nomicon/src/lifetimes.md

# 수명

러스트는 이런 규칙들을 *수명*을 통해서 강제합니다. 수명은 레퍼런스가 유효해야 하는, 이름이 지어진 코드의 지역입니다. 이 지역들은 꽤나 복잡할 수 있는데, 프로그램의 실행 분기에 대응하기 때문입니다. 
또한 이런 실행 분기에는 구멍까지도 있을 수 있는데, 레퍼런스가 다시 쓰이기 전에 재초기화된다면, 이 레퍼런스를 무효화할 수 있기 때문입니다. 
레퍼런스를 포함하는 (또는 포함하는 척하는) 타입도 수명을 붙여서 러스트가 그것을 무효화하는 것을 막게 할 수 있습니다. 

우리의 예제 대부분, 수명은 코드 구역과 대응할 것입니다. 이것은 우리의 예제가 간단하기 때문입니다. 그렇게 대응되지 않는 복잡한 경우는 밑에 서술하겠습니다.

함수 본문 안에서 러스트는 보통 관련된 수명을 명시적으로 쓰게 하지 않습니다. 이것은 지역적인 문맥에서 수명에 대해 말하는 것은 대부분 별로 필요없기 때문입니다: 러스트는 모든 정보를 가지고 있고, 
모든 것들이 가능한 한 최적으로 동작하도록 할 수 있기 때문입니다. 컴파일러가 아니라면 일일히 작성해야 하는 많은 무명 

Within a function body, Rust generally doesn't let you explicitly name the
lifetimes involved. This is because it's generally not really necessary
to talk about lifetimes in a local context; Rust has all the information and
can work out everything as optimally as possible. Many anonymous scopes and
temporaries that you would otherwise have to write are often introduced to
make your code Just Work.

However once you cross the function boundary, you need to start talking about
lifetimes. Lifetimes are denoted with an apostrophe: `'a`, `'static`. To dip
our toes with lifetimes, we're going to pretend that we're actually allowed
to label scopes with lifetimes, and desugar the examples from the start of
this chapter.

Originally, our examples made use of *aggressive* sugar -- high fructose corn
syrup even -- around scopes and lifetimes, because writing everything out
explicitly is *extremely noisy*. All Rust code relies on aggressive inference
and elision of "obvious" things.

One particularly interesting piece of sugar is that each `let` statement
implicitly introduces a scope. For the most part, this doesn't really matter.
However it does matter for variables that refer to each other. As a simple
example, let's completely desugar this simple piece of Rust code:

```rust
let x = 0;
let y = &x;
let z = &y;
```

The borrow checker always tries to minimize the extent of a lifetime, so it will
likely desugar to the following:

<!-- ignore: desugared code -->
```rust,ignore
// NOTE: `'a: {` and `&'b x` is not valid syntax!
'a: {
    let x: i32 = 0;
    'b: {
        // lifetime used is 'b because that's good enough.
        let y: &'b i32 = &'b x;
        'c: {
            // ditto on 'c
            let z: &'c &'b i32 = &'c y; // "a reference to a reference to an i32" (with lifetimes annotated)
        }
    }
}
```

Wow. That's... awful. Let's all take a moment to thank Rust for making this easier.

Actually passing references to outer scopes will cause Rust to infer
a larger lifetime:

```rust
let x = 0;
let z;
let y = &x;
z = y;
```

<!-- ignore: desugared code -->
```rust,ignore
'a: {
    let x: i32 = 0;
    'b: {
        let z: &'b i32;
        'c: {
            // Must use 'b here because the reference to x is
            // being passed to the scope 'b.
            let y: &'b i32 = &'b x;
            z = y;
        }
    }
}
```

## Example: references that outlive referents

Alright, let's look at some of those examples from before:

```rust,compile_fail
fn as_str(data: &u32) -> &str {
    let s = format!("{}", data);
    &s
}
```

desugars to:

<!-- ignore: desugared code -->
```rust,ignore
fn as_str<'a>(data: &'a u32) -> &'a str {
    'b: {
        let s = format!("{}", data);
        return &'a s;
    }
}
```

This signature of `as_str` takes a reference to a u32 with *some* lifetime, and
promises that it can produce a reference to a str that can live *just as long*.
Already we can see why this signature might be trouble. That basically implies
that we're going to find a str somewhere in the scope the reference
to the u32 originated in, or somewhere *even earlier*. That's a bit of a tall
order.

We then proceed to compute the string `s`, and return a reference to it. Since
the contract of our function says the reference must outlive `'a`, that's the
lifetime we infer for the reference. Unfortunately, `s` was defined in the
scope `'b`, so the only way this is sound is if `'b` contains `'a` -- which is
clearly false since `'a` must contain the function call itself. We have therefore
created a reference whose lifetime outlives its referent, which is *literally*
the first thing we said that references can't do. The compiler rightfully blows
up in our face.

