Pass an array of strings into a function without having to specify N (compile time constant)

Question

The following code runs:

fn last_el(arr: [&str; 2]) -> usize {
    arr.len() - 1
}

fn main() {
    let names = ["this", "that"];
    println!("{}", names[last_el(names)]);
}

However it only does so with [&str; 2] and 2 has to match the number of elements in names. For example, the following code fails to compile:

fn last_el(arr: [&str]) -> usize {
    arr.len() - 1
}

fn main(){
    let names = ["this","that"];
    println!("{}", names[last_el(names)]); 
}

How would I write this so that I don't have to specify N?

I understand that arr.len() - 1 is probably less of a headache than trying to write a function that does the same thing, but as far as understanding how functions accept arrays with strings in them, why does the second example fail to compile?

doc.rust-lang.org/std/vec/struct.Vec.html

zerkms
– zerkms

2017-04-03 00:55:43 +00:00
Commented Apr 3, 2017 at 0:55 — zerkms
– zerkms, Commented Apr 3, 2017 at 0:55

Francis Gagné · Accepted Answer · 2023-02-18 02:50:55Z

10

[&str] is an unsized type. You can't manipulate values of unsized types directly, they need to be behind a reference or a pointer. In your case, you should use &[&str] (also called a slice).

fn last_el(arr: &[&str]) -> usize {
    arr.len() - 1
}

fn main() {
    let names = ["this", "that"];
    println!("{}", names[last_el(&names)]);
}

I'll also note that there's a last() method defined on slices. It would be used like this:

fn main() {
    let names = ["this", "that"];
    println!("{}", names.last().unwrap()); 
}

edited Feb 18, 2023 at 2:50

answered Apr 3, 2017 at 0:59

Francis Gagné

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Community · Accepted Answer · 2017-05-23 12:17:34Z

2

And to answer the question you asked:

Pass an array of strings into a function without having to specify N

You cannot:

edited May 23, 2017 at 12:17

CommunityBot

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answered Apr 3, 2017 at 1:11

Shepmaster

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