I'm trying to borrow a hashset inside of a closure of type Filter from the inquire crate
Basically, I want to prevent users from selecting the same option more than once
Here's my code
pub fn getRotors() -> () {
let mut selectedrotors: HashSet<Rotors> = HashSet::new();
let options: Vec<Rotors> = Rotors::iter().collect();
let filter: Filter<Rotors> = &|_, cur_option, _, _| match selectedrotors.contains(cur_option) { // Borrow occurs here
true => false,
false => true,
};
for i in 0..3 {
// Get Rotor from user
let r: Rotors = Select::new("Select a rotor", options.to_vec())
.with_filter(filter)
.prompt()
.unwrap();
selectedrotors.insert(r); // immutable borrow occurs here e0502
// Get char from user
}
()
}
I would expect that selectedrotors.contains to only require immutable access to itself so i'm at a loss here
You're confusing the two borrows. There is an immutable borrow of selectedrotors in your closure and a mutable borrow in the loop. There is an issue because you try to take a mutable borrow in the loop while the immutable borrow from the closure is still active (because the closure will be used in the next loop iteration).
You can fix the issue by declaring the filter inside the loop so that the immutable borrow can be released before taking the mutable one:
pub fn getRotors() -> () {
let mut selectedrotors: HashSet<Rotors> = HashSet::new();
let options: Vec<Rotors> = Rotors::iter().collect();
for i in 0..3 {
let filter: Filter<Rotors> = &|_, cur_option, _, _| match selectedrotors.contains(cur_option) {
true => false,
false => true,
};
// Get Rotor from user
let r: Rotors = Select::new("Select a rotor", options.to_vec())
.with_filter(filter)
.prompt()
.unwrap();
// filter is no longer used here, so its borrow can be released
// allowing the mutable borrow in the next line to proceed.
selectedrotors.insert(r);
// Get char from user
}
()
}
I dont know much about the crate you are using but since you accepted Jmb's answer I would like to point out that your filter is bizarre. Here's a better way:
for i in 0..3 {
// Get Rotor from user
let r = Select::new("Select a rotor", options.to_vec())
.with_filter(|_, cur_option, _, _| !selected_rotors.contains(cur_option))
.prompt()
.unwrap();
selected_rotors.insert(r);
}
Related
I'm new to Rust and still reading the Rust book. Below is my program.
use clap::{App, Arg};
type GenericError = Box<dyn std::error::Error + Send + Sync + 'static>;
type GenericResult<T> = Result<T, GenericError>;
fn main() -> GenericResult<()> {
let matches = App::new("test")
.arg(Arg::new("latency")
.takes_value(true))
.get_matches();
let latency_present = matches.is_present("latency");
let latency = matches.value_of("latency").unwrap_or("1000:10,300:30");
let latency_pairs: Vec<&str> = latency.split(",").collect();
let checker = std::thread::spawn(move || -> GenericResult<()>{
loop {
if latency_present {
for (i, latency_pair) in latency_pairs.iter().enumerate() {
// let latency_pair: Vec<&str> = latency_pair.split(":").collect();
// let latency = latency_pair[0].parse::<f64>().unwrap();
}
}
}
});
checker.join().unwrap()?;
Ok(())
}
When I run it, it tells me this:
error[E0597]: `matches` does not live long enough
--> src\main.rs:14:19
|
14 | let latency = matches.value_of("latency").unwrap_or("1000:10,300:30");
| ^^^^^^^--------------------
| |
| borrowed value does not live long enough
| argument requires that `matches` is borrowed for `'static`
...
30 | }
| - `matches` dropped here while still borrowed
I don't quite understand the error messages here. But I guess it's because I use latency_pairs in the checker thread and latency_pairs could get dropped while checker is still executing. Is my understanding correct? How to fix the error? I tried for (i, latency_pair) in latency_pairs.clone().iter().enumerate() { in order to pass a cloned value for the thread, but it doesn't help.
latency_pairs holds references into latency which in turn references matches. Thus cloning latency_pairs just clones the references into latency and matches.
