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<main>
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<h1 id="迭代器"><a class="header" href="#迭代器">迭代器</a></h1>
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<p>坏男孩最令人头疼，而链表实现中，迭代器就是这样的坏男孩，所以我们放在最后来处理。</p>
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<h2 id="intoiter"><a class="header" href="#intoiter">IntoIter</a></h2>
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<p>由于是转移所有权，因此 <code>IntoIter</code> 一直都是最好实现的:</p>
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<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 IntoIter<T>(List<T>);
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impl<T> List<T> {
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pub fn into_iter(self) -> IntoIter<T> {
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IntoIter(self)
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}
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}
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impl<T> Iterator for IntoIter<T> {
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type Item = T;
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fn next(&mut self) -> Option<Self::Item> {
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self.0.pop_front()
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}
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}
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<span class="boring">}</span></code></pre>
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<p>但是关于双向链表，有一个有趣的事实，它不仅可以从前向后迭代，还能反过来。前面实现的是传统的从前到后，那问题来了，反过来该如何实现呢？</p>
|
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<p>答案是: <code>DoubleEndedIterator</code>，它继承自 <code>Iterator</code>( 通过 <a href="https://course.rs/basic/trait/advance-trait.html?highlight=supertrait#特征定义中的特征约束"><code>supertrait</code></a> )，因此意味着要实现该特征，首先需要实现 <code>Iterator</code>。</p>
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|
|
<p>这样只要为 <code>DoubleEndedIterator</code> 实现 <code>next_back</code> 方法，就可以支持双向迭代了: <code>Iterator</code> 的 <code>next</code> 方法从前往后，而 <code>next_back</code> 从后向前。</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>impl<T> DoubleEndedIterator for IntoIter<T> {
|
|
|
fn next_back(&mut self) -> Option<Self::Item> {
|
|
|
self.0.pop_back()
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>测试下:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>#[test]
|
|
|
fn into_iter() {
|
|
|
let mut list = List::new();
|
|
|
list.push_front(1); list.push_front(2); list.push_front(3);
|
|
|
|
|
|
let mut iter = list.into_iter();
|
|
|
assert_eq!(iter.next(), Some(3));
|
|
|
assert_eq!(iter.next_back(), Some(1));
|
|
|
assert_eq!(iter.next(), Some(2));
|
|
|
assert_eq!(iter.next_back(), None);
|
|
|
assert_eq!(iter.next(), None);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<pre><code class="language-shell">$ cargo test
|
|
|
|
|
|
Running target/debug/lists-5c71138492ad4b4a
|
|
|
|
|
|
running 11 tests
|
|
|
test fourth::test::basics ... ok
|
|
|
test fourth::test::peek ... ok
|
|
|
test fourth::test::into_iter ... ok
|
|
|
test first::test::basics ... ok
|
|
|
test second::test::basics ... ok
|
|
|
test second::test::iter ... ok
|
|
|
test second::test::iter_mut ... ok
|
|
|
test third::test::iter ... ok
|
|
|
