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339 lines
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1 month ago
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<h1 id="无处不在的迭代器"><a class="header" href="#无处不在的迭代器">无处不在的迭代器</a></h1>
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<p>Rust 的迭代器无处不在，直至你在它上面栽了跟头，经过深入调查才发现：哦，原来是迭代器的锅。不信的话，看看这个报错你能想到是迭代器的问题吗: <code>borrow of moved value: words</code>.</p>
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<h2 id="报错的代码"><a class="header" href="#报错的代码">报错的代码</a></h2>
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<p>以下的代码非常简单，用来统计文本中字词的数量，并打印出来：</p>
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<pre><pre class="playground"><code class="language-rust edition2021">fn main() {
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let s = "hello world";
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let mut words = s.split(" ");
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let n = words.count();
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println!("{:?}",words);
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}</code></pre></pre>
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<p>四行代码，行云流水，一气呵成，且看成效：</p>
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<pre><code class="language-console">error[E0382]: borrow of moved value: `words`
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--> src/main.rs:5:21
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3 | let mut words = s.split(" ");
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| --------- move occurs because `words` has type `std::str::Split<'_, &str>`, which does not implement the `Copy` trait
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4 | let n = words.count();
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| ------- `words` moved due to this method call
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5 | println!("{:?}",words);
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| ^^^^^ value borrowed here after move
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</code></pre>
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<p>世事难料，我以为只有的生命周期、闭包才容易背叛革命，没想到一个你浓眉大眼的<code>count</code>方法也背叛革命。从报错来看，是因为<code>count</code>方法拿走了<code>words</code>的所有权，来看看签名：</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>fn count(self) -> usize
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<span class="boring">}</span></code></pre></pre>
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<p>从签名来看，编译器的报错是正确的，但是为什么？为什么一个简单的标准库<code>count</code>方法就敢拿走所有权？</p>
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<h2 id="迭代器回顾"><a class="header" href="#迭代器回顾">迭代器回顾</a></h2>
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<p>在<a href="https://course.rs/advance/functional-programing/iterator.html#%E6%B6%88%E8%B4%B9%E8%80%85%E4%B8%8E%E9%80%82%E9%85%8D%E5%99%A8">迭代器</a>章节中，我们曾经学习过两个概念：迭代器适配器和消费者适配器，前者用于对迭代器中的元素进行操作，最终生成一个新的迭代器，例如<code>map</code>、<code>filter</code>等方法；而后者用于消费掉迭代器，最终产生一个结果，例如<code>collect</code>方法, 一个典型的示例如下：</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>let v1: Vec<i32> = vec![1, 2, 3];
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let v2: Vec<_> = v1.iter().map(|x| x + 1).collect();
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assert_eq!(v2, vec![2, 3, 4]);
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<span class="boring">}</span></code></pre></pre>
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<p>在其中，我们还提到一个细节，消费者适配器会拿走迭代器的所有权，那么这个是否与我们最开始碰到的问题有关系？</p>
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<h2 id="深入调查"><a class="header" href="#深入调查">深入调查</a></h2>
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<p>要解释这个问题，必须要找到<code>words</code>是消费者适配器的证据，因此我们需要深入源码进行查看。</p>
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<p>其实。。也不需要多深，只要进入<code>words</code>的源码，就能看出它属于<code>Iterator</code>特征，那说明<code>split</code>方法产生了一个迭代器？再来看看：</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 fn split<'a, P>(&'a self, pat: P) -> Split<'a, P>
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where
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P: Pattern<'a>,
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//An iterator over substrings of this string slice, separated by characters matched by a pattern.
