How to zip two iterators of unequal length with a default?

You could use the zip_longest provided by the itertools crate.

use itertools::{
    Itertools,
    EitherOrBoth::*,
};

fn main() {
    let num1 = vec![1, 2];
    let num2 = vec![3];

    for pair in num1.iter().rev().zip_longest(num2.iter().rev()) {
        match pair {
            Both(l, r) => println!("({:?}, {:?})", l, r),
            Left(l) => println!("({:?}, 0)", l),
            Right(r) => println!("(0, {:?})", r),
        }
    }
}

Which would produce the following output:

(2, 3)
(1, 0)

Zip will stop as soon as one of iterators stops producing values. If you know which is the longest, you can pad the shorter one with your default value:

use std::iter;

fn main() {
    let longer = vec![1, 2];
    let shorter = vec![3];

    for i in longer
        .iter()
        .rev()
        .zip(shorter.iter().rev().chain(iter::repeat(&0)))
    {
        println!("{:?}", i);
    }
}

If you don't know which is longest, you should use itertools, as Peter Varo suggests.