fold keeps the running state but only hands you the final answer. So you reach for a mut variable plus map, or you give up and collect first. scan is the missing middle: a fold that yields each intermediate step.

The setup is familiar — you want a running total, not just the sum:

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

let mut sum = 0;
let totals: Vec<i32> = nums.iter()
    .map(|&x| { sum += x; sum })
    .collect();
assert_eq!(totals, vec![1, 3, 6, 10, 15]);

It works, but the state lives outside the chain. map is supposed to be pure — leaning on a captured mut makes the iterator harder to refactor and impossible to compose.

scan puts the state inside the chain. You hand it an initial value and a closure that mutates the state and returns each yielded item:

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

let totals: Vec<i32> = nums.iter()
    .scan(0, |sum, &x| {
        *sum += x;
        Some(*sum)
    })
    .collect();
assert_eq!(totals, vec![1, 3, 6, 10, 15]);

No captured state, no second pass. The closure returns Option<U>, so returning None ends the iteration early — handy for “stop when the running total crosses a threshold”:

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let nums = [3, 4, 5, 6, 7];

let until_ten: Vec<i32> = nums.iter()
    .scan(0, |sum, &x| {
        *sum += x;
        (*sum <= 10).then_some(*sum)
    })
    .collect();
assert_eq!(until_ten, vec![3, 7]);

The state isn’t limited to numbers either — pair it with a tuple to track “previous + current” and you’ve got differences in one pass:

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let readings = [10, 13, 12, 18, 20];

let deltas: Vec<i32> = readings.iter()
    .scan(None, |prev, &x| {
        let d = prev.map(|p| x - p);
        *prev = Some(x);
        Some(d)
    })
    .flatten()
    .collect();
assert_eq!(deltas, vec![3, -1, 6, 2]);

Reach for scan whenever you’d otherwise write let mut acc = … outside a map. Same shape, no escapee state.

You insert a placeholder, then index back to fix it up. Two lookups, two bounds checks, one wobble. Vec::insert_mut — stable in 1.95 — hands you the &mut T directly.

The classic dance:

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let mut v = vec![1, 2, 4, 5];
v.insert(2, 0);
v[2] = 3; // recompute the index, second bounds check
assert_eq!(v, [1, 2, 3, 4, 5]);

insert_mut returns the slot:

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let mut v = vec![1, 2, 4, 5];
let slot = v.insert_mut(2, 0);
*slot = 3;
assert_eq!(v, [1, 2, 3, 4, 5]);

This shines when the value is built in pieces — push a default, then fill it in based on where it landed:

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#[derive(Default, Debug, PartialEq)]
struct Job { id: u32, label: String }

let mut jobs: Vec<Job> = vec![Job { id: 1, label: "build".into() }];
let j = jobs.insert_mut(0, Job::default());
j.id = 0;
j.label = "setup".into();

assert_eq!(jobs[0], Job { id: 0, label: String::from("setup") });
assert_eq!(jobs[1].label, "build");

Rust 1.95 grew the whole _mut family alongside Vec::push_mut: Vec::insert_mut, VecDeque::push_front_mut, VecDeque::push_back_mut, VecDeque::insert_mut, LinkedList::push_front_mut, and LinkedList::push_back_mut. Same idea everywhere — the place-it method now returns a mutable reference to the slot it just placed.

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use std::collections::VecDeque;

let mut q: VecDeque<i32> = VecDeque::from([2, 3]);
let head = q.push_front_mut(0);
*head += 1; // now 1
assert_eq!(q, VecDeque::from([1, 2, 3]));

Quietly useful, no churn — just fewer indices floating around.

slice.split_at(i) panics the second i > len. The usual fix is a length check wrapped around the call so you don’t blow up on a bad index. split_at_checked does the same job in one call and hands you an Option.

The classic trap — a single bad index away from a panic:

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let xs = [1, 2, 3, 4];
let (head, tail) = xs.split_at(10); // panics: byte index 10 is out of bounds

The defensive version everyone writes:

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let xs = [1, 2, 3, 4];
let i = 10;

if i <= xs.len() {
    let (head, tail) = xs.split_at(i);
    // ...use head and tail
} else {
    // handle out-of-bounds
}

Two reads of i, one easy off-by-one (< vs <=), and a panic waiting if you ever drop the guard.

