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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 >
< p > 当涉及生命周期时， < / p >
< 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 >
< p > Talk is cheap, 一起来看个例子：< / 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());
let tile = & mut arr[i];
if Tile::Empty == *tile{
return tile;
}
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 我们来看看上面的代码中，< code > loop< / code > 循环有几个引用：< / p >
< ul >
< 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 >
< 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 >
< / 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， < 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 >
< / main >
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