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<main>
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<h1 id="蠢笨编译器之循环生命周期"><a class="header" href="#蠢笨编译器之循环生命周期">蠢笨编译器之循环生命周期</a></h1>
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<p>当涉及生命周期时，Rust 编译器有时会变得不太聪明，如果再配合循环，蠢笨都不足以形容它，不信？那继续跟着我一起看看。</p>
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<h2 id="循环中的生命周期错误"><a class="header" href="#循环中的生命周期错误">循环中的生命周期错误</a></h2>
|
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<p>Talk is cheap, 一起来看个例子：</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>use rand::{thread_rng, Rng};
|
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#[derive(Debug, PartialEq)]
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enum Tile {
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Empty,
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}
|
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fn random_empty_tile(arr: &mut [Tile]) -> &mut Tile {
|
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loop {
|
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let i = thread_rng().gen_range(0..arr.len());
|
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let tile = &mut arr[i];
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if Tile::Empty == *tile{
|
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return tile;
|
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}
|
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}
|
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}
|
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<span class="boring">}</span></code></pre>
|
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<p>我们来看看上面的代码中，<code>loop</code>循环有几个引用：</p>
|
|
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<ul>
|
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<li><code>arr.len()</code>, 一个不可变引用，生命周期随着函数调用的结束而结束</li>
|
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<li><code>tile</code>是可变引用，生命周期在下次循环开始前会结束</li>
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|
</ul>
|
|
|
<p>根据以上的分析，可以得出个初步结论：在同一次循环间各个引用生命周期互不影响，在两次循环间，引用也互不影响。</p>
|
|
|
<p>那就简单了，开心运行，开心。。。报错:</p>
|
|
|
<pre><code class="language-console">error[E0502]: cannot borrow `*arr` as immutable because it is also borrowed as mutable
|
|
|
--> src/main.rs:10:43
|
|
|
|
|
|
|
8 | fn random_empty_tile(arr: &mut [Tile]) -> &mut Tile {
|
|
|
| - let's call the lifetime of this reference `'1`
|
|
|
9 | loop {
|
|
|
10 | let i = thread_rng().gen_range(0..arr.len());
|
|
|
| ^^^ immutable borrow occurs here
|
|
|
11 | let tile = &mut arr[i];
|
|
|
| ----------- mutable borrow occurs here
|
|
|
12 | if Tile::Empty == *tile{
|
|
|
13 | return tile;
|
|
|
| ---- returning this value requires that `arr[_]` is borrowed for `'1`
|
|
|
|
|
|
error[E0499]: cannot borrow `arr[_]` as mutable more than once at a time
|
|
|
--> src/main.rs:11:20
|
|
|
|
|
|
|
8 | fn random_empty_tile(arr: &mut [Tile]) -> &mut Tile {
|
|
|
| - let's call the lifetime of this reference `'1`
|
|
|
...
|
|
|
11 | let tile = &mut arr[i];
|
|
|
| ^^^^^^^^^^^ `arr[_]` was mutably borrowed here in the previous iteration of the loop
|
|
|
12 | if Tile::Empty == *tile{
|
|
|
13 | return tile;
|
|
|
| ---- returning this value requires that `arr[_]` is borrowed for `'1`
|
|
|
</code></pre>
|
|
|
<p>不仅是错误，还是史诗级别的错误！无情刷屏了！只能想办法梳理下：</p>
|
|
|
<ol>
|
|
|
<li><code>arr.len()</code>报错，原因是它借用了不可变引用，但是在紧跟着的<code>&mut arr[i]</code>中又借用了可变引用</li>
|
|
|
<li><code>&mut arr[i]</code>报错，因为在上一次循环中，已经借用过同样的可变引用<code>&mut arr[i]</code></li>
