Per Steve Klabnik\'s writeup in the pre-Rust 1.0 documentation on the difference between String and &str, in Rust you should use &str unless you really
You can create a function that accepts both &[String] and &[&str] using the AsRef trait:
fn test<T: AsRef<str>>(inp: &[T]) {
for x in inp { print!("{} ", x.as_ref()) }
println!("");
}
fn main() {
let vref = vec!["Hello", "world!"];
let vown = vec!["May the Force".to_owned(), "be with you.".to_owned()];
test(&vref);
test(&vown);
}
This is actually impossible without memory allocation1.
Going from String to &str is not just viewing the bits in a different light; String and &str have a different memory layout, and thus going from one to the other requires creating a new object. The same applies to Vec and &[]
Therefore, whilst you can go from Vec<T> to &[T], and thus from Vec<String> to &[String], you cannot directly go from Vec<String> to &[&str]. Your choices are:
&[String]Vec<&str> referencing the first Vec, and convert that into a &[&str]As an example of the allocation:
fn usage(_: &[&str]) {}
fn main() {
let owned = vec![String::new()];
let half_owned: Vec<_> = owned.iter().map(String::as_str).collect();
usage(&half_owned);
}
1 The conversion required is impossible, however using generics and the AsRef<str> bound as shown in @aSpex's answer you get a slightly more verbose function declaration with the flexibility you were asking for.