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Tree traversal in Rust vs Borrow Checker


Tree traversal in Rust vs Borrow Checker

By : user2956230
Date : November 22 2020, 10:54 AM
it helps some times Here's a variant of your first approach, using recursion to avoid borrowing conflicts. The iterative equivalent fails to compile because Rust is too strict when dealing with mutable borrowed pointers to mutable values.
code :
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)
    }
}


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Fighting with the Rust borrow checker

Fighting with the Rust borrow checker


By : Harnil Oza
Date : March 29 2020, 07:55 AM
will help you Firstly, you shouldn't ask for a BufReader<&'static File>. You can't provide one. Ask instead for a BufReader<&'a File> for some lifetime 'a.
code :
fn display_screenful<'a>(rb: &RustBox, fr: BufReader<&'a File>, offset: usize)
fn display_screenful(rb: &RustBox, fr: BufReader<&File>, offset: usize)
fn display_screenful(rb: &RustBox, fr: &mut BufReader<&File>, offset: usize)
When is it necessary to circumvent Rust's borrow checker?

When is it necessary to circumvent Rust's borrow checker?


By : anthonyb
Date : March 29 2020, 07:55 AM
may help you . I'm implementing Conway's game of life to teach myself Rust. The idea is to implement a single-threaded version first, optimize it as much as possible, then do the same for a multi-threaded version. ,
When is it necessary to circumvent Rust's borrow checker?
code :
let mut array = [1, 2];
let a = &mut array[0];
let b = &mut array[1];
let mut array = [1, 2];
let (a, b) = array.split_at_mut(1);
let a = &mut a[0];
let b = &mut b[0];
fn split_at_mut(&mut self, mid: usize) -> (&mut [T], &mut [T]) {
    let len = self.len();
    let ptr = self.as_mut_ptr();

    unsafe {
        assert!(mid <= len);

        (from_raw_parts_mut(ptr, mid),
         from_raw_parts_mut(ptr.offset(mid as isize), len - mid))
    }
}
What is Rust's borrow checker really complaining about here?

What is Rust's borrow checker really complaining about here?


By : Lhoussin
Date : March 29 2020, 07:55 AM
it fixes the issue When you try to access collection[j], the compiler returns a &mut String because that's the type of the vector's elements. When you try to access collection[least_element], the borrow checker doesn't know if least_element != j, and having two mutable references of the same element would be undefined behavior. You can either use std::ops::Index which returns a &&mut String (and it's safe to have two immutable references to the same mutable reference), directly borrowing the elements (&collection[j] < &collection[least_element]) or, if possible, changing the type of collection to Vec<&String> or Vec.
Rust matching and borrow checker

Rust matching and borrow checker


By : schuman
Date : March 29 2020, 07:55 AM
it fixes the issue I keep stumbling on a pattern in my Rust programs that always puts me at odds with the borrow-checker. Consider the following toy example: , Use directly the field those, for example with custom type:
Borrow checker doesn't allow returning a mutable reference from tree traversal function

Borrow checker doesn't allow returning a mutable reference from tree traversal function


By : uhdi
Date : March 29 2020, 07:55 AM
may help you . I need to find the node with maximal value in tree, assuming that subnode's values are always larger than value of owning node, and then modify it: , It is not required to use unsafe:
code :
fn find_max(node: &mut Node, val: usize) -> Option<&mut Node> {
    if node.val < val {
        return None;
    }

    if node.nodes.is_empty() {
        return Some(node);
    }

    let mut max_val = node.val;
    let mut max = None;
    for n in &mut node.nodes {
        if let Some(m) = find_max(n, max_val) {
            max_val = m.val;
            max = Some(m);
        }
    }
    max
}
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