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<h1 id="drop-and-panic-safety"><a class="header" href="#drop-and-panic-safety">Drop and Panic Safety</a></h1>
<p>嘿，你注意到这些注释了吗：</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>// Note that we don't need to mess around with `take` anymore
// because everything is Copy and there are no dtors that will
// run if we mess up... right? :) Riiiight? :)))
<span class="boring">}</span></code></pre></pre>
<p>这对吗？</p>
<p>你忘记你正在读那本书了吗？当然这是错误的（部分上是）。</p>
<p>让我们再次看看 pop_front 内部:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>// Bring the Box back to life so we can move out its value and
// Drop it (Box continues to magically understand this for us).
let boxed_node = Box::from_raw(node.as_ptr());
let result = boxed_node.elem;
// Make the next node into the new front.
self.front = boxed_node.back;
if let Some(new) = self.front {
// Cleanup its reference to the removed node
(*new.as_ptr()).front = None;
} else {
// If the front is now null, then this list is now empty!
debug_assert!(self.len == 1);
self.back = None;
}
self.len -= 1;
result
// Box gets implicitly freed here, knows there is no T.
<span class="boring">}</span></code></pre></pre>
<p>你看到 bug 了吗? 真可怕, 是这一行:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>debug_assert!(self.len == 1);
<span class="boring">}</span></code></pre></pre>
<p>大多数情况下，你不需要考虑或担心恐慌，但一旦你开始编写真正不安全的代码，并在 &quot;invariants(不可变性) &quot;上大做文章，你就需要对恐慌保持高度警惕！</p>
<p>我们必须谈谈 <a href="https://doc.rust-lang.org/nightly/nomicon/exception-safety.html"><em>异常安全</em></a> （又名恐慌安全、解除安全......）。</p>
<p>情况是这样的：在默认情况下，恐慌会被 unwinding。unwind 只是 &quot;让每个函数立即返回 &quot;的一种花哨说法。你可能会想：&quot;好吧，如果每个函数都返回，那么程序就要结束了，何必在乎它呢？&quot;但你错了！</p>
<p>我们必须关注有两个原因：当函数返回时，析构函数会运行，而且可以捕获 unwind。在这两种情况下，代码都可能在恐慌发生后继续运行，因此我们必须非常小心，确保我们的不安全的集合在恐慌发生时始终处于某种一致的状态，因为每次恐慌都是隐式的提前返回！</p>
<p>让我们想一想，到这一行时，我们的集合处于什么状态：</p>
<p>我们将 boxed_node 放在栈上，并从中提取了元素。如果我们在此时返回，Box 将被丢弃，节点将被释放。self.back 仍然指向那个被释放的节点！一旦我们使用 self.back 来处理一些事情，这就可能导致释放后再使用！</p>
<p>有趣的是，这行也有类似的问题，但它要安全得多：</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>self.len -= 1;
<span class="boring">}</span></code></pre></pre>
<p>默认情况下，Rust 会在调试构建时检查上溢和下溢，并在发生时产生恐慌。是的，每一次算术运算都会带来恐慌安全隐患！这行还好，他不会导致内存错误，因为之前已经完成了该做的所有操作。所以调试断言哪行在某种意义上更糟糕，因为它可能将一个小问题升级为关键问题！</p>
<p>在实现过程中，只要我们确保在别人注意到之前修复它们，我们可以临时性的破坏invariants(不可变性)。这实际上是 Rust 的集合所有权和借用系统的 &quot;杀手级应用 &quot;之一：如果一个操作需要一个 <code>&amp;mut Self</code>，那么我们就能保证对我们的集合拥有独占访问权，而且我们可以暂时破坏invariants(不可变性)，因为我们知道没有人能偷偷摸摸地破坏它。</p>
<p>我们有两种方法可以让我们的代码更健壮：</p>
<ul>
<li>更积极地使用 Option::take 这样的操作，因为它们更 &quot;事务性&quot;，更倾向于保留invariants(不可变性)。</li>
<li>放弃 debug_asserts，相信自己能写出更好的测试，并使用专用的 &quot;完整性检查 &quot;函数，而这些函数永远不会在用户代码中运行。</li>
</ul>
<p>原则上，我喜欢第一种方案，但它对双链路列表的实际效果并不好，因为所有内容都是双冗余编码的。Option::take 并不能解决这里的问题，但将 debug_assert 下移一行却可以。不过说真的，为什么要为难我们自己呢？让我们移除那些 debug_asserts，并确保任何可能引起恐慌的事情都发生在我们方法的开头或结尾，而我们在这些地方保持invariants(不可变性)。</p>
<p>这是我们的全部实现：</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>use std::ptr::NonNull;
use std::marker::PhantomData;
pub struct LinkedList&lt;T&gt; {
front: Link&lt;T&gt;,
back: Link&lt;T&gt;,
len: usize,
_boo: PhantomData&lt;T&gt;,
}
type Link&lt;T&gt; = Option&lt;NonNull&lt;Node&lt;T&gt;&gt;&gt;;
struct Node&lt;T&gt; {
front: Link&lt;T&gt;,
back: Link&lt;T&gt;,
elem: T,
}
impl&lt;T&gt; LinkedList&lt;T&gt; {
pub fn new() -&gt; Self {
Self {
front: None,
back: None,
len: 0,
_boo: PhantomData,
}
}
pub fn push_front(&amp;mut self, elem: T) {
// SAFETY: it's a linked-list, what do you want?
unsafe {
let new = NonNull::new_unchecked(Box::into_raw(Box::new(Node {
front: None,
back: None,
elem,
})));
if let Some(old) = self.front {
// Put the new front before the old one
(*old.as_ptr()).front = Some(new);
(*new.as_ptr()).back = Some(old);
} else {
// If there's no front, then we're the empty list and need
// to set the back too.
self.back = Some(new);
}
// These things always happen!
self.front = Some(new);
self.len += 1;
}
}
pub fn pop_front(&amp;mut self) -&gt; Option&lt;T&gt; {
unsafe {
// Only have to do stuff if there is a front node to pop.
self.front.map(|node| {
// Bring the Box back to life so we can move out its value and
// Drop it (Box continues to magically understand this for us).
let boxed_node = Box::from_raw(node.as_ptr());
let result = boxed_node.elem;
// Make the next node into the new front.
self.front = boxed_node.back;
if let Some(new) = self.front {
// Cleanup its reference to the removed node
(*new.as_ptr()).front = None;
} else {
// If the front is now null, then this list is now empty!
self.back = None;
}
self.len -= 1;
result
// Box gets implicitly freed here, knows there is no T.
})
}
}
pub fn len(&amp;self) -&gt; usize {
self.len
}
}
<span class="boring">}</span></code></pre></pre>
<p>这还有什么可以引发恐慌？老实说，要知道这些需要你是 Rust 专家，不过幸好我是！</p>
<p>在这段代码中，我能看到的唯一可能引起恐慌的地方是 <code>Box::new</code>（用于内存不足的情况）和 <code>len</code> 运算。所有这些都在我们方法的最末端或最开始，所以，我们是安全的！</p>
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