Your code would require that latency's type is &'static str but it's actually &'a str where 'a is bound to matches' lifetime.
You can call to_owned() on latency to get an owned value and split the string inside the closure or you can call to_owned() on each of the splits collected in latency_pairs and move that Vec<String> into the closure:
let latency_pairs: Vec<String> = latency.split(",").map(ToOwned::to_owned).collect();
let checker = std::thread::spawn(move || -> GenericResult<()>{
loop {
if latency_present {
for (i, latency_pair) in latency_pairs.iter().enumerate() {
// let latency_pair: Vec<&str> = latency_pair.split(":").collect();
// let latency = latency_pair[0].parse::<f64>().unwrap();
}
}
}
});
If you need to use latency_pairs outside of the closure, you can clone it before moving it into the closure:
let latency_pairs: Vec<String> = latency.split(",").map(ToOwned::to_owned).collect();
let latency_pairs_ = latency_pairs.clone();
let checker = std::thread::spawn(move || -> GenericResult<()>{
loop {
if latency_present {
for (i, latency_pair) in latency_pairs_.iter().enumerate() {
// let latency_pair: Vec<&str> = latency_pair.split(":").collect();
// let latency = latency_pair[0].parse::<f64>().unwrap();
}
}
}
});
println!("{:?}", latency_pairs);
I am learning Rust and I don't quite get why this is not working.
#[derive(Debug)]
struct Node {
value: String,
}
#[derive(Debug)]
pub struct Graph {
nodes: Vec<Box<Node>>,
}
fn mk_node(value: String) -> Node {
Node { value }
}
pub fn mk_graph() -> Graph {
Graph { nodes: vec![] }
}
impl Graph {
fn add_node(&mut self, value: String) {
if let None = self.nodes.iter().position(|node| node.value == value) {
let node = Box::new(mk_node(value));
self.nodes.push(node);
};
}
fn get_node_by_value(&self, value: &str) -> Option<&Node> {
match self.nodes.iter().position(|node| node.value == *value) {
None => None,
Some(idx) => self.nodes.get(idx).map(|n| &**n),
}
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn some_test() {
let mut graph = mk_graph();
graph.add_node("source".to_string());
graph.add_node("destination".to_string());
let source = graph.get_node_by_value("source").unwrap();
let dest = graph.get_node_by_value("destination").unwrap();
graph.add_node("destination".to_string());
}
}
(playground)
This has the error
error[E0502]: cannot borrow `graph` as mutable because it is also borrowed as immutable
--> src/main.rs:50:9
|
47 | let source = graph.get_node_by_value("source").unwrap();
| ----- immutable borrow occurs here
...
50 | graph.add_node("destination".to_string());
| ^^^^^ mutable borrow occurs here
51 | }
| - immutable borrow ends here
This example from Programming Rust is quite similar to what I have but it works:
pub struct Queue {
older: Vec<char>, // older elements, eldest last.
younger: Vec<char>, // younger elements, youngest last.
}
impl Queue {
/// Push a character onto the back of a queue.
pub fn push(&mut self, c: char) {
self.younger.push(c);
}
/// Pop a character off the front of a queue. Return `Some(c)` if there /// was a character to pop, or `None` if the queue was empty.
pub fn pop(&mut self) -> Option<char> {
if self.older.is_empty() {
if self.younger.is_empty() {
return None;
}
// Bring the elements in younger over to older, and put them in // the promised order.
use std::mem::swap;
swap(&mut self.older, &mut self.younger);
self.older.reverse();
}
// Now older is guaranteed to have something. Vec's pop method // already returns an Option, so we're set.
self.older.pop()
}
pub fn split(self) -> (Vec<char>, Vec<char>) {
(self.older, self.younger)
}
}
pub fn main() {
let mut q = Queue {
older: Vec::new(),
younger: Vec::new(),
};
q.push('P');
q.push('D');
assert_eq!(q.pop(), Some('P'));
q.push('X');
let (older, younger) = q.split(); // q is now uninitialized.
