#021 Oct 22, 2022

21. Zip longest

Sometimes there is a need to zip two iterables of various lengths.

If it is known which one is longer, then use the following approach:

First one must be longer.

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let nums1 = [1, 2, 3, 4];
let nums2 = [10, 20];

for (n1, n2) in nums1
    .into_iter()
    .zip(nums2.into_iter().chain(std::iter::repeat(0i32)))
{
    println!("{n1} - {n2}");
}
// Output:
// 1 - 10
// 2 - 20
// 3 - 0
// 4 - 0

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