test third::test::basics ... ok
|
|
|
test second::test::into_iter ... ok
|
|
|
test second::test::peek ... ok
|
|
|
|
|
|
test result: ok. 11 passed; 0 failed; 0 ignored; 0 measured
|
|
|
</code></pre>
|
|
|
<h2 id="iter"><a class="header" href="#iter">Iter</a></h2>
|
|
|
<p>这里又要用到糟糕的 <code>Ref</code>:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>pub struct Iter<'a, T>(Option<Ref<'a, Node<T>>>);
|
|
|
|
|
|
impl<T> List<T> {
|
|
|
pub fn iter(&self) -> Iter<T> {
|
|
|
Iter(self.head.as_ref().map(|head| head.borrow()))
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
</code></pre>
|
|
|
<p>迄今为止一切运行正常，接下来的 <code>next</code> 实现起来会有些麻烦:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>impl<'a, T> Iterator for Iter<'a, T> {
|
|
|
type Item = Ref<'a, T>;
|
|
|
fn next(&mut self) -> Option<Self::Item> {
|
|
|
self.0.take().map(|node_ref| {
|
|
|
self.0 = node_ref.next.as_ref().map(|head| head.borrow());
|
|
|
Ref::map(node_ref, |node| &node.elem)
|
|
|
})
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
error[E0521]: borrowed data escapes outside of closure
|
|
|
--> src/fourth.rs:155:13
|
|
|
|
|
|
|
153 | fn next(&mut self) -> Option<Self::Item> {
|
|
|
| --------- `self` is declared here, outside of the closure body
|
|
|
154 | self.0.take().map(|node_ref| {
|
|
|
155 | self.0 = node_ref.next.as_ref().map(|head| head.borrow());
|
|
|
| ^^^^^^ -------- borrow is only valid in the closure body
|
|
|
| |
|
|
|
| reference to `node_ref` escapes the closure body here
|
|
|
|
|
|
error[E0505]: cannot move out of `node_ref` because it is borrowed
|
|
|
--> src/fourth.rs:156:22
|
|
|
|
|
|
|
153 | fn next(&mut self) -> Option<Self::Item> {
|
|
|
| --------- lifetime `'1` appears in the type of `self`
|
|
|
154 | self.0.take().map(|node_ref| {
|
|
|
155 | self.0 = node_ref.next.as_ref().map(|head| head.borrow());
|
|
|
| ------ -------- borrow of `node_ref` occurs here
|
|
|
| |
|
|
|
| assignment requires that `node_ref` is borrowed for `'1`
|
|
|
156 | Ref::map(node_ref, |node| &node.elem)
|
|
|
| ^^^^^^^^ move out of `node_ref` occurs here
|
|
|
</code></pre>
|
|
|
<p>果然，膝盖又中了一箭。</p>
|
|
|
<p><code>node_ref</code> 活得不够久，跟一般的引用不同，Rust 不允许我们这样分割 <code>Ref</code>，从 <code>head.borrow()</code> 中取出的 <code>Ref</code> 只允许跟 <code>node_ref</code> 活得一样久。</p>
|
|
|
<p>而我们想要的函数是存在的:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>pub fn map_split<U, V, F>(orig: Ref<'b, T>, f: F) -> (Ref<'b, U>, Ref<'b, V>) where
|
|
|
F: FnOnce(&T) -> (&U, &V),
|
|
|
U: ?Sized,
|
|
|
V: ?Sized,
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>喔，这个函数定义的泛型直接晃瞎了我的眼睛。。</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn next(&mut self) -> Option<Self::Item> {
|
|
|
self.0.take().map(|node_ref| {
|
|
|
let (next, elem) = Ref::map_split(node_ref, |node| {
|
|
|
(&node.next, &node.elem)
|
|
|
});
|
|
|
|
|
|
self.0 = next.as_ref().map(|head| head.borrow());
|
|
|
|
|
|
elem
|
|
|
})
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<pre><code class="language-shell">$ cargo build
|
|
|
|
|
|
Compiling lists v0.1.0 (/Users/ABeingessner/dev/temp/lists)
|
|
|
error[E0521]: borrowed data escapes outside of closure
|
|
|
--> src/fourth.rs:159:13
|
|
|
|
|
|
|
153 | fn next(&mut self) -> Option<Self::Item> {
|
|
|
| --------- `self` is declared here, outside of the closure body
|
|
|
...
