reference - What is the purpose of `&` before the loop variable?

Question

What is the purpose of & in the code &i in list? If I remove the &, it produces an error in largest = i, since they have mismatched types (where i is &32 and i is i32). But how does &i convert i into i32?

fn largest(list: &[i32]) -> i32 {
    println!("{:?}", list);
    let mut largest = list[0];
    for &i in list {
        if i > largest {
            largest = i;
        }
    }
    largest
}

fn main() {
    let hey = vec![1, 3, 2, 6, 90, 67, 788, 12, 34, 54, 32];
    println!("The largest number is: {}", largest(&hey));
}

Playground

It seems like it is somehow dereferencing, but then why in the below code, is it not working?

fn main() {
    let mut hey: i32 = 32;
    let x: i32 = 2;
    hey = &&x;
}

It says:

4 |     hey = &&x;
  |           ^^^ expected i32, found &&i32
  |
  = note: expected type `i32`
             found type `&&i32`

Answer 1

So normally when you use for i in list, the loop variable i would be of type &i32.

But when instead you use for &i in list, you are not dereferencing anything, but instead you are using pattern matching to explicitly destructure the reference and that will make i just be of type i32.

See the Rust docs about the for-loop loop variable being a pattern and the reference pattern that we are using here. See also the Rust By Example chapter on destructuring pointers.

 

An other way to solve this, would be to just keep i as it is and then comparing i to a reference to largest, and then dereferencing i before assigning to largest:

fn largest(list: &[i32]) -> i32 {
    println!("{:?}", list);
    let mut largest = list[0];
    for i in list {
        if i > &largest {
            largest = *i;
        }
    }
    largest
}

 

---

fn main() {
    let mut hey: i32 = 32;
    let x: i32 = 2;
    hey = &&x;
}

This simply doesn't work, because here you are assigning hey, which is an i32, to a reference to a reference to an i32. This is quite unrelated to the pattern matching and destructuring in the loop variable case.