|
|
|
<li><code>tile</code>的生命周期跟<code>arr</code>不一致</li>
|
|
|
</ol>
|
|
|
<p>奇了怪了，跟我们之前的分析完全背道而驰，按理来说<code>arr.len()</code>的借用应该在调用后立刻结束，而不是持续到后面的代码行；同时可变借用<code>&mut arr[i]</code>也应该随着每次循环的结束而结束，为什么前后两次循环会因为同一处的引用而报错？</p>
|
|
|
<h2 id="尝试去掉中间变量"><a class="header" href="#尝试去掉中间变量">尝试去掉中间变量</a></h2>
|
|
|
<p>虽然报错复杂，不过可以看出，所有的错误都跟<code>tile</code>这个中间变量有关，我们试着移除它看看：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use rand::{thread_rng, Rng};
|
|
|
|
|
|
#[derive(Debug, PartialEq)]
|
|
|
enum Tile {
|
|
|
Empty,
|
|
|
}
|
|
|
|
|
|
fn random_empty_tile(arr: &mut [Tile]) -> &mut Tile {
|
|
|
loop {
|
|
|
let i = thread_rng().gen_range(0..arr.len());
|
|
|
if Tile::Empty == arr[i] {
|
|
|
return &mut arr[i];
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>见证奇迹的时刻，竟然编译通过了！到底发什么了什么？仅仅移除了中间变量，就编译通过了？是否可以大胆的猜测，因为中间变量，导致编译器变蠢了，因此无法正确的识别引用的生命周期。</p>
|
|
|
<h2 id="循环展开"><a class="header" href="#循环展开">循环展开</a></h2>
|
|
|
<p>如果不使用循环呢？会不会也有这样的错误？咱们试着把循环展开：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use rand::{thread_rng, Rng};
|
|
|
|
|
|
#[derive(Debug, PartialEq)]
|
|
|
enum Tile {
|
|
|
Empty,
|
|
|
}
|
|
|
|
|
|
fn random_empty_tile_2<'arr>(arr: &'arr mut [Tile]) -> &'arr mut Tile {
|
|
|
let len = arr.len();
|
|
|
|
|
|
// First loop iteration
|
|
|
{
|
|
|
let i = thread_rng().gen_range(0..len);
|
|
|
let tile = &mut arr[i]; // Lifetime: 'arr
|
|
|
if Tile::Empty == *tile {
|
|
|
return tile;
|
|
|
}
|
|
|
}
|
|
|
|
|
|
// Second loop iteration
|
|
|
{
|
|
|
let i = thread_rng().gen_range(0..len);
|
|
|
let tile = &mut arr[i]; // Lifetime: 'arr
|
|
|
if Tile::Empty == *tile {
|
|
|
return tile;
|
|
|
}
|
|
|
}
|
|
|
|
|
|
unreachable!()
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>结果，编译器还是不给通过，报的错误几乎一样</p>
|
|
|
<h2 id="深层原因"><a class="header" href="#深层原因">深层原因</a></h2>
|
|
|
<p>令人沮丧的是，我找遍了网上，也没有具体的原因，大家都说这是编译器太笨导致的问题，但是关于深层的原因，也没人能说出个所有然。</p>
|
|
|
<p>因此，我无法在本文中给出为什么编译器会这么笨的真实原因，如果以后有结果，会在这里进行更新。</p>
|
|
|
<p>——2022 年 1 月 13 日更新—––
|
|
|
兄弟们，我带着挖掘出的一些内容回来了，再来看段错误代码先：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>struct A {
|
|
|
a: i32
|
|
|
}
|
|
|
|
|
|
impl A {
|
|
|
fn one(&mut self) -> &i32{
|
|
|
self.a = 10;
|
|
|
&self.a
|
|
|
}
|
|
|
fn two(&mut self) -> &i32 {
|
|
|
loop {
|
|
|
let k = self.one();
|
|
|
if *k > 10i32 {
|
|
|
return k;
|
|
|
}
|
|
|
|
|
|
// 可能存在的剩余代码
|
|
|
// ...
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>我们来逐步深入分析下：</p>
|
|
|
<ol>
|
|
|
<li>首先为<code>two</code>方法增加一下生命周期标识: <code>fn two<'a>(&'a mut self) -> &'a i32 { .. }</code>, 这里根据生命周期的<a href="https://course.rs/basic/lifetime.html#三条消除规则">消除规则</a>添加的</li>
|
|
|
<li>根据生命周期标识可知：<code>two</code>中返回的<code>k</code>的生命周期必须是<code>'a</code></li>
|
|
|
<li>根据第 2 条，又可知：<code>let k = self.one();</code>中对<code>self</code>的借用生命周期也是<code>'a</code></li>
|
|
|
<li>因为<code>k</code>的借用发生在<code>loop</code>循环内，因此它需要小于等于循环的生命周期，但是根据之前的推断，它又要大于等于函数的生命周期<code>'a</code>，而函数的生命周期又大于等于循环生命周期，</li>
|
|
|
</ol>
|
|
|
<p>由上可以推出：<code>let k = self.one();</code>中<code>k</code>的生命周期要大于等于循环的生命周期，又要小于等于循环的生命周期, 唯一满足条件的就是：<code>k</code>的生命周期等于循环生命周期。</p>
|
|
|
<p>但是我们的<code>two</code>方法在循环中对<code>k</code>进行了提前返回，编译器自然会认为存在其它代码，这会导致<code>k</code>的生命周期小于循环的生命周期。</p>
|
|
|
<p>怎么办呢？很简单：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn two(&mut self) -> &i32 {
|
|
|
loop {
|
|
|
let k = self.one();
|
|
|
return k;
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>不要在<code>if</code>分支中返回<code>k</code>，而是直接返回，这样就让它们的生命周期相等了，最终可以顺利编译通过。</p>
|
|
|
<blockquote>