Rust 1.80 stabilised split_at_checked (and split_at_mut_checked), which folds the bounds check into the return type:

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let xs = [1, 2, 3, 4];

assert_eq!(xs.split_at_checked(2), Some((&xs[..2], &xs[2..])));
assert_eq!(xs.split_at_checked(4), Some((&xs[..], &[][..]))); // boundary is fine
assert_eq!(xs.split_at_checked(5), None);                     // would have panicked

Now the bounds check is the API. You get an Option<(&[T], &[T])> and the compiler nudges you to handle the None case:

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fn take_prefix(buf: &[u8], n: usize) -> Option<&[u8]> {
    let (head, _rest) = buf.split_at_checked(n)?;
    Some(head)
}

assert_eq!(take_prefix(b"hello", 3), Some(&b"hel"[..]));
assert_eq!(take_prefix(b"hi", 3), None);

? does the bailout, no manual length check, no panic path. This works on &str too, where the index has to land on a UTF-8 boundary — and it returns None if it doesn’t, instead of panicking.

Updating the largest element in a BinaryHeap shouldn’t take two heap operations. peek_mut lets you mutate it in place and re-heapifies on drop — one sift instead of two.

The pop-and-push dance

Say you keep tasks in a max-heap by priority and you need to bump the top one down. The naive recipe is to pop, change it, and push it back:

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use std::collections::BinaryHeap;

let mut heap = BinaryHeap::from([3, 1, 5, 2, 4]);

// Pull the max out, edit it, push it back.
let top = heap.pop().unwrap();
heap.push(top - 10);

assert_eq!(heap.peek(), Some(&4));

That’s two O(log n) heap operations — a sift-down for pop, a sift-up for push — plus an awkward two-step where the heap briefly forgets it ever had a max.

peek_mut does it in one shot

peek_mut returns a PeekMut guard that derefs to the top element. You mutate it in place, and the heap fixes itself with a single sift-down when the guard is dropped:

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use std::collections::BinaryHeap;

let mut heap = BinaryHeap::from([3, 1, 5, 2, 4]);

if let Some(mut top) = heap.peek_mut() {
    *top -= 10; // 5 becomes -5
}
// PeekMut dropped here — heap re-heapifies once.

assert_eq!(heap.peek(), Some(&4));

One traversal, no temporary owned value, and the heap invariant is restored before the next line of code runs.

Conditionally pop without a re-peek

PeekMut also has an associated pop function. Look at the top, decide whether to keep or remove it, all without peeking twice:

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use std::collections::{BinaryHeap, binary_heap::PeekMut};

let mut heap = BinaryHeap::from([10, 7, 4, 1]);

if let Some(top) = heap.peek_mut() {
    if *top > 5 {
        // Pop only when the top passes a check.
        PeekMut::pop(top);
    }
}

assert_eq!(heap.peek(), Some(&7));

Same shape as Vec::pop_if, but for the heap’s max — pull the top out only when it actually meets your condition.

When to reach for it

Any time you’d write pop().push() to edit the max, peek_mut is the better tool: one heap fixup instead of two, and the borrow stays inside the heap so you don’t shuffle ownership for nothing. Great fit for priority queues that adjust the top entry — schedulers, event loops, top-k trackers.

Need the longest string in a Vec? Don’t sort the whole list to grab the last element — max_by_key walks once, allocates nothing, and returns it directly.

The wasteful way

You want the longest word from a list. The first instinct is often to sort and pick:

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let mut words = vec!["fig", "apple", "kiwi", "blackberry", "pear"];

// Sort the full vec by length, then grab the last element.
words.sort_by_key(|w| w.len());
let longest = words.last().unwrap();

assert_eq!(*longest, "blackberry");

That’s O(n log n) work — and a side effect, since your Vec is now reordered — just to answer “which one is biggest?”.

Enter max_by_key

max_by_key walks the iterator once, tracks the running maximum, and hands you the element it picked:

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let words = vec!["fig", "apple", "kiwi", "blackberry", "pear"];

let longest = words.iter().max_by_key(|w| w.len()).unwrap();

assert_eq!(*longest, "blackberry");

O(n), no allocation, no mutation. The original Vec is untouched and you get a &&str back without cloning anything.

Same energy: min_by_key

The mirror method is just as useful — find the shortest word in one pass:

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let words = vec!["fig", "apple", "kiwi", "blackberry", "pear"];

let shortest = words.iter().min_by_key(|w| w.len()).unwrap();

assert_eq!(*shortest, "fig");

Picking by computed value

The closure can do real work — compute distance, score, age — anything that returns something Ord:

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struct Player { name: &'static str, score: u32 }

let roster = [
    Player { name: "Ferris", score: 80 },
    Player { name: "Corro",  score: 95 },
    Player { name: "Crusty", score: 60 },
];

let top = roster.iter().max_by_key(|p| p.score).unwrap();

assert_eq!(top.name, "Corro");

No need to sort, no need to implement Ord on the struct — just point at the field that decides.

