## References and Borrowing

The issue with the tuple code in Listing 4-5 is that we have to return the
`String` to the calling function so we can still use the `String` after the
call to `calculate_length`, because the `String` was moved into
`calculate_length`.

Here is how you would define and use a `calculate_length` function that has a
reference to an object as a parameter instead of taking ownership of the
value:

<span class="filename">Filename: src/main.rs</span>

```rust
fn main() {
    let s1 = String::from("hello");

    let len = calculate_length(&s1);

    println!("The length of '{}' is {}.", s1, len);
}

fn calculate_length(s: &String) -> usize {
    s.len()
}
```

First, notice that all the tuple code in the variable declaration and the
function return value is gone. Second, note that we pass `&s1` into
`calculate_length` and, in its definition, we take `&String` rather than
`String`.

These ampersands are *references*, and they allow you to refer to some value
without taking ownership of it. Figure 4-5 shows a diagram.

<img alt="&String s pointing at String s1" src="img/trpl04-05.svg" class="center" />

<span class="caption">Figure 4-5: A diagram of `&String s` pointing at `String
s1`</span>

> Note: The opposite of referencing by using `&` is *dereferencing*, which is
> accomplished with the dereference operator, `*`. We’ll see some uses of the
> dereference operator in Chapter 8 and discuss details of dereferencing in
> Chapter 15.

Let’s take a closer look at the function call here:

```rust
# fn calculate_length(s: &String) -> usize {
#     s.len()
# }
let s1 = String::from("hello");

let len = calculate_length(&s1);
```

The `&s1` syntax lets us create a reference that *refers* to the value of `s1`
but does not own it. Because it does not own it, the value it points to will
not be dropped when the reference goes out of scope.

Likewise, the signature of the function uses `&` to indicate that the type of
the parameter `s` is a reference. Let’s add some explanatory annotations:

```rust
fn calculate_length(s: &String) -> usize { // s is a reference to a String
    s.len()
} // Here, s goes out of scope. But because it does not have ownership of what
  // it refers to, nothing happens.
```

The scope in which the variable `s` is valid is the same as any function
parameter’s scope, but we don’t drop what the reference points to when it goes
out of scope because we don’t have ownership. When functions have references as
parameters instead of the actual values, we won’t need to return the values in
order to give back ownership, because we never had ownership.

We call having references as function parameters *borrowing*. As in real life,
if a person owns something, you can borrow it from them. When you’re done, you
have to give it back.

So what happens if we try to modify something we’re borrowing? Try the code in
Listing 4-6. Spoiler alert: it doesn’t work!

<span class="filename">Filename: src/main.rs</span>

```rust,ignore
fn main() {
    let s = String::from("hello");

    change(&s);
}

fn change(some_string: &String) {
    some_string.push_str(", world");
}
```

<span class="caption">Listing 4-6: Attempting to modify a borrowed value</span>

Here’s the error:

```text
error[E0596]: cannot borrow immutable borrowed content `*some_string` as mutable
 --> error.rs:8:5
  |
7 | fn change(some_string: &String) {
  |                        ------- use `&mut String` here to make mutable
8 |     some_string.push_str(", world");
  |     ^^^^^^^^^^^ cannot borrow as mutable
```

Just as variables are immutable by default, so are references. We’re not
allowed to modify something we have a reference to.

### Mutable References

We can fix the error in the code from Listing 4-6 with just a small tweak:

<span class="filename">Filename: src/main.rs</span>

```rust
fn main() {
    let mut s = String::from("hello");

    change(&mut s);
}

fn change(some_string: &mut String) {
    some_string.push_str(", world");
}
```

First, we had to change `s` to be `mut`. Then we had to create a mutable
reference with `&mut s` and accept a mutable reference with `some_string: &mut
String`.

But mutable references have one big restriction: you can only have one mutable
reference to a particular piece of data in a particular scope. This code will
fail:

<span class="filename">Filename: src/main.rs</span>

```rust,ignore
let mut s = String::from("hello");

let r1 = &mut s;
let r2 = &mut s;
```

Here’s the error:

```text
error[E0499]: cannot borrow `s` as mutable more than once at a time
 --> borrow_twice.rs:5:19
  |
4 |     let r1 = &mut s;
  |                   - first mutable borrow occurs here
5 |     let r2 = &mut s;
  |                   ^ second mutable borrow occurs here
6 | }
  | - first borrow ends here
```

This restriction allows for mutation but in a very controlled fashion. It’s
something that new Rustaceans struggle with, because most languages let you
mutate whenever you’d like.

