reference - 使用可变引用迭代递归结构并返回最后一个有效引用

Question

我正在尝试向下递归节点结构，修改它们，然后返回我到达的最后一个 Node。我使用 example in the non-lexical lifetimes RFC 解决了循环中可变引用的问题.如果我尝试返回对最后一个 Node 的可变引用，我会得到一个 use of moved value 错误:

#[derive(Debug)]
struct Node {
    children: Vec<Node>,
}

impl Node {
    fn new(children: Vec<Self>) -> Self {
        Self { children }
    }
    fn get_last(&mut self) -> Option<&mut Node> {
        self.children.last_mut()
    }
}

fn main() {
    let mut root = Node::new(vec![Node::new(vec![])]);

    let current = &mut root;

    println!("Final: {:?}", get_last(current));
}


fn get_last(mut current: &mut Node) -> &mut Node {
    loop {
        let temp = current;
        println!("{:?}", temp);

        match temp.get_last() {
            Some(child) => { current = child },
            None => break,
        }
    }

    current
}

给出这个错误

error[E0382]: use of moved value: `*current`
  --> test.rs:51:5
   |
40 |         let temp = current;
   |             ---- value moved here
...
51 |     current
   |     ^^^^^^^ value used here after move
   |
   = note: move occurs because `current` has type `&mut Node`, which does not implement the `Copy` trait

如果我返回临时值而不是中断，我会得到错误cannot borrow as mutable more than once。

fn get_last(mut current: &mut Node) -> &mut Node {
    loop {
        let temp = current;
        println!("{:?}", temp);

        match temp.get_last() {
            Some(child) => { current = child },
            None => return temp,
        }
    }
}

error[E0499]: cannot borrow `*temp` as mutable more than once at a time
  --> test.rs:47:28
   |
43 |         match temp.get_last() {
   |               ---- first mutable borrow occurs here
...
47 |             None => return temp,
   |                            ^^^^ second mutable borrow occurs here
48 |         }
49 |     }
   |     - first borrow ends here

如何使用可变引用遍历结构并返回最后一个 Node？我进行了搜索，但没有找到针对此特定问题的任何解决方案。

我不能使用 Obtaining a mutable reference by iterating a recursive structure因为它给我一个借用不止一次的错误:

fn get_last(mut current: &mut Node) -> &mut Node {
    loop {
        let temp = current;
        println!("{:?}", temp);

        match temp.get_last() {
            Some(child) => current = child,
            None => current = temp,
        }
    }
    current
}

Answer 1

这确实不同于Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time .如果我们查看那里的答案，稍微修改一下，我们可以看到它匹配一个值，并且能够返回在终端案例中匹配的值。也就是说，返回值是一个Option:

fn back(&mut self) -> &mut Option<Box<Node>> {
    let mut anchor = &mut self.root;

    loop {
        match {anchor} {
            &mut Some(ref mut node) => anchor = &mut node.next,
            other => return other, // transferred ownership to here
        }
    }
}

您的情况在两个方面很复杂:

1. 缺少non-lexical lifetimes .

2. 在一种情况下(有 child )而不是在另一种情况下(没有 child )您想要获取可变引用并“放弃”这一事实。这在概念上与此相同:

   fn maybe_identity<T>(_: T) -> Option<T> { None }
   
   fn main() {
       let name = String::from("vivian");
   
       match maybe_identity(name) {
           Some(x) => println!("{}", x),
           None => println!("{}", name),
       }
   }
   

   编译器无法判断 None 情况可以(非常理论上)继续使用 name。

直接的解决方案是明确编码这个“找回” Action 。我们创建一个枚举，在没有 child 的情况下返回 &mut self，一个返回该枚举的辅助方法，并重写主要方法以使用辅助方法:

enum LastOrNot<'a> {
    Last(&'a mut Node),
    NotLast(&'a mut Node),
}

impl Node {
    fn get_last_or_self(&mut self) -> LastOrNot<'_> {
        match self.children.is_empty() {
            false => LastOrNot::Last(self.children.last_mut().unwrap()),
            true => LastOrNot::NotLast(self),
        }
    }

    fn get_last(mut current: &mut Node) -> &mut Node {
        loop {
            match { current }.get_last_or_self() {
                LastOrNot::Last(child) => current = child,
                LastOrNot::NotLast(end) => return end,
            }
        }
    }
}

请注意，我们正在使用 Returning a reference from a HashMap or Vec causes a borrow to last beyond the scope it's in? 中公开的所有技术和 Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time .

---

正在进行中 reimplementation of NLL ，我们可以稍微简化 get_last_or_self 以避免 bool 值:

fn get_last_or_self(&mut self) -> LastOrNot<'_> {
    match self.children.last_mut() {
        Some(l) => LastOrNot::Last(l),
        None => LastOrNot::NotLast(self),
    }
}

Polonius 的最终版本应该允许将整个问题简化为非常的简单形式:

fn get_last(mut current: &mut Node) -> &mut Node {
    while let Some(child) = current.get_last() {
        current = child;
    }

    current
}

另见:

• Returning a reference from a HashMap or Vec causes a borrow to last beyond the scope it's in?
• Cannot obtain a mutable reference when iterating a recursive structure: cannot borrow as mutable more than once at a time