# Lifetimes

Rust enforces these rules through *lifetimes*. Lifetimes are named
regions of code that a reference must be valid for. Those regions
may be fairly complex, as they correspond to paths of execution
in the program. There may even be holes in these paths of execution,
as it's possible to invalidate a reference as long as it's reinitialized
before it's used again. Types which contain references (or pretend to)
may also be tagged with lifetimes so that Rust can prevent them from
being invalidated as well.

In most of our examples, the lifetimes will coincide with scopes. This is
because our examples are simple. The more complex cases where they don't
coincide are described below.

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;
        }
    }
}
```

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 this reference is
            // being passed to that scope.
            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` 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.
