How do I create two new mutable slices from one slice? - slice

Question

I would like to take a mutable slice and copy the contents into two new mutable slices. Each slice being one half of the original.
My attempt #1:
let my_list: &mut [u8] = &mut [0, 1, 2, 3, 4, 5];
let list_a: &mut [u8] = my_list[0..3].clone();
let list_b: &mut [u8] = my_list[3..6].clone();
println!("{:?}", my_list);
println!("{:?}", list_a);
println!("{:?}", list_b);
Output:
error: no method named `clone` found for type `[u8]` in the current scope
--> src/main.rs:3:43
|
3 | let list_a: &mut [u8] = my_list[0..3].clone();
| ^^^^^
error: no method named `clone` found for type `[u8]` in the current scope
--> src/main.rs:4:43
|
4 | let list_b: &mut [u8] = my_list[3..6].clone();
| ^^^^^
My attempt #2:
let my_list: &mut [u8] = &mut [0, 1, 2, 3, 4, 5];
let list_a: &mut [u8] = my_list[0..3].to_owned();
let list_b: &mut [u8] = my_list[3..6].to_owned();
println!("{:?}", my_list);
println!("{:?}", list_a);
println!("{:?}", list_b);
Output:
error[E0308]: mismatched types
--> src/main.rs:12:29
|
12 | let list_a: &mut [u8] = my_list[0..3].to_owned();
| ^^^^^^^^^^^^^^^^^^^^^^^^ expected &mut [u8], found struct `std::vec::Vec`
|
= note: expected type `&mut [u8]`
found type `std::vec::Vec<u8>`
= help: try with `&mut my_list[0..3].to_owned()`
error[E0308]: mismatched types
--> src/main.rs:13:29
|
13 | let list_b: &mut [u8] = my_list[3..6].to_owned();
| ^^^^^^^^^^^^^^^^^^^^^^^^ expected &mut [u8], found struct `std::vec::Vec`
|
= note: expected type `&mut [u8]`
found type `std::vec::Vec<u8>`
= help: try with `&mut my_list[3..6].to_owned()`
I can use two Vec<u8> and just loop over the input and push cloned values I guess, but I was hoping there was a nicer way to do this:
extern crate rand;
use rand::{thread_rng, Rng};
fn main() {
let my_list: &mut [u8] = &mut [0; 100];
thread_rng().fill_bytes(my_list);
let list_a = &mut Vec::new();
let list_b = &mut Vec::new();
for i in 0..my_list.len() {
if i < my_list.len() / 2 {
list_a.push(my_list[i].clone());
} else {
list_b.push(my_list[i].clone());
}
}
println!("{:?}", list_a.as_slice());
println!("{:?}", list_b.as_slice());
println!("{:?}", my_list);
}

Answer 1

The split_at and split_at_mut methods will give you two slices, which you can then copy or even safely use without copying if borrow checker allows.
let (list_a, list_b) = my_list.split_at_mut(my_list.len()/2)

Answer 2

You can build vectors from slices directly by cloning the elements using multiple methods:
Vec::to_vec
From / Into
ToOwned
fn main() {
let my_list: &mut [u8] = &mut [0, 1, 2, 3, 4, 5];
let mut vec1 = my_list[0..2].to_vec();
let mut vec2: Vec<u8> = my_list[2..4].into();
let mut vec3 = my_list[2..6].to_owned();
println!("{:?}", vec1);
println!("{:?}", vec2);
}
Your original problem was caused because all of these return a Vec but you were attempting to claim that it was a slice, equivalent to:
let thing: &mut [u8] = Vec::new();

Answer 3

You could chain two iterators over the slices.
let my_list: &mut [u8] = &mut [0, 1, 2, 3, 4, 5];
let mut slices = my_list[0..3].iter().chain(my_list[3..6].iter());
for e in slices {}
chain will iterate over the first iterator, then the second.
To create new lists:
let my_list: &mut [u8] = &mut [0, 1, 2, 3, 4, 5];
let mut a: Vec<u8> = my_list[0..3].iter().cloned().collect();
let mut b: Vec<u8> = my_list[3..6].iter().cloned().collect();