assert_eq!(older, vec!['D']);
assert_eq!(younger, vec!['X']);
}
A MRE of your problem can be reduced to this:
// This applies to the version of Rust this question
// was asked about; see below for updated examples.
fn main() {
let mut items = vec![1];
let item = items.last();
items.push(2);
}
error[E0502]: cannot borrow `items` as mutable because it is also borrowed as immutable
--> src/main.rs:4:5
|
3 | let item = items.last();
| ----- immutable borrow occurs here
4 | items.push(2);
| ^^^^^ mutable borrow occurs here
5 | }
| - immutable borrow ends here
You are encountering the exact problem that Rust was designed to prevent. You have a reference pointing into the vector and are attempting to insert into the vector. Doing so might require that the memory of the vector be reallocated, invalidating any existing references. If that happened and you used the value in item, you'd be accessing uninitialized memory, potentially causing a crash.
In this particular case, you aren't actually using item (or source, in the original) so you could just... not call that line. I assume you did that for some reason, so you could wrap the references in a block so that they go away before you try to mutate the value again:
fn main() {
let mut items = vec![1];
{
let item = items.last();
}
items.push(2);
}
This trick is no longer needed in modern Rust because non-lexical lifetimes have been implemented, but the underlying restriction still remains — you cannot have a mutable reference while there are other references to the same thing. This is one of the rules of references covered in The Rust Programming Language. A modified example that still does not work with NLL:
let mut items = vec![1];
let item = items.last();
items.push(2);
println!("{:?}", item);
In other cases, you can copy or clone the value in the vector. The item will no longer be a reference and you can modify the vector as you see fit:
fn main() {
let mut items = vec![1];
let item = items.last().cloned();
items.push(2);
}
If your type isn't cloneable, you can transform it into a reference-counted value (such as Rc or Arc) which can then be cloned. You may or may not also need to use interior mutability:
struct NonClone;
use std::rc::Rc;
fn main() {
let mut items = vec![Rc::new(NonClone)];
let item = items.last().cloned();
items.push(Rc::new(NonClone));
}
this example from Programming Rust is quite similar
No, it's not, seeing as how it doesn't use references at all.
See also
Cannot borrow `*x` as mutable because it is also borrowed as immutable
Pushing something into a vector depending on its last element
Why doesn't the lifetime of a mutable borrow end when the function call is complete?
How should I restructure my graph code to avoid an "Cannot borrow variable as mutable more than once at a time" error?
Why do I get the error "cannot borrow x as mutable more than once"?
Why does Rust want to borrow a variable as mutable more than once at a time?
Try to put your immutable borrow inside a block {...}.
This ends the borrow after the block.
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn some_test() {
let mut graph = mk_graph();
graph.add_node("source".to_string());
graph.add_node("destination".to_string());
{
let source = graph.get_node_by_value("source").unwrap();
let dest = graph.get_node_by_value("destination").unwrap();
}
graph.add_node("destination".to_string());
}
}
So for anyone else banging their head against this problem and wanting a quick way out - use clones instead of references. Eg I'm iterating this list of cells and want to change an attribute so I first copy the list:
let created = self.cells
.into_iter()
.map(|c| {
BoardCell {
x: c.x,
y: c.y,
owner: c.owner,
adjacency: c.adjacency.clone(),
}
})
.collect::<Vec<BoardCell>>();
And then modify the values in the original by looping the copy:
for c in created {
self.cells[(c.x + c.y * self.size) as usize].adjacency[dir] = count;
}
Using Vec<&BoardCell> would just yield this error. Not sure how Rusty this is but hey, it works.
I am learning Rust and I don't quite get why this is not working.