Tiebreaks: last one wins

When two elements share the same key, max_by_key returns the last one and min_by_key returns the first:

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let v = vec![(1, 'a'), (3, 'b'), (3, 'c'), (2, 'd')];

let max = v.iter().max_by_key(|(k, _)| *k).unwrap();
let min = v.iter().min_by_key(|(k, _)| *k).unwrap();

assert_eq!(*max, (3, 'c')); // last 3
assert_eq!(*min, (1, 'a')); // first 1

Worth knowing if your data has duplicates — pick the iteration direction so the right element wins.

When to reach for it

If you catch yourself sorting and then grabbing first(), last(), or [0], stop. min_by_key and max_by_key are the targeted tools: one pass, zero allocations, and the result type is exactly the element you wanted.

Need to swap a chunk of a Vec for a different number of items? Most people reach for drain plus extend and a bunch of bookkeeping. Vec::splice does the whole thing in one call — and even hands back what it removed.

The clunky way

You want to replace v[1..4] with two different values. Without splice, you end up juggling indices:

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let mut v = vec![1, 2, 3, 4, 5];

// Remove the middle, save the tail, then stitch it back together.
let tail: Vec<_> = v.drain(4..).collect();
v.truncate(1);
v.extend([10, 20]);
v.extend(tail);

assert_eq!(v, [1, 10, 20, 5]);

Three steps, one temporary Vec, and you didn’t even capture what was removed. Easy to off‑by‑one, easy to get wrong.

Enter splice

splice takes a range and an iterator and swaps them in place — the replacement can be any length:

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let mut v = vec![1, 2, 3, 4, 5];

v.splice(1..4, [10, 20]);

assert_eq!(v, [1, 10, 20, 5]);

One call. No tail buffer. Works whether the replacement is shorter, longer, or the same size as the range.

It returns the removed items too

splice is a draining iterator — collect it and you get the elements that were taken out:

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let mut v = vec![10, 20, 30, 40, 50];

let removed: Vec<_> = v.splice(1..4, [99]).collect();

assert_eq!(removed, [20, 30, 40]);
assert_eq!(v, [10, 99, 50]);

Perfect for “swap this section, but undo it later” patterns.

Insert without removing

Pass an empty range and splice becomes a multi‑item insert:

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let mut v = vec![1, 5];

v.splice(1..1, [2, 3, 4]);

assert_eq!(v, [1, 2, 3, 4, 5]);

That’s far nicer than calling insert in a loop and watching every push shift the tail again.

When to reach for it

Any time you’d write drain(..) followed by extend(..) or a tower of insert calls, try splice first. One method, one allocation pattern, and the removed items come along for free.

Called iter.take(n).collect() and watched the borrow checker swallow the whole iterator? by_ref lets you consume a chunk and keep iterating from where you stopped.

The trap: adapters consume their iterator

Most iterator adapters take self, not &mut self. The moment you write iter.take(2), you have moved iter into the new Take adapter. The original binding is gone:

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let v = vec![1, 2, 3, 4, 5];
let mut iter = v.into_iter();

let _first_two: Vec<i32> = iter.take(2).collect();
// let rest: Vec<i32> = iter.collect();
// ^^ error: use of moved value: `iter`

So even though there are clearly elements left, the variable that points at them is no longer usable.

The fix: iter.by_ref()

Iterator::by_ref returns a &mut Self — and &mut I itself implements Iterator whenever I: Iterator. Adapters chained off by_ref() consume from the underlying iterator without moving it:

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let v = vec![1, 2, 3, 4, 5];
let mut iter = v.into_iter();

let first_two: Vec<i32> = iter.by_ref().take(2).collect();
let rest: Vec<i32> = iter.collect();

assert_eq!(first_two, [1, 2]);
assert_eq!(rest, [3, 4, 5]);

Same iterator, two phases, no clones, no indexing.

A real use: parse a header, then a body

The pattern shines when the front of a stream uses different rules than the rest. Pull lines until you hit a blank one, then hand the same iterator to the body parser:

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let input = "Subject: hi\nFrom: a@b\n\nbody line 1\nbody line 2";
let mut lines = input.lines();

let headers: Vec<&str> = lines
    .by_ref()
    .take_while(|l| !l.is_empty())
    .collect();

let body: Vec<&str> = lines.collect();

assert_eq!(headers, ["Subject: hi", "From: a@b"]);
assert_eq!(body, ["body line 1", "body line 2"]);

Without by_ref, take_while would have consumed lines whole and the body would be unreachable.

