% References and Borrowing

This is the second of three sections presenting Rust’s ownership system. This is one of
Rust’s most distinct and compelling features, with which Rust developers should
become quite acquainted. Ownership is how Rust achieves its largest goal,
memory safety. There are a few distinct concepts, each with its own
chapter:

* [ownership][ownership], the key concept
* borrowing, which you’re reading now
* [lifetimes][lifetimes], an advanced concept of borrowing

These three chapters are related, and in order. You’ll need all three to fully
understand the ownership system.

[ownership]: ownership.html
[lifetimes]: lifetimes.html

# Meta

Before we get to the details, two important notes about the ownership system.

Rust has a focus on safety and speed. It accomplishes these goals through many
‘zero-cost abstractions’, which means that in Rust, abstractions cost as little
as possible in order to make them work. The ownership system is a prime example
of a zero-cost abstraction. All of the analysis we’ll talk about in this guide
is _done at compile time_. You do not pay any run-time cost for any of these
features.

However, this system does have a certain cost: learning curve. Many new users
to Rust experience something we like to call ‘fighting with the borrow
checker’, where the Rust compiler refuses to compile a program that the author
thinks is valid. This often happens because the programmer’s mental model of
how ownership should work doesn’t match the actual rules that Rust implements.
You probably will experience similar things at first. There is good news,
however: more experienced Rust developers report that once they work with the
rules of the ownership system for a period of time, they fight the borrow
checker less and less.

With that in mind, let’s learn about borrowing.

# Borrowing

At the end of the [ownership][ownership] section, we had a nasty function that looked
like this:

```rust
fn foo(v1: Vec<i32>, v2: Vec<i32>) -> (Vec<i32>, Vec<i32>, i32) {
    // do stuff with v1 and v2

    // hand back ownership, and the result of our function
    (v1, v2, 42)
}

let v1 = vec![1, 2, 3];
let v2 = vec![1, 2, 3];

let (v1, v2, answer) = foo(v1, v2);
```

This is not idiomatic Rust, however, as it doesn’t take advantage of borrowing. Here’s
the first step:

```rust
fn foo(v1: &Vec<i32>, v2: &Vec<i32>) -> i32 {
    // do stuff with v1 and v2

    // return the answer
    42
}

let v1 = vec![1, 2, 3];
let v2 = vec![1, 2, 3];

let answer = foo(&v1, &v2);

// we can use v1 and v2 here!
```

A more concrete example:

```rust
fn main() {
    // Don't worry if you don't understand how `fold` works, the point here is that an immutable reference is borrowed.
    fn sum_vec(v: &Vec<i32>) -> i32 {
        return v.iter().fold(0, |a, &b| a + b);
    }
    // Borrow two vectors and sum them.
    // This kind of borrowing does not allow mutation to the borrowed.
    fn foo(v1: &Vec<i32>, v2: &Vec<i32>) -> i32 {
        // do stuff with v1 and v2
        let s1 = sum_vec(v1);
        let s2 = sum_vec(v2);
        // return the answer
        s1 + s2
    }

    let v1 = vec![1, 2, 3];
    let v2 = vec![4, 5, 6];

    let answer = foo(&v1, &v2);
    println!("{}", answer);
}
```

Instead of taking `Vec<i32>`s as our arguments, we take a reference:
`&Vec<i32>`. And instead of passing `v1` and `v2` directly, we pass `&v1` and
`&v2`. We call the `&T` type a ‘reference’, and rather than owning the resource,
it borrows ownership. A binding that borrows something does not deallocate the
resource when it goes out of scope. This means that after the call to `foo()`,
we can use our original bindings again.

References are immutable, like bindings. This means that inside of `foo()`,
the vectors can’t be changed at all:

```rust,ignore
fn foo(v: &Vec<i32>) {
     v.push(5);
}

let v = vec![];

foo(&v);
```

will give us this error:

```text
error: cannot borrow immutable borrowed content `*v` as mutable
v.push(5);
^
```

Pushing a value mutates the vector, and so we aren’t allowed to do it.

# &mut references

There’s a second kind of reference: `&mut T`. A ‘mutable reference’ allows you
to mutate the resource you’re borrowing. For example:

```rust
let mut x = 5;
{
    let y = &mut x;
    *y += 1;
}
println!("{}", x);
```

This will print `6`. We make `y` a mutable reference to `x`, then add one to
the thing `y` points at. You’ll notice that `x` had to be marked `mut` as well.
If it wasn’t, we couldn’t take a mutable borrow to an immutable value.

You'll also notice we added an asterisk (`*`) in front of `y`, making it `*y`,
this is because `y` is a `&mut` reference. You'll need to use astrisks to
access the contents of a reference as well.

Otherwise, `&mut` references are like references. There _is_ a large
difference between the two, and how they interact, though. You can tell
something is fishy in the above example, because we need that extra scope, with
the `{` and `}`. If we remove them, we get an error:

```text
error: cannot borrow `x` as immutable because it is also borrowed as mutable
    println!("{}", x);
                   ^
note: previous borrow of `x` occurs here; the mutable borrow prevents
subsequent moves, borrows, or modification of `x` until the borrow ends
        let y = &mut x;
                     ^
note: previous borrow ends here
fn main() {

}
^
```

As it turns out, there are rules.