#[derive(Debug)]
struct Node {
value: String,
}
#[derive(Debug)]
pub struct Graph {
nodes: Vec<Box<Node>>,
}
fn mk_node(value: String) -> Node {
Node { value }
}
pub fn mk_graph() -> Graph {
Graph { nodes: vec![] }
}
impl Graph {
fn add_node(&mut self, value: String) {
if let None = self.nodes.iter().position(|node| node.value == value) {
let node = Box::new(mk_node(value));
self.nodes.push(node);
};
}
fn get_node_by_value(&self, value: &str) -> Option<&Node> {
match self.nodes.iter().position(|node| node.value == *value) {
None => None,
Some(idx) => self.nodes.get(idx).map(|n| &**n),
}
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn some_test() {
let mut graph = mk_graph();
graph.add_node("source".to_string());
graph.add_node("destination".to_string());
let source = graph.get_node_by_value("source").unwrap();
let dest = graph.get_node_by_value("destination").unwrap();
graph.add_node("destination".to_string());
}
}
(playground)
This has the error
error[E0502]: cannot borrow `graph` as mutable because it is also borrowed as immutable
--> src/main.rs:50:9
|
47 | let source = graph.get_node_by_value("source").unwrap();
| ----- immutable borrow occurs here
...
50 | graph.add_node("destination".to_string());
| ^^^^^ mutable borrow occurs here
51 | }
| - immutable borrow ends here
This example from Programming Rust is quite similar to what I have but it works:
pub struct Queue {
older: Vec<char>, // older elements, eldest last.
younger: Vec<char>, // younger elements, youngest last.
}
impl Queue {
/// Push a character onto the back of a queue.
pub fn push(&mut self, c: char) {
self.younger.push(c);
}
/// Pop a character off the front of a queue. Return `Some(c)` if there /// was a character to pop, or `None` if the queue was empty.
pub fn pop(&mut self) -> Option<char> {
if self.older.is_empty() {
if self.younger.is_empty() {
return None;
}
// Bring the elements in younger over to older, and put them in // the promised order.
use std::mem::swap;
swap(&mut self.older, &mut self.younger);
self.older.reverse();
}
// Now older is guaranteed to have something. Vec's pop method // already returns an Option, so we're set.
self.older.pop()
}
pub fn split(self) -> (Vec<char>, Vec<char>) {
(self.older, self.younger)
}
}
pub fn main() {
let mut q = Queue {
older: Vec::new(),
younger: Vec::new(),
};
q.push('P');
q.push('D');
assert_eq!(q.pop(), Some('P'));
q.push('X');
let (older, younger) = q.split(); // q is now uninitialized.
assert_eq!(older, vec!['D']);
assert_eq!(younger, vec!['X']);
}
A MRE of your problem can be reduced to this:
// This applies to the version of Rust this question
// was asked about; see below for updated examples.
fn main() {
let mut items = vec![1];
let item = items.last();
items.push(2);
}
error[E0502]: cannot borrow `items` as mutable because it is also borrowed as immutable
--> src/main.rs:4:5
|
3 | let item = items.last();
| ----- immutable borrow occurs here
4 | items.push(2);
| ^^^^^ mutable borrow occurs here
5 | }
| - immutable borrow ends here
You are encountering the exact problem that Rust was designed to prevent. You have a reference pointing into the vector and are attempting to insert into the vector. Doing so might require that the memory of the vector be reallocated, invalidating any existing references. If that happened and you used the value in item, you'd be accessing uninitialized memory, potentially causing a crash.
In this particular case, you aren't actually using item (or source, in the original) so you could just... not call that line. I assume you did that for some reason, so you could wrap the references in a block so that they go away before you try to mutate the value again:
fn main() {
let mut items = vec![1];
{
let item = items.last();
}
items.push(2);
}
This trick is no longer needed in modern Rust because non-lexical lifetimes have been implemented, but the underlying restriction still remains — you cannot have a mutable reference while there are other references to the same thing. This is one of the rules of references covered in The Rust Programming Language. A modified example that still does not work with NLL:
let mut items = vec![1];
let item = items.last();
items.push(2);
println!("{:?}", item);
In other cases, you can copy or clone the value in the vector. The item will no longer be a reference and you can modify the vector as you see fit:
fn main() {
let mut items = vec![1];
let item = items.last().cloned();
items.push(2);
}
If your type isn't cloneable, you can transform it into a reference-counted value (such as Rc or Arc) which can then be cloned. You may or may not also need to use interior mutability:
struct NonClone;
use std::rc::Rc;
fn main() {
let mut items = vec![Rc::new(NonClone)];
let item = items.last().cloned();
items.push(Rc::new(NonClone));
}
this example from Programming Rust is quite similar
No, it's not, seeing as how it doesn't use references at all.
