Lifetimes in Function Calls

In addition to borrowing its arguments, a function can return a borrowed value:

#[derive(Debug)]
struct Point(i32, i32);

fn left_most<'a>(p1: &'a Point, p2: &'a Point) -> &'a Point {
    if p1.0 < p2.0 { p1 } else { p2 }
}

fn main() {
    let p1: Point = Point(10, 10);
    let p2: Point = Point(20, 20);
    let p3: &Point = left_most(&p1, &p2);
    println!("p3: {p3:?}");
}
  • 'a is a generic parameter, it is inferred by the compiler.
  • Lifetimes start with ' and 'a is a typical default name.
  • Read &'a Point as “a borrowed Point which is valid for at least the lifetime a”.
    • The at least part is important when parameters are in different scopes.

In the above example, try the following:

  • Move the declaration of p2 and p3 into a new scope ({ ... }), resulting in the following code:

    #[derive(Debug)]
struct Point(i32, i32);

fn left_most<'a>(p1: &'a Point, p2: &'a Point) -> &'a Point {
    if p1.0 < p2.0 { p1 } else { p2 }
}

fn main() {
    let p1: Point = Point(10, 10);
    let p3: &Point;
    {
        let p2: Point = Point(20, 20);
        p3 = left_most(&p1, &p2);
    }
    println!("p3: {p3:?}");
}

    Note how this does not compile since p3 outlives p2.

  • Reset the workspace and change the function signature to fn left_most<'a, 'b>(p1: &'a Point, p2: &'a Point) -> &'b Point. This will not compile because the relationship between the lifetimes 'a and 'b is unclear.

  • Another way to explain it:

    • Two references to two values are borrowed by a function and the function returns another reference.
    • It must have come from one of those two inputs (or from a global variable).
    • Which one is it? The compiler needs to know, so at the call site the returned reference is not used for longer than a variable from where the reference came from.