借用

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:?}");
}

add 函式會「借用」兩個點，並傳回新的點。
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This slide should take about 10 minutes.

這張投影片會複習第 1 天講解過的參照，並略微延伸討論何謂函式引數和回傳值。

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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

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