可变借用超出范围但不能重新借用

Question

当第一个可变借用似乎超出范围时，我无法理解为什么我不能第二次使用 v：

fn get_or_insert(v: &mut Vec<Option<i32>>, index: usize, default: i32) -> &mut i32 {
    if let Some(entry) = v.get_mut(index) { // <-- first borrow here
        if let Some(value) = entry.as_mut() {
            return value;
        }
    }

    // error[E0502]: cannot borrow `*v` as immutable because it is also borrowed as mutable    
    while v.len() <= index {                // <-- compiler error here
        v.push(None);
    }

    // error[E0499]: cannot borrow `*v` as mutable more than once at a time
    let entry = v.get_mut(index).unwrap();  // <-- compiler error here
    *entry = Some(default);
    entry.as_mut().unwrap()
}

playground link

是我的变量范围有误，还是借用检查器保护了我免受我看不到的东西的影响？

编辑：启用 NLL 的错误消息非常好：

error[E0502]: cannot borrow `*v` as immutable because it is also borrowed as mutable
  --> src/main.rs:10:11
   |
3  | fn get_or_insert(v: &mut Vec<Option<i32>>, index: usize, default: i32) -> &mut i32 {
   |                     - let's call the lifetime of this reference `'1`
4  |     if let Some(entry) = v.get_mut(index) {
   |                          - mutable borrow occurs here
5  |         if let Some(value) = entry.as_mut() {
6  |             return value;
   |                    ----- returning this value requires that `*v` is borrowed for `'1`
...
10 |     while v.len() <= index {
   |           ^ immutable borrow occurs here

Answer 1

关键是，即使使用 NLL，返回值的生命周期也会跨越整个函数。在确定v 引用是否可在下方代码中访问时，并未考虑函数在第 4 行提前返回的事实。

@Stargateur 建议的解决方法是在访问元素之前根据需要增大向量：

fn get_or_insert(v: &mut Vec<Option<i32>>, index: usize, value: i32) -> &mut i32 {
    if v.len() < index {
        v.resize(index + 1, None);
    }
    v[index].get_or_insert(value)
}

playground link

Here's where I used the technique in the final code.