Tree 实现可变树结构
我正在尝试动态构建一棵树,并在下降、叶子和备份期间修改树的部分内容。我相信我对如何在Rust做这样的事情有一个基本的误解。这是我的密码:Tree 实现可变树结构,tree,rust,mutable,Tree,Rust,Mutable,我正在尝试动态构建一棵树,并在下降、叶子和备份期间修改树的部分内容。我相信我对如何在Rust做这样的事情有一个基本的误解。这是我的密码: struct Node { children: Vec<Node>, data: usize, } impl Node { pub fn new() -> Node { Node { children: vec![], data: 0,
struct Node {
children: Vec<Node>,
data: usize,
}
impl Node {
pub fn new() -> Node {
Node {
children: vec![],
data: 0,
}
}
pub fn expand(&mut self) {
self.children = vec![Node::new(), Node::new()];
}
pub fn is_leaf(&self) -> bool {
self.children.len() == 0
}
}
pub fn main() {
let mut root = Node::new();
for _ in 0..10 {
let mut node = &mut root;
let mut path = vec![];
// Descend and potential modify the node in the process
while !node.is_leaf() {
let index = 0;
path.push(index);
node = &mut node.children[index];
}
// Do something to the leaf node
node.expand();
// Do something during "backup" (in my case it doesn't matter
// in which order the modification is happening).
node = &mut root;
for &i in path.iter() {
node.data += 1;
node = &mut node.children[i];
}
}
}
这里同时发生了一些事情。简单的答案是:您正试图为同一个项目创建多个可变借用。Rust禁止您创建多个借用,即使您没有尝试修改它们(因为这比试图正式证明您的程序是正确的要容易) 由于您基本上试图以命令式的方式实现递归函数,因此我建议您采用更具功能性的方法来解决问题。我将循环中的逻辑移到了递归函数中,该函数直接在
节点上实现
struct Node {
children: Vec<Node>,
data: usize,
}
impl Node {
pub fn new() -> Node {
Node {
children: vec!(),
data: 0
}
}
pub fn expand(&mut self) {
self.children = vec!(Node::new(), Node::new());
}
pub fn is_leaf(&self) -> bool {
self.children.len() == 0
}
fn expand_leaf_and_inc(&mut self) {
if self.is_leaf() {
self.expand();
} else {
let index = 0;
self.children[index].expand_leaf_and_inc();
}
self.data += 1
}
}
pub fn main() {
let mut root = Node::new();
for _ in 0..10 {
root.expand_leaf_and_inc();
}
}
在上一个版本中,这是一个非常巧妙的技巧,在path
的赋值过程中,您引入了一个新的作用域。我知道这一点!我想得太迫切了。我仍然需要接受锈的某种功能性质。
let mut root = Node::new();
for _ in 0..10 {
let mut path = vec![];
{
let mut node = &mut root;
// Descend and potential modify the node in the process
while !node.is_leaf() {
let index = 0;
path.push(index);
node = &mut {node}.children[index];
}
// Do something to the leaf node
node.expand();
}
// Do something during "backup" (in my case it doesn't matter
// in which order the modification is happening).
let mut node = &mut root;
for &i in path.iter() {
node.data += 1;
node = &mut {node}.children[i];
}
}