|
|
|
<p>如果一个引用值从函数的某个路径提前返回了，那么该借用必须要在函数的所有返回路径都合法</p>
|
|
|
</blockquote>
|
|
|
<h2 id="解决方法"><a class="header" href="#解决方法">解决方法</a></h2>
|
|
|
<p>虽然不能给出原因，但是我们可以看看解决办法，在上面，<strong>移除中间变量</strong>和<strong>消除代码分支</strong>都是可行的方法，还有一种方法就是将部分引用移到循环外面.</p>
|
|
|
<h4 id="引用外移"><a class="header" href="#引用外移">引用外移</a></h4>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn random_empty_tile(arr: &mut [Tile]) -> &mut Tile {
|
|
|
let len = arr.len();
|
|
|
let mut the_chosen_i = 0;
|
|
|
loop {
|
|
|
let i = rand::thread_rng().gen_range(0..len);
|
|
|
let tile = &mut arr[i];
|
|
|
if Tile::Empty == *tile {
|
|
|
the_chosen_i = i;
|
|
|
break;
|
|
|
}
|
|
|
}
|
|
|
&mut arr[the_chosen_i]
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>在上面代码中，我们只在循环中保留一个可变引用，剩下的<code>arr.len</code>和返回值引用，都移到循环外面，顺利通过编译.</p>
|
|
|
<h2 id="一个更复杂的例子"><a class="header" href="#一个更复杂的例子">一个更复杂的例子</a></h2>
|
|
|
<p>再来看一个例子，代码会更复杂，但是原因几乎相同：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::collections::HashMap;
|
|
|
|
|
|
enum Symbol {
|
|
|
A,
|
|
|
}
|
|
|
|
|
|
pub struct SymbolTable {
|
|
|
scopes: Vec<Scope>,
|
|
|
current: usize,
|
|
|
}
|
|
|
|
|
|
struct Scope {
|
|
|
parent: Option<usize>,
|
|
|
symbols: HashMap<String, Symbol>,
|
|
|
}
|
|
|
|
|
|
impl SymbolTable {
|
|
|
pub fn get_mut(&mut self, name: &String) -> &mut Symbol {
|
|
|
let mut current = Some(self.current);
|
|
|
|
|
|
while let Some(id) = current {
|
|
|
let scope = self.scopes.get_mut(id).unwrap();
|
|
|
if let Some(symbol) = scope.symbols.get_mut(name) {
|
|
|
return symbol;
|
|
|
}
|
|
|
|
|
|
current = scope.parent;
|
|
|
}
|
|
|
|
|
|
panic!("Value not found: {}", name);
|
|
|
}
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>运行后报错如下：</p>
|
|
|
<pre><code class="language-console">error[E0499]: cannot borrow `self.scopes` as mutable more than once at a time
|
|
|
--> src/main.rs:22:25
|
|
|
|
|
|
|
18 | pub fn get_mut(&mut self, name: &String) -> &mut Symbol {
|
|
|
| - let's call the lifetime of this reference `'1`
|
|
|
...
|
|
|
22 | let scope = self.scopes.get_mut(id).unwrap();
|
|
|
| ^^^^^^^^^^^ `self.scopes` was mutably borrowed here in the previous iteration of the loop
|
|
|
23 | if let Some(symbol) = scope.symbols.get_mut(name) {
|
|
|
24 | return symbol;
|
|
|
| ------ returning this value requires that `self.scopes` is borrowed for `'1`
|
|
|
</code></pre>
|
|
|
<p>对于上述代码，只需要将返回值修改下，即可通过编译：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn get_mut(&mut self, name: &String) -> &mut Symbol {
|
|
|
let mut current = Some(self.current);
|
|
|
|
|
|
while let Some(id) = current {
|
|
|
let scope = self.scopes.get_mut(id).unwrap();
|
|
|
if scope.symbols.contains_key(name) {
|
|
|
return self.scopes.get_mut(id).unwrap().symbols.get_mut(name).unwrap();
|
|
|
}
|
|
|
|
|
|
current = scope.parent;
|
|
|
}
|
|
|
|
|
|
panic!("Value not found: {}", name);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>其中的关键就在于返回的时候，新建一个引用，而不是使用中间状态的引用。</p>
|
|
|
<h2 id="新借用检查器-polonius"><a class="header" href="#新借用检查器-polonius">新借用检查器 Polonius</a></h2>
|
|
|
<p>针对现有借用检查器存在的各种问题，Rust 团队正在研发一个全新的借用检查器，名曰<a href="https://github.com/rust-lang/polonius"><code>polonius</code></a>,但是目前它仍然处在开发阶段，如果想在自己项目中使用，需要在<code>rustc/RUSTFLAGS</code>中增加标志<code>-Zpolonius</code>，但是可能会导致编译速度变慢，或者引入一些新的编译错误。</p>
|
|
|
<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
|
|
|
<p>编译器不是万能的，它也会迷茫，也会犯错。</p>
|
|
|
<p>因此我们在循环中使用引用类型时要格外小心，特别是涉及可变引用，这种情况下，最好的办法就是避免中间状态，或者在返回时避免使用中间状态。</p>
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