当返回对范围之外的值的可变引用的不可变引用时,为什么在范围结束时删除了可变引用？

Question

fn main() {
    // block1: fails
    {
        let mut m = 10;

        let n = {
            let b = &&mut m;
            &**b // just returning b fails
        };

        println!("{:?}", n);
    }

    // block2: passes
    {
        let mut m = 10;

        let n = {
            let b = &&m;
            &**b // just returning b fails here too
        };

        println!("{:?}", n);
    }
}

block1因错误而失败:

error[E0597]: borrowed value does not live long enough
  --> src/main.rs:7:22
   |
7  |             let b = &&mut m;
   |                      ^^^^^^ temporary value does not live long enough
8  |             &**b // just returning b fails
9  |         };
   |         - temporary value dropped here while still borrowed
...
12 |     }
   |     - temporary value needs to live until here

假设内部不可变引用扩展到block2范围之外,我是否正确,而在block1中,即使存在外部引用,内部可变引用也总是被删除？

Answer 1

这里将可变借用视为非Copy结构(S在下面的代码片段中)就足够了,它取得了引用值的所有权.该模型代表了可变借入的独有性质.

基于此模型的推理:在块2中n是对原始的引用m,而在块1中n最终将成为m对可变借入所拥有的副本的引用.在两个块中,内部引用在let-block的末尾被删除,但是仅在块1中这会导致问题,因为在块1中,n当删除此内部引用时,引用的目标仍然由内部引用所拥有.

struct S { m: i32 }
let mut m = 10;

let n = {
    let s = S { m };
    let b = &s;
    &(*b).m
}; // s is dropped

println!("{:?}", n);

在上面的代码片段中s取得了所有权的副本m.该引用n将指向n删除时s丢弃的副本- 不允许.如果m是非复制,m将被移入s,这将具有相同的含义.

在块2中,原件m直接借用而不复制它.如果强制复制,则会出现与block1相同的错误:

let mut m = 10;

let n = {
    let m2 = m;
    let mut a = &m2;
    let b = &a;
    &**b
};

println!("{:?}", n);