The benefit of having this restriction is that Rust can prevent data races at
compile time. A *data race* is similar to a race condition and happens when
these three behaviors occur:

* Two or more pointers access the same data at the same time.
* At least one of the pointers is being used to write to the data.
* There’s no mechanism being used to synchronize access to the data.

Data races cause undefined behavior and can be difficult to diagnose and fix
when you’re trying to track them down at runtime; Rust prevents this problem
from happening because it won’t even compile code with data races!

As always, we can use curly brackets to create a new scope, allowing for
multiple mutable references, just not *simultaneous* ones:

```rust
let mut s = String::from("hello");

{
    let r1 = &mut s;

} // r1 goes out of scope here, so we can make a new reference with no problems.

let r2 = &mut s;
```

A similar rule exists for combining mutable and immutable references. This code
results in an error:

```rust,ignore
let mut s = String::from("hello");

let r1 = &s; // no problem
let r2 = &s; // no problem
let r3 = &mut s; // BIG PROBLEM
```

Here’s the error:

```text
error[E0502]: cannot borrow `s` as mutable because it is also borrowed as
immutable
 --> borrow_thrice.rs:6:19
  |
4 |     let r1 = &s; // no problem
  |               - immutable borrow occurs here
5 |     let r2 = &s; // no problem
6 |     let r3 = &mut s; // BIG PROBLEM
  |                   ^ mutable borrow occurs here
7 | }
  | - immutable borrow ends here
```

Whew! We *also* cannot have a mutable reference while we have an immutable one.
Users of an immutable reference don’t expect the values to suddenly change out
from under them! However, multiple immutable references are okay because no one
who is just reading the data has the ability to affect anyone else’s reading of
the data.

Even though these errors may be frustrating at times, remember that it’s the
Rust compiler pointing out a potential bug early (at compile time rather than
at runtime) and showing you exactly where the problem is. Then you don’t have
to track down why your data isn’t what you thought it was.

### Dangling References

In languages with pointers, it’s easy to erroneously create a *dangling
pointer*, a pointer that references a location in memory that may have been
given to someone else, by freeing some memory while preserving a pointer to
that memory. In Rust, by contrast, the compiler guarantees that references will
never be dangling references: if you have a reference to some data, the
compiler will ensure that the data will not go out of scope before the
reference to the data does.

Let’s try to create a dangling reference, which Rust will prevent with a
compile-time error:

<span class="filename">Filename: src/main.rs</span>

```rust,ignore
fn main() {
    let reference_to_nothing = dangle();
}

fn dangle() -> &String {
    let s = String::from("hello");

    &s
}
```

Here’s the error:

```text
error[E0106]: missing lifetime specifier
 --> dangle.rs:5:16
  |
5 | fn dangle() -> &String {
  |                ^ expected lifetime parameter
  |
  = help: this function's return type contains a borrowed value, but there is
  no value for it to be borrowed from
  = help: consider giving it a 'static lifetime
```

This error message refers to a feature we haven’t covered yet: *lifetimes*.
We’ll discuss lifetimes in detail in Chapter 10. But, if you disregard the
parts about lifetimes, the message does contain the key to why this code is a
problem:

```text
this function's return type contains a borrowed value, but there is no value
for it to be borrowed from.
```

Let’s take a closer look at exactly what’s happening at each stage of our
`dangle` code:

```rust,ignore
fn dangle() -> &String { // dangle returns a reference to a String

    let s = String::from("hello"); // s is a new String

    &s // we return a reference to the String, s
} // Here, s goes out of scope, and is dropped. Its memory goes away.
  // Danger!
```

Because `s` is created inside `dangle`, when the code of `dangle` is finished,
`s` will be deallocated. But we tried to return a reference to it. That means
this reference would be pointing to an invalid `String` That’s no good! Rust
won’t let us do this.

The solution here is to return the `String` directly:

```rust
fn no_dangle() -> String {
    let s = String::from("hello");

    s
}
```

This works without any problems. Ownership is moved out, and nothing is
deallocated.

### The Rules of References

Let’s recap what we’ve discussed about references:

* At any given time, you can have *either* (but not both of) one mutable
  reference or any number of immutable references.
* References must always be valid.

Next, we’ll look at a different kind of reference: slices.
