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:?}"); }
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.
'a
를 left_most
에 적절하게 추가합니다.
fn left_most<'a>(p1: &'a Point, p2: &'a Point) -> &'a Point {
즉, "a
보다 오래 지속되는 p1과 p2가 있으면 반환 값은 최소한 'a
동안 유지됩니다.
일반적으로 수명은 다음 슬라이드에 설명된 대로 생략될 수 있습니다.