Borrowing a Value

As we saw before, instead of transferring ownership when calling a function, you can let a function borrow the value:

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

fn add(p1: &Point, p2: &Point) -> Point {
    Point(p1.0 + p2.0, p1.1 + p2.1)
}

fn main() {
    let p1 = Point(3, 4);
    let p2 = Point(10, 20);
    let p3 = add(&p1, &p2);
    println!("{p1:?} + {p2:?} = {p3:?}");
}
  • The add function borrows two points and returns a new point.
  • The caller retains ownership of the inputs.
This slide should take about 10 minutes.

This slide is a review of the material on references from day 1, expanding slightly to include function arguments and return values.

More to Explore

Notes on stack returns:

  • Demonstrate that the return from add is cheap because the compiler can eliminate the copy operation. Change the above code to print stack addresses and run it on the Playground or look at the assembly in Godbolt. In the “DEBUG” optimization level, the addresses should change, while they stay the same when changing to the “RELEASE” setting:

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

fn add(p1: &Point, p2: &Point) -> Point {
    let p = Point(p1.0 + p2.0, p1.1 + p2.1);
    println!("&p.0: {:p}", &p.0);
    p
}

pub fn main() {
    let p1 = Point(3, 4);
    let p2 = Point(10, 20);
    let p3 = add(&p1, &p2);
    println!("&p3.0: {:p}", &p3.0);
    println!("{p1:?} + {p2:?} = {p3:?}");
}

  • The Rust compiler can do return value optimization (RVO).

  • In C++, copy elision has to be defined in the language specification because constructors can have side effects. In Rust, this is not an issue at all. If RVO did not happen, Rust will always perform a simple and efficient memcpy copy.