wrestling with borrow checker - rust

Question

I'm new to Rust. As a learning exercise I'm trying to make a basic binary tree. This is what I have so far:
fn main() {
let data = vec![6,1,2,3,4,5];
let mut root = Node::<i32> { value: data[0], left: None, right: None };
for val in data {
createAndInsert::<i32>(&root, val);
}
println!("Root value: {}", root.value);
}
fn createAndInsert<T: PartialOrd>(mut root: &Node<T>, value: T) {
let mut n = Node::<T> { value: value, left: None, right: None };
insert::<T>(&root, &n);
}
fn insert<T: PartialOrd>(mut curr: &Node<T>, new: &Node<T>) {
if new.value > curr.value {
match curr.right {
Some(ref n) => insert(n, new),
None => curr.right = Some(Box::new(*new))
}
} else {
match curr.left {
Some(ref n) => insert(n, new),
None => curr.left = Some(Box::new(*new))
}
}
}
struct Node<T: PartialOrd> {
value: T,
left: Option<Box<Node<T>>>,
right: Option<Box<Node<T>>>,
}
The compiler errors I'm getting:
test.rs:21:48: 21:52 error: cannot move out of borrowed content
test.rs:21 None => curr.right = Some(Box::new(*new))
^~~~
test.rs:26:47: 26:51 error: cannot move out of borrowed content
test.rs:26 None => curr.left = Some(Box::new(*new))
^~~~
test.rs:21:21: 21:54 error: cannot assign to immutable field `curr.right`
test.rs:21 None => curr.right = Some(Box::new(*new))
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
test.rs:26:21: 26:53 error: cannot assign to immutable field `curr.left`
test.rs:26 None => curr.left = Some(Box::new(*new))
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
error: aborting due to 4 previous errors
I've got myself tangled in all the refs and muts and &'s and *'s and I'm not sure how to get out. Where am I going wrong?

Answer 1

You have two problems:
Cannot move out of borrowed context: see Cannot move out of borrowed content when borrowing a generic type for an explanation.
Cannot assign to immutable field: you only have a &Node<T>; to modify the Node you need a &mut Node<T>. mut curr in the pattern merely makes the binding mutable, meaning that you can assign a new value to curr. You can’t, however, modify the contents of what curr refers to. Propagate the &-to-&mut conversion throughout the code and it’ll work.

Answer 2

Since you are new to Rust it might help to see how I would have written it:
struct Node<T> {
value: T,
left: Option<Box<Node<T>>>,
right: Option<Box<Node<T>>>,
}
impl<T> Node<T> {
fn new(x: T) -> Node<T> {
Node { value: x, left: None, right: None }
}
fn boxed(x: T) -> Box<Node<T>> {
Box::new(Node::new(x))
}
}
fn insert<T: PartialOrd>(root: &mut Option<Box<Node<T>>>, new: Box<Node<T>>) {
if let Some(ref mut rbx) = *root {
if new.value < rbx.value {
insert(&mut rbx.left, new);
} else {
insert(&mut rbx.right, new);
}
} else {
*root = Some(new);
}
}
fn main() {
let data = vec![6,1,2,3,4,5];
let mut root = None;
for val in data {
insert(&mut root, Node::boxed(val));
}
println!("Root value: {}", root.unwrap().value);
}
I realize that this is more of an exercise but keep in mind that this kind of data structure should not grow beyond a certain tree depth since it might otherwise cause the stack to overflow when the nodes are recursivly deallocated.