함수 호출에서의 수명

A reference has a lifetime, which must not “outlive” the value it refers to. This is verified by the borrow checker.

The lifetime can be implicit - this is what we have seen so far. Lifetimes can also be explicit: &'a Point, &'document str. 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”.

Lifetimes are always inferred by the compiler: you cannot assign a lifetime yourself. Explicit lifetime annotations create constraints where there is ambiguity; the compiler verifies that there is a valid solution.

수명은 함수에 값을 전달하고 함수에서 값을 반환하는 경우를 고려할 때 더 복잡해집니다.

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

fn left_most(p1: &Point, p2: &Point) -> &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 = left_most(&p1, &p2); // p3의 수명은 어떻게 되나요?
    println!("p3: {p3:?}");
}

This slide should take about 10 minutes.

In this example, the compiler does not know what lifetime to infer for p3. Looking inside the function body shows that it can only safely assume that p3’s lifetime is the shorter of p1 and p2. But just like types, Rust requires explicit annotations of lifetimes on function arguments and return values.

'a를 left_most에 적절하게 추가합니다.

fn left_most<'a>(p1: &'a Point, p2: &'a Point) -> &'a Point {

즉, “a보다 오래 지속되는 p1과 p2가 있으면 반환 값은 최소한 'a 동안 유지됩니다.

일반적으로 수명은 다음 슬라이드에 설명된 대로 생략될 수 있습니다.

Comprehensive Rust 🦀

함수 호출에서의 수명