How do you replace the value of a mutable variable by taking ownership of it? - collections

Question

I am working with a LinkedList and I want to remove all elements which do not pass a test. However, I am running into the error cannot move out of borrowed content.
From what I understand, this is because I am working with &mut self, so I do not have the right to invalidate (i.e. move) one of the contained values even for a moment to construct a new list of its values.
In C++/Java, I would simply iterate the list and remove any elements which match a criteria. As there is no remove that I have yet found, I have interpreted it as an iterate, filter, and collect.
The goal is to avoid creating a temporary list, cloning values, and needing take self and return a "new" object. I have constructed an example which produces the same error. Playground.
use std::collections::LinkedList;
#[derive(Debug)]
struct Example {
list: LinkedList<i8>,
// Other stuff here
}
impl Example {
pub fn default() -> Example {
let mut list = LinkedList::new();
list.push_back(-5);
list.push_back(3);
list.push_back(-1);
list.push_back(6);
Example { list }
}
// Simmilar idea, but with creating a new list
pub fn get_positive(&self) -> LinkedList<i8> {
self.list.iter()
.filter(|&&x| x > 0)
.map(|x| x.clone())
.collect()
}
// Now, attempt to filter the elements without cloning anything
pub fn remove_negative(&mut self) {
self.list = self.list.into_iter()
.filter(|&x| x > 0)
.collect()
}
}
fn main() {
let mut e = Example::default();
println!("{:?}", e.get_positive());
println!("{:?}", e);
}
In my actual case, I cannot simply consume the wrapping object because it needs to be referenced from different places and contains other important values.
In my research, I found some unsafe code which leads me to question if a safe function could be constructed to perform this action in a similar way to std::mem::replace.

Answer 1

You can std::mem::swap your field with a temp, and then replace it with your modified list like this. The big downside is the creation of the new LinkedList. I don't know how expensive that is.
pub fn remove_negative(&mut self) {
let mut temp = LinkedList::new();
std::mem::swap(&mut temp, &mut self.list);
self.list = temp.into_iter()
.filter(|&x| x > 0)
.collect();
}

Answer 2

If the goal is not clone you may use a reference-counting pointer: the clone method on Rc increments the reference counter.
use std::collections::LinkedList;
use std::rc::Rc;
#[derive(Debug)]
struct Example {
list: LinkedList<Rc<i8>>,
// ...
}
impl Example {
pub fn default() -> Example {
let mut list = LinkedList::new();
list.push_back(Rc::new(-5));
list.push_back(Rc::new(3));
list.push_back(Rc::new(-1));
list.push_back(Rc::new(6));
Example { list }
}
// Simmilar idea, but with creating a new list
pub fn get_positive(&self) -> LinkedList<Rc<i8>> {
self.list.iter()
.filter(|&x| x.as_ref() > &0)
.map(|x| x.clone())
.collect()
}
// Now, attempt to filter the elements without cloning anything
pub fn remove_negative(&mut self) {
self.list = self.list.iter()
.filter(|&x| x.as_ref() > &0)
.map(|x| x.clone())
.collect()
}
}
fn main() {
let mut e = Example::default();
e.remove_negative();
println!("{:?}", e.get_positive());
println!("{:?}", e);
}