# The Rules

Here are the rules for borrowing in Rust:

First, any borrow must last for a scope no greater than that of the owner.
Second, you may have one or the other of these two kinds of borrows, but not
both at the same time:

* one or more references (`&T`) to a resource,
* exactly one mutable reference (`&mut T`).


You may notice that this is very similar to, though not exactly the same as,
the definition of a data race:

> There is a ‘data race’ when two or more pointers access the same memory
> location at the same time, where at least one of them is writing, and the
> operations are not synchronized.

With references, you may have as many as you’d like, since none of them are
writing. However, as we can only have one `&mut` at a time, it is impossible to
have a data race. This is how Rust prevents data races at compile time: we’ll
get errors if we break the rules.

With this in mind, let’s consider our example again.

## Thinking in scopes

Here’s the code:

```rust,ignore
fn main() {
    let mut x = 5;
    let y = &mut x;

    *y += 1;

    println!("{}", x);
}
```

This code gives us this error:

```text
error: cannot borrow `x` as immutable because it is also borrowed as mutable
    println!("{}", x);
                   ^
```

This is because we’ve violated the rules: we have a `&mut T` pointing to `x`,
and so we aren’t allowed to create any `&T`s. It's one or the other. The note
hints at how to think about this problem:

```text
note: previous borrow ends here
fn main() {

}
^
```

In other words, the mutable borrow is held through the rest of our example. What
we want is for the mutable borrow by `y` to end so that the resource can be
returned to the owner, `x`. `x` can then provide a immutable borrow to `println!`.
In Rust, borrowing is tied to the scope that the borrow is valid for. And our
scopes look like this:

```rust,ignore
fn main() {
    let mut x = 5;

    let y = &mut x;    // -+ &mut borrow of x starts here
                       //  |
    *y += 1;           //  |
                       //  |
    println!("{}", x); // -+ - try to borrow x here
}                      // -+ &mut borrow of x ends here
                       
```

The scopes conflict: we can’t make an `&x` while `y` is in scope.

So when we add the curly braces:

```rust
let mut x = 5;

{
    let y = &mut x; // -+ &mut borrow starts here
    *y += 1;        //  |
}                   // -+ ... and ends here

println!("{}", x);  // <- try to borrow x here
```

There’s no problem. Our mutable borrow goes out of scope before we create an
immutable one. So scope is the key to seeing how long a borrow lasts for.

## Issues borrowing prevents

Why have these restrictive rules? Well, as we noted, these rules prevent data
races. What kinds of issues do data races cause? Here are a few.

### Iterator invalidation

One example is ‘iterator invalidation’, which happens when you try to mutate a
collection that you’re iterating over. Rust’s borrow checker prevents this from
happening:

```rust
let mut v = vec![1, 2, 3];

for i in &v {
    println!("{}", i);
}
```

This prints out one through three. As we iterate through the vector, we’re
only given references to the elements. And `v` is itself borrowed as immutable,
which means we can’t change it while we’re iterating:

```rust,ignore
let mut v = vec![1, 2, 3];

for i in &v {
    println!("{}", i);
    v.push(34);
}
```

Here’s the error:

```text
error: cannot borrow `v` as mutable because it is also borrowed as immutable
    v.push(34);
    ^
note: previous borrow of `v` occurs here; the immutable borrow prevents
subsequent moves or mutable borrows of `v` until the borrow ends
for i in &v {
          ^
note: previous borrow ends here
for i in &v {
    println!(“{}”, i);
    v.push(34);
}
^
```

We can’t modify `v` because it’s borrowed by the loop.

### Use after free

References must not live longer than the resource they refer to. Rust will
check the scopes of your references to ensure that this is true.

If Rust didn’t check this property, we could accidentally use a reference
which was invalid. For example:

```rust,ignore
let y: &i32;
{
    let x = 5;
    y = &x;
}

println!("{}", y);
```

We get this error:

```text
error: `x` does not live long enough
    y = &x;
         ^
note: reference must be valid for the block suffix following statement 0 at
2:16...
let y: &i32;
{
    let x = 5;
    y = &x;
}

note: ...but borrowed value is only valid for the block suffix following
statement 0 at 4:18
    let x = 5;
    y = &x;
}
```

In other words, `y` is only valid for the scope where `x` exists. As soon as
`x` goes away, it becomes invalid to refer to it. As such, the error says that
the borrow ‘doesn’t live long enough’ because it’s not valid for the right
amount of time.

The same problem occurs when the reference is declared _before_ the variable it
refers to. This is because resources within the same scope are freed in the
opposite order they were declared:

```rust,ignore
let y: &i32;
let x = 5;
y = &x;

println!("{}", y);
```

We get this error:

```text
error: `x` does not live long enough
y = &x;
     ^
note: reference must be valid for the block suffix following statement 0 at
2:16...
    let y: &i32;
    let x = 5;
    y = &x;

    println!("{}", y);
}

note: ...but borrowed value is only valid for the block suffix following
statement 1 at 3:14
    let x = 5;
    y = &x;

    println!("{}", y);
}
```

In the above example, `y` is declared before `x`, meaning that `y` lives longer
than `x`, which is not allowed.
