Is there a builtin identity function in Rust?

Question

I have a vector of Options and I want to filter only the Somes. I use filter_map with identity:

let v = vec![Some(1), None, Some(2)];
for i in v.into_iter().filter_map(|o| o) {
    println!("{}", i);
}

Is there a builtin function permitting to write something like filter_map(identity)?

Note that `identity` is 4 more characters than your original example; I'd expect people would be too lazy to type it out ^_^. — Shepmaster, Aug 30 '17 at 20:18
@Shepmaster : As with your answer, F# calls it `id`, so there's precedent for that in the stdlib at least. — ildjarn, Aug 30 '17 at 22:37
Same with Haskell. [`id`](https://www.haskell.org/hoogle/?hoogle=id)s everywhere! — Mateen Ulhaq, May 10 '18 at 01:27

score 18 · Accepted Answer · edited Mar 04 '19 at 03:05

18

edited Mar 04 '19 at 03:05

ændrük

answered Sep 26 '18 at 20:29

Boiethios

Shepmaster · Answer 2 · 2020-05-19T14:52:17.247

12

Answering your question

After Rust 1.33, see the sibling answer.

Before Rust 1.33, there is no such function in stable Rust. You can create your own:

fn id<T>(v: T) -> T { v }

Although most people just inline the code, as you did.

After Rust 1.29, use Iterator::flatten:

let v = vec![Some(1), None, Some(2)];
for i in v.into_iter().flatten() {
    println!("{}", i);
}

Before Rust 1.29, use Iterator::flat_map:

let v = vec![Some(1), None, Some(2)];
for i in v.into_iter().flat_map(|o| o) {
    println!("{}", i);
}

See also:

edited May 19 '20 at 14:52

answered Aug 30 '17 at 20:15

Shepmaster

I made another one when the item is not moved: `fn id_deref(v: &Option) -> Option<&T> {v.as_ref()}` – Boiethios Aug 31 '17 at 08:59
3

@Boiethios I'd be very hesitant to call that anything like "identity" considering it changes the value. I'd also not call it "deref" because that means something in Rust which isn't what that function does. Also, that function is redundant; you can just say `Option::as_ref` instead. – Shepmaster Aug 31 '17 at 12:54