See also
Cannot borrow `*x` as mutable because it is also borrowed as immutable
Pushing something into a vector depending on its last element
Why doesn't the lifetime of a mutable borrow end when the function call is complete?
How should I restructure my graph code to avoid an "Cannot borrow variable as mutable more than once at a time" error?
Why do I get the error "cannot borrow x as mutable more than once"?
Why does Rust want to borrow a variable as mutable more than once at a time?
Try to put your immutable borrow inside a block {...}.
This ends the borrow after the block.
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn some_test() {
let mut graph = mk_graph();
graph.add_node("source".to_string());
graph.add_node("destination".to_string());
{
let source = graph.get_node_by_value("source").unwrap();
let dest = graph.get_node_by_value("destination").unwrap();
}
graph.add_node("destination".to_string());
}
}
So for anyone else banging their head against this problem and wanting a quick way out - use clones instead of references. Eg I'm iterating this list of cells and want to change an attribute so I first copy the list:
let created = self.cells
.into_iter()
.map(|c| {
BoardCell {
x: c.x,
y: c.y,
owner: c.owner,
adjacency: c.adjacency.clone(),
}
})
.collect::<Vec<BoardCell>>();
And then modify the values in the original by looping the copy:
for c in created {
self.cells[(c.x + c.y * self.size) as usize].adjacency[dir] = count;
}
Using Vec<&BoardCell> would just yield this error. Not sure how Rusty this is but hey, it works.
I am still learning Rust and when trying to implement Dikjstra as part of a training project, I encountered this peculiar catch. First I define a HashMap:
let mut dist: HashMap<Node, usize> = HashMap::new();
And later:
let state = State { node: next_node.clone(), cost: cost + 1 };
let current_dist = dist.get(&state.node);
if (current_dist == None) || (state.cost < *current_dist.unwrap()) {
dist.insert(state.node.clone(), state.cost);
heap.push(state);
}
Which yields a compile error because dist.get triggers a immutable borrow which stays in scope until after the if ... {...} statement, and in particular when I dist.insert, asking for a mutable borrow.
I think I miss a pattern or a keyword allowing me this type of process. For now I tried a drop at the beginning of the if scope, and other current_dist evaluation such as
let current_dist;
{
current_dist = dist.get(&state.node);
}
or
let current_dist = {|| dist.get(&state.node)}();
but the end of scope of the immutable borrow still happen after the if statement.
After non-lexical lifetimes
Since non-lexical lifetimes are now enabled, the original code compiles. That being said, you should still use the entry API for efficiency, otherwise you have to hash the key multiple times:
use std::collections::hash_map::Entry;
use std::collections::HashMap;
fn main() {
let mut dist: HashMap<u8, u8> = HashMap::new();
let cost = 21;
match dist.entry(42) {
Entry::Vacant(entry) => {
entry.insert(42);
}
Entry::Occupied(mut entry) => {
if *entry.get() < cost {
entry.insert(42);
}
}
}
}
Before non-lexical lifetimes
because dist.get triggers a mutable borrow
No, it's just an immutable borrow:
pub fn get<Q: ?Sized>(&self, k: &Q) -> Option<&V>
where
K: Borrow<Q>,
Q: Hash + Eq,
I tried a drop
Explicit drops do not affect lifetimes.
let current_dist;
{
current_dist = dist.get(&state.node);
}
Here you aren't fooling anyone. If the compiler was confused by this, it wouldn't be very good. This still has a borrow to the HashMap, there's just some extra blocks scattered about.
let current_dist = {|| dist.get(&state.node)}();
Same here. Returning the reference from a closure is still returning a reference. You really cannot easily trick the compiler into thinking that your reference to the HashMap doesn't exist.