Gotcha: take_while peeks one past

take_while reads the first non-matching element to know when to stop — and that element is gone from the underlying iterator. With by_ref you’ll see this directly:

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let mut nums = (1..).into_iter();

let small: Vec<i32> = nums.by_ref().take_while(|&n| n < 3).collect();
let next = nums.next();

assert_eq!(small, [1, 2]);
assert_eq!(next, Some(4)); // 3 was eaten by take_while

If you need to keep the boundary element, reach for Peekable instead.

When to use it

Any time you want to apply an adapter to part of an iterator and continue afterwards: chunked parsing, header/body splits, “take the first N then process the rest”, consuming until a sentinel. by_ref() turns “I moved my iterator and now I can’t use it” into a single extra method call.

Trying to copy a slice of a Vec back onto the end of itself? v.extend_from_slice(&v[..k]) won’t compile — &v and &mut v collide. extend_from_within is the one-call answer.

The borrow checker says no

It looks innocent: take the first few bytes of a buffer and append them to the end. The naïve write doesn’t even reach the type checker without complaint:

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let mut buf: Vec<u8> = vec![0xCA, 0xFE, 0xBA, 0xBE];
// buf.extend_from_slice(&buf[..2]);
// ^^ error: cannot borrow `buf` as immutable because it is also borrowed as mutable

extend_from_slice needs &mut self, and &buf[..2] is an immutable borrow of the same Vec. Two borrows, one of them mutable, same value — instant rejection.

The usual workarounds bloat the call site. Either make a temporary copy, or reach for unsafe and a raw pointer dance:

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let mut buf: Vec<u8> = vec![0xCA, 0xFE, 0xBA, 0xBE];

let head: Vec<u8> = buf[..2].to_vec();   // extra allocation
buf.extend_from_slice(&head);

assert_eq!(buf, [0xCA, 0xFE, 0xBA, 0xBE, 0xCA, 0xFE]);

A whole heap allocation just to copy two bytes inside a vector you already own. There’s a better way.

extend_from_within — one call, no temp

Vec::extend_from_within takes a range into self and appends those elements. No second buffer, no unsafe, no fight with the borrow checker:

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let mut buf: Vec<u8> = vec![0xCA, 0xFE, 0xBA, 0xBE];
buf.extend_from_within(..2);

assert_eq!(buf, [0xCA, 0xFE, 0xBA, 0xBE, 0xCA, 0xFE]);

It accepts any RangeBounds<usize> — ..k, i..j, i..=j, .. — so you can grab whatever slice you want, including the whole Vec:

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let mut v = vec![1, 2, 3];
v.extend_from_within(..);   // double the contents

assert_eq!(v, [1, 2, 3, 1, 2, 3]);

It requires T: Clone, and clones each element exactly once. For Copy types like primitives it lowers to a single memcpy.

A real use: building patterns

It’s especially nice for building repeating or growing patterns where each step depends on the previous one — think DSP buffers, simple test fixtures, or doubling tricks:

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let mut data = vec![1u32];
for _ in 0..4 {
    data.extend_from_within(..); // double in place
}

assert_eq!(data.len(), 16);
assert!(data.iter().all(|&x| x == 1));

When to reach for it

Any time you’d write v.extend_from_slice(&v[range]) and the borrow checker stops you, swap in v.extend_from_within(range). Stable since Rust 1.53, exists for both Vec<T> and VecDeque<T>, and quietly turns a frustrating compile error into a one-liner.

Writing if condition { Some(value) } else { None } for the hundredth time? bool::then_some collapses that whole pattern into a single chainable call.

The familiar boilerplate

You have a bool and you want to lift it into an Option. The naive form works, but it’s noisy and breaks chains:

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fn even_double(n: i32) -> Option<i32> {
    if n % 2 == 0 {
        Some(n * 2)
    } else {
        None
    }
}

assert_eq!(even_double(4), Some(8));
assert_eq!(even_double(5), None);

Three lines and an if/else for what is conceptually “if true, wrap it.” It also doesn’t slot into iterator chains — you’d have to wrap it in a closure.

then_some: the eager form

bool::then_some(value) returns Some(value) if the bool is true, None otherwise:

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fn even_double(n: i32) -> Option<i32> {
    (n % 2 == 0).then_some(n * 2)
}

assert_eq!(even_double(4), Some(8));
assert_eq!(even_double(5), None);

One line, no branches in sight. The intent reads exactly like the English: if even, then some n * 2.

then: the lazy form

then_some always evaluates its argument — fine for cheap values, wasteful for expensive ones. Use then(|| ...) when the value should only be computed when the condition holds:

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fn parse_if_digits(s: &str) -> Option<u64> {
    s.chars().all(|c| c.is_ascii_digit())
        .then(|| s.parse().unwrap())
}

assert_eq!(parse_if_digits("42"), Some(42));
assert_eq!(parse_if_digits("4a2"), None);

The parse only runs when the string is all digits. With then_some(s.parse().unwrap()) you’d unwrap on every call — instant panic on bad input.

