How to perform assignment to part of Vec instead of using loop

What I'd like to do is to assign the same value to part of a Vec.
Given that:
let mut a = vec![0;10];
I'd like to assign 1 to a[..3], assign 2 to a[3..9] and assign 3 to a[9..].
Is there any solution instead of using loop?

let mut a = vec![0; 10];
a[..3].copy_from_slice(&[1; 3]);
a[3..9].copy_from_slice(&[2; 6]);
a[9] = 3;

But it looks strange for me... looks like much unnecessary work is done here. I would prefer just:

let a = vec![1, 1, 1, 2, 2, 2, 2, 2, 2, 4];
1 Like

Actually, what I described is just a concrete example.
Which part should be assigned to what value is known at runtime.
Assuming that

let arr1 = vec![2,3,1,3,2,4,6,9,2];
let arr2 = vec![2,1,4,3,9,6];

I need to create a Vec arr3 by sorting arr1 in relative order of arr2, which should be


I get amount of each element in arr2 by a HashMap, then I need to assign values to parts of arr3.


while loop in line 37-40 is my solution.

You can replace that while loop with for loop. Anyway, is there a reason to avoid loop here?

for i in 0..cnt {
  ret[bet +i] = elem;

Perhaps you would find using a helper function cleaner?

fn assign_all<T: Copy>(slice: &mut [T], value: T) {
    for ptr in slice {
        *ptr = value;

fn main() {
    let mut a = vec![0; 10];
    assign_all(&mut a[..3], 1);
    assign_all(&mut a[3..9], 2);
    assign_all(&mut a[9..], 3);

Thanks for your replies.
It seems that avoiding loop here is not appropriate.

how about this, you only need iterate arr2

let mut a = vec![];
for k in arr2 {
    let v = map.get(k).unwrap();

It's elegant, but how about their time complexity especially for its extending empty Vec other than allocating space first?

just use with_capacity method on Vec.

let mut a = Vec::with_capacity(n);

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