You need to use a block to constrain how long the borrow exists. the simplest transformation is something akin to:
use std::collections::HashMap;
fn main() {
let mut dist: HashMap<u8, u8> = HashMap::new();
let do_it = {
let current_dist = dist.get(&42);
current_dist == None || true
};
if do_it {
dist.insert(42, 42);
}
}
This isn't the prettiest, but some combinators can clean it up:
use std::collections::HashMap;
fn main() {
let mut dist: HashMap<u8, u8> = HashMap::new();
let cost = 21;
if dist.get(&42).map_or(true, |&val| val < cost) {
dist.insert(42, 42);
}
}
Note that now there's no more implicit panic from the unwrap call.
See also:
How to update-or-insert on a Vec?
I'm having difficulty getting the borrow checker working for a simple iterative linked list builder.
fn main() {
let v = vec![1,5,3,8,12,56,1230,2,1];
let nodes = Vec::<Node>::with_capacity(v.len());
let mut root: Option<&mut Box<Node>> = None;
let mut prev: &Option<&mut Box<Node>> = &None;
for i in v {
let curr = Some(&mut Box::new(Node { value: i, next: None }));
match *prev {
Some(ref mut p) => {
p.next = curr;
prev = &mut p.next;
},
None => {
root = curr;
prev = &mut root;
}
}
}
}
struct Node<'a> {
value: i32,
next: Option<&'a mut Box<Node<'a>>>,
}
The errors I'm receiving when I try to compile:
linked_list.rs:8:30: 8:69 error: borrowed value does not live long enough
linked_list.rs:8 let curr = Some(&mut Box::new(Node { value: i, next: None }));
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:4:49: 20:2 note: reference must be valid for the block suffix following statement 2 at 4:48...
linked_list.rs:4 let mut root: Option<&mut Box<Node>> = None;
linked_list.rs:5 let mut prev: &Option<&mut Box<Node>> = &None;
linked_list.rs:6
linked_list.rs:7 for i in v {
linked_list.rs:8 let curr = Some(&mut Box::new(Node { value: i, next: None }));
linked_list.rs:9 match *prev {
...
linked_list.rs:8:9: 8:71 note: ...but borrowed value is only valid for the statement at 8:8
linked_list.rs:8 let curr = Some(&mut Box::new(Node { value: i, next: None }));
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
linked_list.rs:8:9: 8:71 help: consider using a `let` binding to increase its lifetime
linked_list.rs:8 let curr = Some(&mut Box::new(Node { value: i, next: None }));
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
linked_list.rs:10:18: 10:27 error: cannot borrow immutable anonymous field `(prev:core::option::Some).0` as mutable
linked_list.rs:10 Some(ref mut p) => {
^~~~~~~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:15:17: 15:28 error: cannot assign to `root` because it is borrowed
linked_list.rs:15 root = curr;
^~~~~~~~~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:16:29: 16:33 note: borrow of `root` occurs here
linked_list.rs:16 prev = &mut root;
^~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:16:29: 16:33 error: cannot borrow `root` as mutable more than once at a time
linked_list.rs:16 prev = &mut root;
^~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:16:29: 16:33 note: previous borrow of `root` occurs here; the mutable borrow prevents subsequent moves, borrows, or modification of `root` until the borrow ends
linked_list.rs:16 prev = &mut root;
^~~~
note: in expansion of for loop expansion
linked_list.rs:7:5: 19:6 note: expansion site
linked_list.rs:20:2: 20:2 note: previous borrow ends here
linked_list.rs:1 fn main() {
...
linked_list.rs:20 }
^
error: aborting due to 4 previous errors
What I'm trying to go for is fairly simple. We iterate through a Vec, creating a new node on each iteration. If prev is None this must be the start, so we make the root variable take ownership of that first node. If it's not, we update the previous node's next value to point to this node.
