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@ -279,6 +279,80 @@ use(a); // | |
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- [项目地址](https://github.com/rust-lang/polonius)
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- [具体介绍](http://smallcultfollowing.com/babysteps/blog/2018/04/27/an-alias-based-formulation-of-the-borrow-checker/)
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## Reborrow再借用
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学完`NLL`后,我们就有了一定的基础,可以继续学习关于借用和生命周期的一个高级内容:**再借用**。
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先来看一段代码:
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```rust
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#[derive(Debug)]
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struct Point {
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x: i32,
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y: i32,
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}
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impl Point {
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fn move_to(&mut self, x: i32, y: i32) {
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self.x = x;
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self.y = y;
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}
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}
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fn main() {
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let mut p = Point { x: 0, y: 0 };
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let r = &mut p;
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let rr: &Point = &*r;
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println!("{:?}", rr);
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r.move_to(10, 10);
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println!("{:?}", r);
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}
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```
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以上代码,大家可能会觉得可变引用`r`和不可变引用`rr`同时存在会报错吧?但是事实上并不会,原因在于`rr`是对`r`的再借用。
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对于再借用而言,`rr`在借用时不会破坏借用规则,但是你不能在它的生命周期内再使用原来的借用`r`,来看看对上段代码的分析:
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```rust
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fn main() {
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let mut p = Point { x: 0, y: 0 };
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let r = &mut p;
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// reborrow! 此时对`r`的再借用不会导致跟上面的借用冲突
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let rr: &Point = &*r;
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// 再借用`rr`最后一次使用发生在这里,在它的生命周期中,我们并没有使用原来的借用`r`,因此不会报错
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println!("{:?}", rr);
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// 再借用结束后,才去使用原来的借用`r`
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r.move_to(10, 10);
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println!("{:?}", r);
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}
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```
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再来看一个例子:
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```rust
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use std::vec::Vec;
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fn read_length(strings: &mut Vec<String>) -> usize {
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strings.len()
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}
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```
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如上所示,函数体内对参数的二次借用也是典型的Reborrow场景。
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那么下面让我们来做件坏事, 破坏这条规则,使其报错:
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```rust
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fn main() {
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let mut p = Point { x: 0, y: 0 };
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let r = &mut p;
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let rr: &Point = &*r;
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r.move_to(10, 10);
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println!("{:?}", rr);
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println!("{:?}", r);
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}
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```
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果然,破坏永远比重建简单:) 只需要在`rr`再借用的生命周期内使用一次原来的借用`r`即可!
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## 生命周期消除规则补充
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在上一节中,我们介绍了三大基础生命周期消除规则,实际上,随着Rust的版本进化,该规则也在不断演进,这里再介绍几个常见的消除规则:
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