107. Vec::splice — Replace a Range and Keep the Old Values

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.