I'm new to Rust so I'm not sure where I'm going wrong. My interpretation is that the borrow checker isn't handling this well. It can't infer that the None branch in the match, containing the 'root' assignment, will only ever be called once, causing the two errors about root being borrowed twice. Am I correct?
Is this approach possible in Rust? Is there a more idiomatic way to do this sort of thing?
(A recursive approach is probably much easier but I'd like to complete an iterative one as a learning exercise.)
First of all, you should probably make sure you've read and understood the Rust Book chapters on Ownership and References and Borrowing. Your immediate problem is that you're borrowing things that aren't owned by anything, and will thus just disappear. You also have other problems like trying to mutate through an immutable pointer.
Let's get something that does, at least, work:
fn main() {
let v = vec![1,5,3,8,12,56,1230,2,1];
let mut root: Option<Box<Node>> = None;
for i in v.into_iter().rev() {
root = Some(Box::new(Node { value: i, next: root }));
}
println!("root: {}",
root.map(|n| n.to_string()).unwrap_or(String::from("None")));
}
struct Node {
value: i32,
next: Option<Box<Node>>,
}
impl std::fmt::Display for Node {
fn fmt(&self, fmt: &mut std::fmt::Formatter) -> Result<(), std::fmt::Error> {
let mut cur = Some(self);
let mut first = true;
try!(write!(fmt, "["));
while let Some(node) = cur {
if !first { try!(write!(fmt, ", ")); }
first = false;
try!(write!(fmt, "{}", node.value));
cur = node.next.as_ref().map(|n| &**n);
}
try!(write!(fmt, "]"));
Ok(())
}
}
This constructs a list and shows how you can iteratively display it. Note the complete lack of borrows in the construction code.
I have cheated somewhat, in that I've iterated the vector backwards to construct the list.
The problem with the original code is that, even if you strip out everything that isn't necessary, down to something like this:
let v = vec![1,5,3,8,12,56,1230,2,1];
let mut v = v.into_iter();
let mut root: Option<Box<Node>> = None;
if let Some(i) = v.next() {
root = Some(Box::new(Node { value: i, next: None }));
let mut prev: &mut Box<Node> = root.as_mut().unwrap();
for i in v {
let curr = Some(Box::new(Node { value: i, next: None }));
prev.next = curr;
prev = prev.next.as_mut().unwrap();
}
}
You still end up in a situation where the compiler sees you mutating a thing you've borrowed by a second path. It's not quite smart enough to realise that re-assigning prev doesn't actually create any aliases. On the other hand, if you break the loop into an equivalent recursion:
if let Some(i) = v.next() {
root = Some(Box::new(Node { value: i, next: None }));
fn step<It>(prev: &mut Box<Node>, mut v: It) where It: Iterator<Item=i32> {
if let Some(i) = v.next() {
let curr = Some(Box::new(Node { value: i, next: None }));
prev.next = curr;
step(prev.next.as_mut().unwrap(), v)
}
}
step(root.as_mut().unwrap(), v);
}
Then it's totally fine with it. Sadly, even with optimisations turned on, Rust doesn't perform tail call elimination in this case. So between borrow checker limitations and a lack of guaranteed tail call elimination, this design might be impossible to do in safe code.
I've run into this problem myself; loops and &mut pointers don't always play nicely with one another. You can work around this by switching to RefCell, with its associated runtime cost, although this then complicates iterating over such a list in a loop. Another alternative is to use usizes instead of pointers, and have all the nodes allocated into a Vec somewhere, although that introduces bounds checking overhead.
Failing all that, there's unsafe code, which lets you write more or less exactly what you would write in another language like C or C++, but without Rust's usual safety guarantees.
At the end of the day, writing data structures that are not just wrappers around an existing data structure in safe Rust without overhead is borderline impossible. It's why the fundamental data structures in Rust are all written using some amount of unsafe code.