Rust vs Borrow Checker 中的树遍历

Question

我正在尝试在 Rust 中实现一个树结构，遍历它并修改它，但我遇到了借用检查器的问题。我的设置或多或少如下：

#![feature(slicing_syntax)]

use std::collections::HashMap;

#[deriving(PartialEq, Eq, Hash)]
struct Id {
    id: int,  // let’s pretend it’s that
}

struct Node {
    children: HashMap<Id, Box<Node>>,
    decoration: String,
    // other fields
}

struct Tree {
   root: Box<Node>
}

impl Tree {
    /// Traverse the nodes along the specified path.
    /// Return the node at which traversal stops either because the path is exhausted
    /// or because there are no more nodes matching the path.
    /// Also return any remaining steps in the path that did not have matching nodes.
    fn traverse_path<'p>(&mut self, mut path: &'p [Id]) -> (&mut Box<Node>, &'p [Id]) {
        let mut node = &mut self.root;
        loop {
            match node.children.get_mut(&path[0]) {
                Some(child_node) => {
                    path = path[1..];
                    node = child_node;
                },
                None => {
                    break;
                }
            }
        }
        (node, path)
    }
}

我在这里有可变引用，因为我希望能够改变方法返回的节点。例如，add 方法会调用 traverse_path，然后为路径中没有匹配节点的剩余部分添加节点。

这会产生以下错误：

s.rs:28:19: 28:32 error: cannot borrow `node.children` as mutable more than once at a time
s.rs:28             match node.children.get_mut(&path[0]) {
                          ^~~~~~~~~~~~~
s.rs:28:19: 28:32 note: previous borrow of `node.children` occurs here; the mutable borrow prevents subsequent moves, borrows, or modification of `node.children` until the borrow ends
s.rs:28             match node.children.get_mut(&path[0]) {
                          ^~~~~~~~~~~~~
s.rs:39:6: 39:6 note: previous borrow ends here
s.rs:25     fn traverse_path<'p>(&mut self, mut path: &'p [Id]) -> (&mut Box<Node>, &'p [Id]) {
...
s.rs:39     }
            ^
s.rs:31:21: 31:38 error: cannot assign to `node` because it is borrowed
s.rs:31                     node = child_node;
                            ^~~~~~~~~~~~~~~~~
s.rs:28:19: 28:32 note: borrow of `node` occurs here
s.rs:28             match node.children.get_mut(&path[0]) {
                          ^~~~~~~~~~~~~
s.rs:38:10: 38:14 error: cannot borrow `*node` as mutable more than once at a time
s.rs:38         (node, path)
                 ^~~~
s.rs:28:19: 28:32 note: previous borrow of `node.children` occurs here; the mutable borrow prevents subsequent moves, borrows, or modification of `node.children` until the borrow ends
s.rs:28             match node.children.get_mut(&path[0]) {
                          ^~~~~~~~~~~~~
s.rs:39:6: 39:6 note: previous borrow ends here
s.rs:25     fn traverse_path<'p>(&mut self, mut path: &'p [Id]) -> (&mut Box<Node>, &'p [Id]) {
...
s.rs:39     }
            ^
error: aborting due to 3 previous errors

我明白为什么借用检查器不喜欢这段代码，但我不知道如何使它工作。

我还尝试了使用迭代器的替代实现，使用如下代码：

struct PathIter<'a> {
    path: &'a [Id],
    node: &'a mut Box<Node>
}
impl<'a> Iterator<Box<Node>> for PathIter<'a> {
    fn next(&mut self) -> Option<Box<Node>> {
        let child = self.node.get_child(&self.path[0]);
        if child.is_some() {
            self.path = self.path[1..];
            self.node = child.unwrap();
        }
        child
    }
}

这里的错误最终与生命周期相关：

src/http_prefix_tree.rs:147:27: 147:53 error: cannot infer an appropriate lifetime for autoref due to conflicting requirements
src/http_prefix_tree.rs:147     let child = self.node.get_child(&self.path[0]);
                                                  ^~~~~~~~~~~~~~~~~~~~~~~~~~
src/http_prefix_tree.rs:146:3: 153:4 help: consider using an explicit lifetime parameter as shown: fn next(&'a mut self) -> Option<Box<Node>>
src/http_prefix_tree.rs:146   fn next(&mut self) -> Option<Box<Node>> {
src/http_prefix_tree.rs:147     let child = self.node.get_child(&self.path[0]);
src/http_prefix_tree.rs:148     if child.is_some() {
src/http_prefix_tree.rs:149       self.path = self.path[1..];
src/http_prefix_tree.rs:150       self.node = child.unwrap();
src/http_prefix_tree.rs:151     }

我感兴趣的另一件事是收集匹配节点的decoration 字段的值，并在路径完全耗尽时显示这些值。我的第一个想法是从节点到其父节点的反向链接，但我发现的唯一例子是 DList 中的 Rawlink，这让我感到害怕。我的下一个希望是迭代器实现（如果我能让它工作的话）自然而然地适合这样的事情。这是正确的追求吗？

Answer 1

这是第一种方法的变体，使用递归来避免借用冲突。迭代等价物无法编译，因为 Rust 在处理指向可变值的可变借用指针时过于严格。

impl Node {
    fn traverse_path<'p>(&mut self, mut path: &'p [Id]) -> (&mut Node, &'p [Id]) { // '
        if self.children.contains_key(&path[0]) {
            self.children[path[0]].traverse_path(path[1..])
        } else {
            (self, path)
        }
    }
}

impl Tree {
    /// Traverse the nodes along the specified path.
    /// Return the node at which traversal stops either because the path is exhausted
    /// or because there are no more nodes matching the path.
    /// Also return any remaining steps in the path that did not have matching nodes.
    fn traverse_path<'p>(&mut self, mut path: &'p [Id]) -> (&mut Node, &'p [Id]) { // '
        self.root.traverse_path(path)
    }
}

请注意，我已将返回类型从 &mut Box<Node> 更改为 &mut Node；您无需向用户透露您在实施中使用了Box。另外，查看Node::traverse_path 如何首先使用contains_key() 检查映射中是否存在值，然后使用索引检索该值。这意味着该值会被查找两次，但这是我发现无需不安全代码即可完成这项工作的唯一方法。

P.S.：您可以将Tree 中的root 更改为Node，而不是Box<Node>。