|
|
|
|
@ -0,0 +1,237 @@
|
|
|
|
|
# 蠢笨编译器之循环生命周期
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
## 循环中的生命周期错误
|
|
|
|
|
Talk is cheap, 一起来看个例子：
|
|
|
|
|
```rust
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
我们来看看上面的代码中，`loop`循环有几个引用：
|
|
|
|
|
|
|
|
|
|
- `arr.len()`, 一个不可变引用，生命周期随着函数调用的结束而结束
|
|
|
|
|
- `tile`是可变引用，生命周期在下次循环开始前会结束
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
那就简单了，开心运行，开心。。。报错:
|
|
|
|
|
```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`
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
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：
|
|
|
|
|
|
|
|
|
|
1. `arr.len()`报错，原因是它借用了不可变引用，但是在紧跟着的`&mut arr[i]`中又借用了可变引用
|
|
|
|
|
2. `&mut arr[i]`报错，因为在上一次循环中，已经借用过同样的可变引用`&mut arr[i]`
|
|
|
|
|
3. `tile`的生命周期跟`arr`不一致
|
|
|
|
|
|
|
|
|
|
奇了怪了，跟我们之前的分析完全背道而驰，按理来说`arr.len()`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`&mut arr[i]`也应该随着每次循环的结束而结束，为什么会前后两次循环会因为同一处的引用而报错？
|
|
|
|
|
|
|
|
|
|
## 尝试去掉中间变量
|
|
|
|
|
虽然报错复杂，不过可以看出，所有的错误都跟`tile`这个中间变量有关，我们试着移除它看看：
|
|
|
|
|
```rust
|
|
|
|
|
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];
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
## 循环展开
|
|
|
|
|
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：
|
|
|
|
|
```rust
|
|
|
|
|
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!()
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
结果，编译器还是不给通过，报的错误几乎一样
|
|
|
|
|
|
|
|
|
|
## 深层原因
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
## 解决办法
|
|
|
|
|
虽然不能给出原因，但是我们可以看看解决办法，在上面，**移除中间变量**是一种办法，还有一种办法就是将部分引用移到循环外面.
|
|
|
|
|
|
|
|
|
|
#### 引用外移
|
|
|
|
|
```rust
|
|
|
|
|
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]
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
在上面代码中，我们只在循环中保留一个可变引用，剩下的`arr.len`和返回值引用，都移到循环外面，顺利通过编译.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
## 一个更复杂的例子
|
|
|
|
|
再来看一个例子，代码会更复杂，但是原因几乎相同：
|
|
|
|
|
```rust
|
|
|
|
|
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);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
运行后报错如下：
|
|
|
|
|
```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`
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
对于上述代码，只需要将返回值修改下，即可通过编译：
|
|
|
|
|
```rust
|
|
|
|
|
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);
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
|
|
|
|
## 新编译器Polonius
|
|
|
|
|
针对现有编译器存在的各种问题，Rust团队正在研发一个全新的编译器，名曰[`polonius`](https://github.com/rust-lang/polonius),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`rustc/RUSTFLAGS`中增加标志`-Zpolonius`，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。
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
## 总结
|
|
|
|
|
编译器不是万能的，它也会迷茫，也会犯错。
|
|
|
|
|
|
|
|
|
|
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。
|
|
|
|
|
|
sunface
3 years ago