To make this more clear, we can expand the example:

<!-- ignore: desugared code -->
```rust,ignore
fn as_str<'a>(data: &'a u32) -> &'a str {
    'b: {
        let s = format!("{}", data);
        return &'a s
    }
}

fn main() {
    'c: {
        let x: u32 = 0;
        'd: {
            // An anonymous scope is introduced because the borrow does not
            // need to last for the whole scope x is valid for. The return
            // of as_str must find a str somewhere before this function
            // call. Obviously not happening.
            println!("{}", as_str::<'d>(&'d x));
        }
    }
}
```

Shoot!

Of course, the right way to write this function is as follows:

```rust
fn to_string(data: &u32) -> String {
    format!("{}", data)
}
```

We must produce an owned value inside the function to return it! The only way
we could have returned an `&'a str` would have been if it was in a field of the
`&'a u32`, which is obviously not the case.

(Actually we could have also just returned a string literal, which as a global
can be considered to reside at the bottom of the stack; though this limits
our implementation *just a bit*.)

## Example: aliasing a mutable reference

How about the other example:

```rust,compile_fail
let mut data = vec![1, 2, 3];
let x = &data[0];
data.push(4);
println!("{}", x);
```

<!-- ignore: desugared code -->
```rust,ignore
'a: {
    let mut data: Vec<i32> = vec![1, 2, 3];
    'b: {
        // 'b is as big as we need this borrow to be
        // (just need to get to `println!`)
        let x: &'b i32 = Index::index::<'b>(&'b data, 0);
        'c: {
            // Temporary scope because we don't need the
            // &mut to last any longer.
            Vec::push(&'c mut data, 4);
        }
        println!("{}", x);
    }
}
```

The problem here is a bit more subtle and interesting. We want Rust to
reject this program for the following reason: We have a live shared reference `x`
to a descendant of `data` when we try to take a mutable reference to `data`
to `push`. This would create an aliased mutable reference, which would
violate the *second* rule of references.

However this is *not at all* how Rust reasons that this program is bad. Rust
doesn't understand that `x` is a reference to a subpath of `data`. It doesn't
understand `Vec` at all. What it *does* see is that `x` has to live for `'b` in
order to be printed. The signature of `Index::index` subsequently demands that
the reference we take to `data` has to survive for `'b`. When we try to call
`push`, it then sees us try to make an `&'c mut data`. Rust knows that `'c` is
contained within `'b`, and rejects our program because the `&'b data` must still
be alive!

Here we see that the lifetime system is much more coarse than the reference
semantics we're actually interested in preserving. For the most part, *that's
totally ok*, because it keeps us from spending all day explaining our program
to the compiler. However it does mean that several programs that are totally
correct with respect to Rust's *true* semantics are rejected because lifetimes
are too dumb.

## The area covered by a lifetime

A reference (sometimes called a *borrow*) is *alive* from the place it is
created to its last use. The borrowed value needs to outlive only borrows that
are alive. This looks simple, but there are a few subtleties.

The following snippet compiles, because after printing `x`, it is no longer
needed, so it doesn't matter if it is dangling or aliased (even though the
variable `x` *technically* exists to the very end of the scope).

```rust
let mut data = vec![1, 2, 3];
let x = &data[0];
println!("{}", x);
// This is OK, x is no longer needed
data.push(4);
```

However, if the value has a destructor, the destructor is run at the end of the
scope. And running the destructor is considered a use ‒ obviously the last one.
So, this will *not* compile.

```rust,compile_fail
#[derive(Debug)]
struct X<'a>(&'a i32);

impl Drop for X<'_> {
    fn drop(&mut self) {}
}

let mut data = vec![1, 2, 3];
let x = X(&data[0]);
println!("{:?}", x);
data.push(4);
// Here, the destructor is run and therefore this'll fail to compile.
```

One way to convince the compiler that `x` is no longer valid is by using `drop(x)` before `data.push(4)`.

Furthermore, there might be multiple possible last uses of the borrow, for
example in each branch of a condition.

```rust
# fn some_condition() -> bool { true }
let mut data = vec![1, 2, 3];
let x = &data[0];

if some_condition() {
    println!("{}", x); // This is the last use of `x` in this branch
    data.push(4);      // So we can push here
} else {
    // There's no use of `x` in here, so effectively the last use is the
    // creation of x at the top of the example.
    data.push(5);
}
```

And a lifetime can have a pause in it. Or you might look at it as two distinct
borrows just being tied to the same local variable. This often happens around
loops (writing a new value of a variable at the end of the loop and using it for
the last time at the top of the next iteration).

```rust
let mut data = vec![1, 2, 3];
// This mut allows us to change where the reference points to
let mut x = &data[0];

println!("{}", x); // Last use of this borrow
data.push(4);
x = &data[3]; // We start a new borrow here
println!("{}", x);
```

Historically, Rust kept the borrow alive until the end of scope, so these
examples might fail to compile with older compilers. Also, there are still some
corner cases where Rust fails to properly shorten the live part of the borrow
and fails to compile even when it looks like it should. These'll be solved over
time.