生命周期的局限

让我们来看以下代码:

#[derive(Debug)]
struct Foo;

impl Foo {
    fn mutate_and_share(&mut self) -> &Self { &*self }
    fn share(&self) {}
}

fn main() {
    let mut foo = Foo;
    let loan = foo.mutate_and_share();
    foo.share();
    println!("{:?}", loan);
}

人们可能期望它能被编译成功,我们调用mutate_and_share,它可以暂时可变借用foo,但随后只返回一个共享引用。因此我们期望foo.share()能够成功,因为foo不应该被可变借用。

然而,当我们试图编译它时:

error[E0502]: cannot borrow `foo` as immutable because it is also borrowed as mutable
  --> src/main.rs:12:5
   |
11 |     let loan = foo.mutate_and_share();
   |                --- mutable borrow occurs here
12 |     foo.share();
   |     ^^^ immutable borrow occurs here
13 |     println!("{:?}", loan);

这是为啥?好吧,我们得到的推理和上一节例 2完全一样。我们对程序进行解语法糖后,可以得到如下结果:

struct Foo;

impl Foo {
    fn mutate_and_share<'a>(&'a mut self) -> &'a Self { &'a *self }
    fn share<'a>(&'a self) {}
}

fn main() {
    'b: {
        let mut foo: Foo = Foo;
        'c: {
            let loan: &'c Foo = Foo::mutate_and_share::<'c>(&'c mut foo);
            'd: {
                Foo::share::<'d>(&'d foo);
            }
            println!("{:?}", loan);
        }
    }
}

由于loan的生命周期和mutate_and_share的签名,生命周期系统被迫将&mut foo扩展为'c的生命周期。然后当我们试图调用share时,它看到我们试图别名&'c mut foo,然后就炸了!

根据我们真正关心的引用语义,这个程序显然是正确的,但是生命周期系统太蠢了(原话是粗糙),无法处理这个问题。

不正确地缩减借用

下面的代码无法编译成功,因为 Rust 不明白这个借用已经不需要了,所以保守地退回到使用整个作用域。不过不用担心,这个问题最终会得到解决:


#![allow(unused)]
fn main() {
use std::collections::HashMap;
use std::hash::Hash;
fn get_default<'m, K, V>(map: &'m mut HashMap<K, V>, key: K) -> &'m mut V
where
    K: Clone + Eq + Hash,
    V: Default,
{
    match map.get_mut(&key) {
        Some(value) => value,
        None => {
            map.insert(key.clone(), V::default());
            map.get_mut(&key).unwrap()
        }
    }
}
}

由于所施加的生命周期限制,&mut map的生命周期与其他可变的借用重叠,导致编译错误:

error[E0499]: cannot borrow `*map` as mutable more than once at a time
  --> src/main.rs:12:13
   |
4  |   fn get_default<'m, K, V>(map: &'m mut HashMap<K, V>, key: K) -> &'m mut V
   |                  -- lifetime `'m` defined here
...
9  |       match map.get_mut(&key) {
   |       -     --- first mutable borrow occurs here
   |  _____|
   | |
10 | |         Some(value) => value,
11 | |         None => {
12 | |             map.insert(key.clone(), V::default());
   | |             ^^^ second mutable borrow occurs here
13 | |             map.get_mut(&key).unwrap()
14 | |         }
15 | |     }
   | |_____- returning this value requires that `*map` is borrowed for `'m`