235. Iterator::rposition — Find the Last Match Without Reversing-and-Subtracting

Need the index of the last element that matches? The reflex is iter().rev().position(...) then len - 1 - i — and that arithmetic is a classic off-by-one. rposition searches from the back and hands you the real forward index.

The reversed-index trap

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let bytes = [b'a', b'/', b'b', b'/', b'c'];

// "index of the last slash" — the fragile way
let from_end = bytes.iter().rev().position(|&b| b == b'/').unwrap();
let idx = bytes.len() - 1 - from_end;

assert_eq!(idx, 3);

position on a reversed iterator counts from the end, so you have to flip it back with len - 1 - i. Get the - 1 wrong and you’re off by one.

rposition does the flip for you

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let bytes = [b'a', b'/', b'b', b'/', b'c'];

let idx = bytes.iter().rposition(|&b| b == b'/');

assert_eq!(idx, Some(3)); // real index, counted from the front

rposition walks from the back but returns the index in the original, forward order — no arithmetic, and None when nothing matches instead of a panic on the empty case.

It short-circuits from the right

Just as position stops at the first match from the front, rposition stops at the first match from the back — so it only scans the tail it needs:

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let nums = [1, 2, 3, 4, 5, 6];

// last even number
let idx = nums.iter().rposition(|&n| n % 2 == 0);
assert_eq!(idx, Some(5)); // stopped immediately at 6

The requirement

rposition needs a DoubleEndedIterator (so it can walk backwards) that is also an ExactSizeIterator (so it knows the length to report the front index). Slices, Vec, and arrays give you both. A lazy adapter like filter isn’t ExactSizeIterator, so index a slice or collect first.

For string byte or substring searches, str::rfind is the more direct tool — but for an arbitrary predicate over any exact-size sequence, rposition is the one to reach for.