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351 lines
61 KiB
351 lines
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2 years ago
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<main>
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<h1 id="peek"><a class="header" href="#peek">Peek</a></h1>
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<p><code>push</code> 和 <code>pop</code> 的防不胜防的编译报错着实让人出了些冷汗，下面来看看轻松的，至少在之前的链表中是很轻松的 :)</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>pub fn peek_front(&self) -> Option<&T> {
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self.head.as_ref().map(|node| {
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&node.elem
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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>额...好像被人发现我是复制黏贴的了，赶紧换一个:</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>pub fn peek_front(&self) -> Option<&T> {
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self.head.as_ref().map(|node| {
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// BORROW!!!!
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&node.borrow().elem
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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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<pre><code class="language-shell">$ cargo build
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error[E0515]: cannot return value referencing temporary value
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--> src/fourth.rs:66:13
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66 | &node.borrow().elem
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| ^ ----------^^^^^
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| | |
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| | temporary value created here
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| returns a value referencing data owned by the current function
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</code></pre>
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<p>从报错可以看出，原因是我们引用了局部的变量并试图在函数中返回。为了解释这个问题，先来看看 <code>borrow</code> 的定义:</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>fn borrow<'a>(&'a self) -> Ref<'a, T>
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fn borrow_mut<'a>(&'a self) -> RefMut<'a, T>
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<span class="boring">}
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</span></code></pre></pre>
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<p>这里返回的并不是 <code>&T</code> 或 <code>&mut T</code>，而是一个 <a href="https://doc.rust-lang.org/std/cell/struct.Ref.html"><code>Ref</code></a> 和 <a href="https://doc.rust-lang.org/std/cell/struct.RefMut.html"><code>RefMut</code></a>，那么它们是什么？说白了，它们就是在借用到的引用外包裹了一层。而且 <code>Ref</code> 和 <code>RefMut</code> 分别实现了 <code>Deref</code> 和 <code>DerefMut</code>，在绝大多数场景中，我们都可以像使用 <code>&T</code> 一样去使用它们。</p>
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<p>只能说是成是败都赖萧何，恰恰就因为这一层包裹，导致生命周期改变了，也就是 <code>Ref</code> 和内部引用的生命周期不再和 <code>RefCell</code> 相同，而 <code>Ref</code> 的生命周期是什么，相信大家都能看得出来，因此就造成了局部引用的问题。</p>
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<p>事实上，这是必须的，如果内部的引用和外部的 <code>Ref</code> 生命周期不一致，那该如何管理？当 <code>Ref</code> 因超出作用域被 <code>drop</code> 时，内部的引用怎么办？</p>
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<p>现在该怎么办？我们只想要一个引用，现在却多了一个 <code>Ref</code> 拦路虎。等等，如果我们不返回 <code>&T</code> 而是返回 <code>Ref</code> 呢？</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::cell::{Ref, RefCell};
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pub fn peek_front(&self) -> Option<Ref<T>> {
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self.head.as_ref().map(|node| {
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node.borrow()
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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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<pre><code class="language-shell">$ cargo build
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error[E0308]: mismatched types
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--> src/fourth.rs:64:9
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64 | / self.head.as_ref().map(|node| {
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65 | | node.borrow()
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66 | | })
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| |__________^ expected type parameter, found struct `fourth::Node`
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= note: expected type `std::option::Option<std::cell::Ref<'_, T>>`
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found type `std::option::Option<std::cell::Ref<'_, fourth::Node<T>>>`
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</code></pre>
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<p>嗯，类型不匹配了，要返回的是 <code>Ref<T></code> 但是获取的却是 <code>Ref<Node<T>></code>，那么现在看上去有两个选择：</p>
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<ul>
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<li>抛弃这条路，换一条重新开始</li>
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<li>一条路走到死，最终通过更复杂的实现来解决</li>
|
||
|
|
</ul>
|
||
|
|
<p>但是，仔细想想，这两个选择都不是我们想要的，那没办法了，只能继续深挖，看看有没有其它解决办法。啊哦，还真发现了一只野兽：</p>
|
||
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
||
|
|
<span class="boring">#![allow(unused)]
|
||
|
|
</span><span class="boring">fn main() {
|
||
|
|
</span>map<U, F>(orig: Ref<'b, T>, f: F) -> Ref<'b, U>
|
||
|
|
where F: FnOnce(&T) -> &U,
|
||
|
|
U: ?Sized
|
||
|
|
<span class="boring">}
|
||
|
|
</span></code></pre></pre>
|
||
|
|
<p>就像在 <code>Result</code> 和 <code>Option</code> 上使用 <code>map</code> 一样，我们还能在 <code>Ref</code> 上使用 <code>map</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 peek_front(&self) -> Option<Ref<T>> {
|
||
|
|
self.head.as_ref().map(|node| {
|
||
|
|
Ref::map(node.borrow(), |node| &node.elem)
|
||
|
|
})
|
||
|
|
}
|
||
|
|
<span class="boring">}
|
||
|
|
</span></code></pre></pre>
|
||
|
|
<pre><code class="language-shell">$ cargo build
|
||
|
|
</code></pre>
|
||
|
|
<p>Gooood! 本章节的编译错误可以说是多个链表中最难解决的之一，依然被我们成功搞定了！</p>
|
||
|
|
<p>下面来写下测试用例，需要注意的是 <code>Ref</code> 不能被直接比较，因此我们需要先利用 <code>Deref</code> 解引用出其中的值，再进行比较。</p>
|
||
|
|
<pre><pre class="playground"><code class="language-rust edition2021">
|
||
|
|
<span class="boring">#![allow(unused)]
|
||
|
|
</span><span class="boring">fn main() {
|
||
|
|
</span>#[test]
|
||
|
|
fn peek() {
|
||
|
|
let mut list = List::new();
|
||
|
|
assert!(list.peek_front().is_none());
|
||
|
|
list.push_front(1); list.push_front(2); list.push_front(3);
|
||
|
|
|
||
|
|
assert_eq!(&*list.peek_front().unwrap(), &3);
|
||
|
|
}
|
||
|
|
<span class="boring">}
|
||
|
|
</span></code></pre></pre>
|
||
|
|
<pre><code class="language-shell">$ cargo test
|
||
|
|
|
||
|
|
Running target/debug/lists-5c71138492ad4b4a
|
||
|
|
|
||
|
|
running 10 tests
|
||
|
|
test first::test::basics ... ok
|
||
|
|
test fourth::test::basics ... ok
|
||
|
|
test second::test::basics ... ok
|
||
|
|
test fourth::test::peek ... ok
|
||
|
|
test second::test::iter_mut ... ok
|
||
|
|
test second::test::into_iter ... ok
|
||
|
|
test third::test::basics ... ok
|
||
|
|
test second::test::peek ... ok
|
||
|
|
test second::test::iter ... ok
|
||
|
|
test third::test::iter ... ok
|
||
|
|
|
||
|
|
test result: ok. 10 passed; 0 failed; 0 ignored; 0 measured
|
||
|
|
</code></pre>
|
||
|
|
<p>终于可以把文章开头的冷汗擦拭干净了，忘掉这个章节吧，让我来养你...哦不对，让我们开始一段真正轻松的章节。</p>
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