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<span class="boring">}</span></code></pre></pre>
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<p>还真是，从代码注释来看，<code>Split</code>就是一个迭代器类型，用来迭代被分隔符隔开的子字符串集合。</p>
|
||
|
|
<p>真相大白了，<code>split</code>产生一个迭代器，而<code>count</code>方法是一个消费者适配器，用于消耗掉前者产生的迭代器，最终生成字词统计的结果。</p>
|
||
|
|
<p>本身问题不复杂，但是在<strong>解决方法上，可能还有点在各位客官的意料之外</strong>，且看下文。</p>
|
||
|
|
<h2 id="最-rusty-的解决方法"><a class="header" href="#最-rusty-的解决方法">最 rusty 的解决方法</a></h2>
|
||
|
|
<p>你可能会想用<code>collect</code>来解决这个问题，先收集成一个集合，然后进行统计。当然此方法完全可行，但是很不<code>rusty</code>(很符合 rust 规范、潮流的意思)，以下给出最<code>rusty</code>的解决方案：</p>
|
||
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
||
|
|
</span><span class="boring">fn main() {
|
||
|
|
</span>let words = s.split(",");
|
||
|
|
let n = words.clone().count();
|
||
|
|
<span class="boring">}</span></code></pre></pre>
|
||
|
|
<p>在继续之前，我得先找一个地方藏好，因为俺有一个感觉，烂西红柿正在铺天盖地的呼啸而来，伴随而来的是读者的正义呵斥：
|
||
|
|
<strong>你管<code>clone</code>叫最好、最<code>rusty</code>的解决方法？？</strong></p>
|
||
|
|
<p>大家且听我慢慢道来，事实上，在 Rust 中<code>clone</code>不总是性能低下的代名词，因为<code>clone</code>的行为完全取决于它的具体实现。</p>
|
||
|
|
<h4 id="迭代器的clone代价"><a class="header" href="#迭代器的clone代价">迭代器的<code>clone</code>代价</a></h4>
|
||
|
|
<p>对于迭代器而言，它其实并不需要持有数据才能进行迭代，事实上它包含一个引用，该引用指向了保存在堆上的数据，而迭代器自身的结构是保存在栈上。</p>
|
||
|
|
<p>因此对迭代器的<code>clone</code>仅仅是复制了一份栈上的简单结构，性能非常高效，例如:</p>
|
||
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
||
|
|
</span><span class="boring">fn main() {
|
||
|
|
</span>pub struct Split<'a, T: 'a, P>
|
||
|
|
where
|
||
|
|
P: FnMut(&T) -> bool,
|
||
|
|
{
|
||
|
|
// Used for `SplitWhitespace` and `SplitAsciiWhitespace` `as_str` methods
|
||
|
|
pub(crate) v: &'a [T],
|
||
|
|
pred: P,
|
||
|
|
// Used for `SplitAsciiWhitespace` `as_str` method
|
||
|
|
pub(crate) finished: bool,
|
||
|
|
}
|
||
|
|
|
||
|
|
impl<T, P> Clone for Split<'_, T, P>
|
||
|
|
where
|
||
|
|
P: Clone + FnMut(&T) -> bool,
|
||
|
|
{
|
||
|
|
fn clone(&self) -> Self {
|
||
|
|
Split { v: self.v, pred: self.pred.clone(), finished: self.finished }
|
||
|
|
}
|
||
|
|
}
|
||
|
|
<span class="boring">}</span></code></pre></pre>
|
||
|
|
<p>以上代码实现了对<code>Split</code>迭代器的克隆，可以看出，底层的的数组<code>self.v</code>并没有被克隆而是简单的复制了一个引用，依然指向了底层的数组<code>&[T]</code>，因此这个克隆非常高效。</p>
|
||
|
|
<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
|
||
|
|
<p>看起来是无效借用导致的错误，实际上是迭代器被消费了导致的问题，这说明 Rust 编译器虽然会告诉你错误原因，但是这个原因不总是根本原因。我们需要一双慧眼和勤劳的手，来挖掘出这个宝藏，最后为己所用。</p>
|
||
|
|
<p>同时，克隆在 Rust 中也并不总是<strong>bad guy</strong>的代名词，有的时候我们可以大胆去使用，当然前提是了解你的代码场景和具体的<code>clone</code>实现，这样你也能像文中那样作出非常<code>rusty</code>的选择。</p>
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