'How do I find the index of an element in an array, vector or slice?
I need to find an index of an element in a vector of strings. This is what I got so far:
fn main() {
let test: Vec<String> = vec![
"one".to_string(),
"two".to_string(),
"three".to_string(),
"four".to_string(),
];
let index: i32 = test
.iter()
.enumerate()
.find(|&r| r.1.to_string() == "two".to_string())
.unwrap()
.0;
}
It produces an error:
error[E0308]: mismatched types
--> src/main.rs:9:22
|
9 | let index: i32 = test
| ______________________^
10 | | .iter()
11 | | .enumerate()
12 | | .find(|&r| r.1.to_string() == "two".to_string())
13 | | .unwrap()
14 | | .0;
| |__________^ expected i32, found usize
I assume that's because enumerate()
returns a tuple of (usize, _)
(correct me if I'm wrong), but how do I convert usize
to i32
here? If there is a better approach, I'm open to suggestions.
Solution 1:[1]
TLDR Use an iterator with the position
method, the Rust docs shows a good example.
No, it's because indices are usize
, not i32
. In fact, i32
is completely inappropriate for this purpose; it may not be large enough, and there's no reason for it to be signed. Just use usize
.
Some other notes: calling to_string()
is not free, and you don't need it for the comparison; you can compare string slices just fine!
Also, if you really want to turn a usize
into an i32
, you can do that with a cast: x as i32
, though this will not produce an error on over- or under-flow (i.e. the result may be negative).
All that said, as noted in Mathieu David's answer, there's a position
method on iterators that does what you want.
Solution 2:[2]
I think you should look at the position
method instead.
fn main() {
let test = vec!["one", "two", "three"];
let index = test.iter().position(|&r| r == "two").unwrap();
println!("{}", index);
}
You can test it here.
Note that this works for any iterator, so it can be used for vectors, arrays, and slices, all of which produce iterators.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | gustafbstrom |
Solution 2 | Cirelli94 |