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tree - "Borrowed value does not live long enough"在 B-Tree 节点枚举上匹配以获取引用时

转载 作者:行者123 更新时间:2023-11-29 08:29:37 26 4
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我正在尝试在 Rust 中实现 B 树,并且我正在努力解决如何管理与 match 语句交互的生命周期。我是 Rust 的新手,虽然我认为我了解生命周期是什么以及它们如何工作,但我无法弄清楚如何让它们做我想做的事。

我的数据结构是这样的:

enum Node<T: Key> {
Internal(InternalNode<T>),
Leaf(LeafNode<T>),
}

enum NodeRef<'a, T: 'a + Key> {
Internal(&'a InternalNode<T>),
Leaf(&'a LeafNode<T>),
}

enum NodeRefMut<'a, T: 'a + Key> {
Internal(&'a mut InternalNode<T>),
Leaf(&'a mut LeafNode<T>),
}

struct InternalNode<T: Key> {
keys: Vec<T>,
children: Vec<Box<Node<T>>>,
num_keys: usize,
}

struct LeafNode<T: Key> {
keys: Vec<T>,
num_keys: usize,
}

pub struct BTree<T: Key> {
num_keys: usize,
root: Node<T>
}

在 B 树插入中,您需要插入一个节点,然后保留对您插入的节点的引用,以便进行平衡和拆分。这不是关于算法的问题,但这是一个说明问题的代码片段:

pub fn insert(&mut self, key: T) -> bool {
let (_inserted_at, success) = match self.root {
Node::Internal(ref mut node) =>
BTree::insert_at_internal_node(&mut node, key),

Node::Leaf(ref mut node) =>
BTree::insert_at_leaf_node(&mut node, key),
};

panic!() // snip
}

fn insert_at_leaf_node<'a>(leaf: &'a mut LeafNode<T>, key: T) -> (NodeRefMut<'a, T>, bool) {
panic!() // snip
}

fn insert_at_internal_node<'a>(internal: &'a mut InternalNode<T>, key: T) -> (NodeRefMut<'a, T>, bool) {
panic!() // snip
}

然后我们得到以下我不知道如何解决的相当明显的错误:

   error[E0597]: `node` does not live long enough
--> src/trees/mod.rs:102:53
|
102 | BTree::insert_at_internal_node(&mut node, key),
| ^^^^ borrowed value does not live long enough
...
106 | };
| - `node` dropped here while still borrowed
...
109 | }
| - borrowed value needs to live until here

error[E0597]: `node` does not live long enough
--> src/trees/mod.rs:105:49
|
105 | BTree::insert_at_leaf_node(&mut node, key),
| ^^^^ borrowed value does not live long enough
106 | };
| - `node` dropped here while still borrowed
...
109 | }
| - borrowed value needs to live until here

本质上,我们需要一个对我们插入的叶节点的可变引用,insert_at 方法可以轻松生成它。但是,当我执行 match 语句时,它会生成一个对节点的可变引用,该节点的生命周期与 match 语句一样长(因此生成的引用不能超过 match block ,从而违背了目的).

目前我只想可变地借用所有东西(我有一个 &mut self 是有原因的)但我不知道该怎么做。

最佳答案

TL;DR — 从调用中删除 &mut:

let (_inserted_at, success) = match self.root {
Node::Internal(ref mut node) => BTree::insert_at_internal_node(node, key),
Node::Leaf(ref mut node) => BTree::insert_at_leaf_node(node, key),
};

您的问题可以简化为(并且可能进一步简化,但这仍然具有原始的一般形状):

enum Node {
Internal(InternalNode),
}

enum NodeRefMut<'a> {
Internal(&'a mut InternalNode),
}

struct InternalNode;

fn insert(mut root: Node) {
let _inserted_at = match root {
Node::Internal(ref mut node) => internal(&mut node),
};
}

fn internal<'a>(_internal: &'a mut InternalNode) -> NodeRefMut<'a> {
unimplemented!()
}

如果你print out the type of node ,您会看到,通过在模式中使用 ref mut,您已经创建了一个 &mut 节点:

Node::Internal(ref mut node) => {
let _: () = node;
internal(&mut node)
}
error[E0308]: mismatched types
--> src/main.rs:14:21
|
14 | let _: () = node;
| ^^^^ expected (), found mutable reference
|
= note: expected type `()`
found type `&mut InternalNode`

当您获取第二个可变引用时,您获取的是对仅在匹配臂范围内的变量的可变引用。该变量的生命周期正是错误消息所指出的:匹配臂本身。

有趣的是,此错误仅在您从 match 返回值时发生。如果你不这样做,你会得到一个不同的错误:

error[E0596]: cannot borrow immutable local variable `node` as mutable
--> src/main.rs:17:27
|
17 | internal(&mut node)
| ^^^^
| |
| cannot reborrow mutably
| try removing `&mut` here

因为变量 node 本身确实是不可变的!

关于tree - "Borrowed value does not live long enough"在 B-Tree 节点枚举上匹配以获取引用时,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49709624/

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