周期不够长呢?我们试着来手动标注下:</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(&'a mut self, elem: T) {
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<p>当当当当,成功通过编译:</p>
|
||
|
<pre><code class="language-shell">$ cargo build
|
||
|
|
||
|
warning: field is never used: `elem`
|
||
|
--> src/fifth.rs:9:5
|
||
|
|
|
||
|
9 | elem: T,
|
||
|
| ^^^^^^^
|
||
|
|
|
||
|
= note: #[warn(dead_code)] on by default
|
||
|
</code></pre>
|
||
|
<p>这个错误可以称之为错误之王,但是我们依然成功的解决了它,太棒了!再来实现下 <code>pop</code>:</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(&'a mut self) -> Option<T> {
|
||
|
self.head.take().map(|head| {
|
||
|
let head = *head;
|
||
|
self.head = head.next;
|
||
|
|
||
|
if self.head.is_none() {
|
||
|
self.tail = None;
|
||
|
}
|
||
|
|
||
|
head.elem
|
||
|
})
|
||
|
}
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<p>看起来不错,写几个测试用例溜一溜:</p>
|
||
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
||
|
</span><span class="boring">fn main() {
|
||
|
</span>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);
|
||
|
}
|
||
|
}
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<pre><code class="language-shell">$ 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
|
||
|
</code></pre>
|
||
|
<p>🙀🙀🙀,震惊!但编译器真的没错,因为都是我们刚才那个标记惹的祸。</p>
|
||
|
<p>我们为 <code>self</code> 标记了 <code>'a</code>,意味着在 <code>'a</code> 结束前,无法再去使用 <code>self</code>,大家可以自己推断下 <code>'a</code> 的生命周期是多长。</p>
|
||
|
<p>那么该怎么办?回到老路 <code>RefCell</code> 上?显然不可能,那只能祭出大杀器:裸指针。</p>
|
||
|
<blockquote>
|
||
|
<p>事实上,上文的问题主要是自引用引起的,感兴趣的同学可以查看<a href="https://course.rs/advance/circle-self-ref/intro.html">这里</a>深入阅读。</p>
|
||
|
</blockquote>
|
||
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
||
|
</span><span class="boring">fn main() {
|
||
|
</span>pub struct List<T> {
|
||
|
head: Link<T>,
|
||
|
tail: *mut Node<T>, // DANGER DANGER
|
||
|
}
|
||
|
|
||
|
type Link<T> = Option<Box<Node<T>>>;
|
||
|
|
||
|
struct Node<T> {
|
||
|
elem: T,
|
||
|
next: Link<T>,
|
||
|
}
|
||
|
<span class="boring">}</span></code></pre></pre>
|
||
|
<p>如上所示,当使用裸指针后, <code>head</code> 和 <code>tail</code> 就不会形成自引用的问题,也不再违反 Rust 严苛的借用规则。</p>
|
||
|
<blockquote>
|
||
|
<p>注意!当前的实现依然是有严重问题的,在后面我们会修复</p>
|
||
|
</blockquote>
|
||
|
<p>果然,编程的最高境界就是回归本质:使用 C 语言的东东。</p>
|
||
|
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