|
|
|
159 | self.0 = next.as_ref().map(|head| head.borrow());
|
|
|
| ^^^^^^ ---- borrow is only valid in the closure body
|
|
|
| |
|
|
|
| reference to `next` escapes the closure body here
|
|
|
</code></pre>
|
|
|
<p>额，借用的内容只允许在闭包体中使用，看起来我们还是得用 <code>Ref::map</code> 来解决问题:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn next(&mut self) -> Option<Self::Item> {
|
|
|
self.0.take().map(|node_ref| {
|
|
|
let (next, elem) = Ref::map_split(node_ref, |node| {
|
|
|
(&node.next, &node.elem)
|
|
|
});
|
|
|
|
|
|
self.0 = if next.is_some() {
|
|
|
Some(Ref::map(next, |next| &**next.as_ref().unwrap()))
|
|
|
} else {
|
|
|
None
|
|
|
};
|
|
|
|
|
|
elem
|
|
|
})
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<pre><code class="language-shell">error[E0308]: mismatched types
|
|
|
--> src/fourth.rs:162:22
|
|
|
|
|
|
|
162 | Some(Ref::map(next, |next| &**next.as_ref().unwrap()))
|
|
|
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `fourth::Node`, found struct `std::cell::RefCell`
|
|
|
|
|
|
|
= note: expected type `std::cell::Ref<'_, fourth::Node<_>>`
|
|
|
found type `std::cell::Ref<'_, std::cell::RefCell<fourth::Node<_>>>`
|
|
|
</code></pre>
|
|
|
<p>晕, 多了一个 <code>RefCell</code> ，随着我们的对链表的逐步深入，<code>RefCell</code> 的代码嵌套变成了不可忽视的问题。</p>
|
|
|
<p>看起来我们已经无能为力了，只能试着去摆脱 <code>RefCell</code> 了。<code>Rc</code> 怎么样？我们完全可以对 <code>Rc</code> 进行完整的克隆:</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>pub struct Iter<T>(Option<Rc<Node<T>>>);
|
|
|
|
|
|
impl<T> List<T> {
|
|
|
pub fn iter(&self) -> Iter<T> {
|
|
|
Iter(self.head.as_ref().map(|head| head.clone()))
|
|
|
}
|
|
|
}
|
|
|
|
|
|
impl<T> Iterator for Iter<T> {
|
|
|
type Item =
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>等等，那现在返回的是什么？<code>&T</code> 还是 <code>Ref<T></code> ?</p>
|
|
|
<p>两者都不是，现在我们的 <code>Iter</code> 已经没有生命周期了：无论是 <code>&T</code> 还是 <code>Ref<T></code> 都需要我们在 <code>next</code> 之前声明好生命周期。但是我们试图从 <code>Rc</code> 中取出来的值其实是迭代器的引用。</p>
|
|
|
<p>也可以通过对 <code>Rc</code> 进行 map 获取到 <code>Rc<T></code>？但是标准库并没有给我们提供相应的功能，第三方倒是有<a href="https://crates.io/crates/owning_ref">一个</a>。</p>
|
|
|
<p>但是，即使这么做了，还有一个更大的坑在等着：一个会造成迭代器不合法的可怕幽灵。事实上，之前我们对于迭代器不合法是免疫的，但是一旦迭代器产生 <code>Rc</code>，那它们就不再会借用链表。这意味着人们可以在持有指向链表内部的指针时，还可以进行 <code>push</code> 和 <code>pop</code> 操作。</p>
|
|
|
<p>严格来说，<code>push</code> 问题不大，因为链表两端的增长不会对我们正在关注的某个子链表造成影响。</p>
|
|
|
<p>但是 <code>pop</code> 就是另一个故事了，如果在我们关注的子链表之外 <code>pop</code>, 那问题不大。但是如果是 <code>pop</code> 一个正在引用的子链表中的节点呢？那一切就完了，特别是，如果大家还试图去 unwrap <code>try_unwrap</code> 返回的 <code>Result</code> ，会直接造成整个程序的 <code>panic</code>。</p>
|
|
|
<p>仔细想一想，好像也不错，程序一切正常，除非去 <code>pop</code> 我们正在引用的节点，最美的是，就算遇到这种情况，程序也会直接崩溃，提示我们错误的发生。</p>
|
|
|
<p>其实我们大部分的努力都是为了实现隐藏的细节和优雅的 API，典型的二八原则，八成时间花在二成的细节上。但是如果不关心这些细节，可以接受自己的平凡的话，那把节点简单的到处传递就行。</p>
|
|
|
<p>总之，可以看出，内部可变性非常适合写一个安全性的应用程序，但是如果是安全性高的库，那内部可变性就有些捉襟见肘了。</p>
|
|
|
<p>最终，我选择了放弃，不再实现 <code>Iter</code> 和 <code>IterMut</code>，也许努力下，可以实现，但是。。。不愉快，算了。</p>
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