借用

您可以不必在呼叫函式時轉移所有權，而是讓函式「借用」值：

#[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:?}");
}

add 函式會「借用」兩個點，並傳回新的點。
呼叫端會保留輸入內容的所有權。

有關堆疊回傳的注意事項：

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:?}");
}

Rust 編譯器可以執行回傳值最佳化 (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.

Comprehensive Rust 🦀

借用