Where it really shines: filter_map chains

The big payoff is in iterator pipelines. Filter and transform in one step, no closure-returning-Option boilerplate:

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

let doubled_evens: Vec<i32> = nums
    .iter()
    .filter_map(|&n| (n % 2 == 0).then_some(n * 2))
    .collect();

assert_eq!(doubled_evens, [4, 8, 12]);

Compare to the if/else version inside the closure — it works, but it’s twice as wide and the else None arm adds zero information.

Watch the eager trap

then_some evaluates its argument unconditionally — which means an “obvious” guard like this is actually a panic:

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fn first<T: Copy>(items: &[T]) -> Option<T> {
    // BAD: items[0] runs even when the slice is empty.
    // (!items.is_empty()).then_some(items[0])

    // GOOD: defer the indexing into the closure.
    (!items.is_empty()).then(|| items[0])
}

assert_eq!(first(&[10, 20, 30]), Some(10));
assert_eq!(first::<i32>(&[]), None);

The bool short-circuits, the closure doesn’t. Whenever the value can panic, unwrap, or just costs real work, reach for then, not then_some.

When to reach for which

Use then_some(value) when the value is already computed or trivially cheap — a literal, a copy, a field access. Use then(|| value) whenever building the value costs something or could panic. Both are stable (then since Rust 1.50, then_some since 1.62) and both read cleaner than the if/else they replace.

Calling Vec::remove(i) shifts every element after i left by one — that’s O(n) per call. If you don’t care about order, swap_remove does the same job in O(1).

The hidden cost of Vec::remove

remove takes the value out of the vector, but to keep the rest in order it has to shift every following element left by one slot. On a long vector, removing near the front is a giant memcpy:

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let mut v = vec![1, 2, 3, 4, 5];
let removed = v.remove(1);

assert_eq!(removed, 2);
assert_eq!(v, [1, 3, 4, 5]);

Inside a loop — “remove every item that matches X” — that O(n) cost compounds into O(n²). A tight loop that should be linear silently becomes quadratic.

swap_remove: trade order for speed

swap_remove(i) removes the element at index i by swapping it with the last element, then popping the tail. No shifting, no memcpy chain — just one swap and a pop:

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let mut v = vec![1, 2, 3, 4, 5];
let removed = v.swap_remove(1);

assert_eq!(removed, 2);
// 5 took 2's spot — order is gone, but the op was O(1).
assert_eq!(v, [1, 5, 3, 4]);

If your Vec is really a set — entities in a game world, pending jobs in a worker pool, items keyed by id elsewhere — order rarely matters. swap_remove is free.

Removing many items in place

The classic mistake: iterate forward calling remove. swap_remove lets you do it without the O(n²) blow-up. Walk by index and don’t bump the cursor when you remove:

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let mut nums = vec![1, 2, 3, 4, 5, 6, 7, 8];

let mut i = 0;
while i < nums.len() {
    if nums[i] % 2 == 0 {
        nums.swap_remove(i);
        // Don't increment — a new element is now at position i.
    } else {
        i += 1;
    }
}

nums.sort(); // canonicalize for the assert
assert_eq!(nums, [1, 3, 5, 7]);

For bulk predicate-based filtering, Vec::retain stays cleaner (and is also O(n)). Reach for swap_remove when you have a specific index you want gone in O(1).

With structs

swap_remove returns the removed value, so you can hand it off to another collection on the way out:

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#[derive(PartialEq, Debug)]
struct Job { id: u32 }

let mut queue = vec![
    Job { id: 1 },
    Job { id: 2 },
    Job { id: 3 },
    Job { id: 4 },
];

let pulled = queue.swap_remove(0);

assert_eq!(pulled, Job { id: 1 });
assert_eq!(queue.len(), 3);

When to reach for it

Use swap_remove whenever you’d write remove(i) and the surrounding code doesn’t depend on element order. Order-preserving removal still needs remove, but most “remove one item from this Vec” code is order-agnostic — and swap_remove turns an O(n) pothole into a one-line O(1) win. Stable since Rust 1.0, one of those quietly-essential APIs that’s easy to forget exists.