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<h1 id="drop-and-panic-safety"><a class="header" href="#drop-and-panic-safety">Drop and Panic Safety</a></h1>
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<p>嘿,你注意到这些注释了吗:</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>// Note that we don't need to mess around with `take` anymore
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// because everything is Copy and there are no dtors that will
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// run if we mess up... right? :) Riiiight? :)))
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<span class="boring">}</span></code></pre></pre>
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<p>这对吗?</p>
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<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>
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<p>让我们再次看看 pop_front 内部:</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>// Bring the Box back to life so we can move out its value and
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// Drop it (Box continues to magically understand this for us).
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let boxed_node = Box::from_raw(node.as_ptr());
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let result = boxed_node.elem;
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// Make the next node into the new front.
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self.front = boxed_node.back;
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if let Some(new) = self.front {
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// Cleanup its reference to the removed node
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(*new.as_ptr()).front = None;
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} else {
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// If the front is now null, then this list is now empty!
|
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debug_assert!(self.len == 1);
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self.back = None;
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}
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self.len -= 1;
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result
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// Box gets implicitly freed here, knows there is no T.
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<span class="boring">}</span></code></pre></pre>
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<p>你看到 bug 了吗? 真可怕, 是这一行:</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>debug_assert!(self.len == 1);
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<span class="boring">}</span></code></pre></pre>
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<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 "invariants(不可变性) "上大做文章,你就需要对恐慌保持高度警惕!</p>
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<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>
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<p>情况是这样的:在默认情况下,恐慌会被 unwinding。unwind 只是 "让每个函数立即返回 "的一种花哨说法。你可能会想:"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?"但你错了!</p>
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<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>
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<p>让我们想一想,到这一行时,我们的集合处于什么状态:</p>
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<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>
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<p>有趣的是,这行也有类似的问题,但它要安全得多:</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>self.len -= 1;
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<span class="boring">}</span></code></pre></pre>
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<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>
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<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 "杀手级应用 "之一:如果一个操作需要一个 <code>&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>
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<p>我们有两种方法可以让我们的代码更健壮:</p>
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<ul>
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<li>更积极地使用 Option::take 这样的操作,因为它们更 "事务性",更倾向于保留invariants(不可变性)。</li>
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<li>放弃 debug_asserts,相信自己能写出更好的测试,并使用专用的 "完整性检查 "函数,而这些函数永远不会在用户代码中运行。</li>
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</ul>
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<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>
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<p>这是我们的全部实现:</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::ptr::NonNull;
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use std::marker::PhantomData;
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pub struct LinkedList<T> {
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front: Link<T>,
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back: Link<T>,
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len: usize,
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_boo: PhantomData<T>,
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}
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type Link<T> = Option<NonNull<Node<T>>>;
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struct Node<T> {
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front: Link<T>,
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back: Link<T>,
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elem: T,
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}
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impl<T> LinkedList<T> {
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|
pub fn new() -> Self {
|
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|
Self {
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|
front: None,
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|
back: None,
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len: 0,
|
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|
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|
_boo: PhantomData,
|
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|
}
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}
|
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|
pub fn push_front(&mut self, elem: T) {
|
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|
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|
// SAFETY: it's a linked-list, what do you want?
|
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|
unsafe {
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|
let new = NonNull::new_unchecked(Box::into_raw(Box::new(Node {
|
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|
front: None,
|
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|
back: None,
|
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|
elem,
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})));
|
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|
if let Some(old) = self.front {
|
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|
// Put the new front before the old one
|
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|
(*old.as_ptr()).front = Some(new);
|
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|
(*new.as_ptr()).back = Some(old);
|
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} else {
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|
// If there's no front, then we're the empty list and need
|
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// to set the back too.
|
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|
self.back = Some(new);
|
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|
}
|
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|
|
// These things always happen!
|
|
|
|
|
self.front = Some(new);
|
|
|
|
|
self.len += 1;
|
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|
|
|
}
|
|
|
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|
}
|
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|
pub fn pop_front(&mut self) -> Option<T> {
|
|
|
|
|
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(&self) -> 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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