# Mutability

Mutability, the ability to change something, works a bit differently in Rust than in other languages. The first aspect of mutability is its non-default status:

fn main() { let x = 5; x = 6; // error! }

    let x = 5;
    x = 6; // error!

We can introduce mutability with the `mut` keyword:

fn main() { let mut x = 5; x = 6; // no problem! }

    let mut x = 5;

    x = 6; // no problem!

This is a mutable variable binding. When a binding is mutable, it means you’re allowed to change what the binding points to. So in the above example, it’s not so much that the value at `x` is changing, but that the binding changed from one `i32` to another.

You can also create a reference to it, using `&x`, but if you want to use the reference to change it, you will need a mutable reference:

fn main() { let mut x = 5; let y = &mut x; }

    let mut x = 5;
    let y = &mut x;

`y` is an immutable binding to a mutable reference, which means that you can’t bind 'y' to something else (`y = &mut z`), but `y` can be used to bind `x` to something else (`*y = 5`). A subtle distinction.

Of course, if you need both:

fn main() { let mut x = 5; let mut y = &mut x; }

    let mut x = 5;
    let mut y = &mut x;

Now `y` can be bound to another value, and the value it’s referencing can be changed.

It’s important to note that `mut` is part of a pattern, so you can do things like this:

fn main() { let (mut x, y) = (5, 6); fn foo(mut x: i32) { } }

    let (mut x, y) = (5, 6);

    fn foo(mut x: i32) {

Note that here, the `x` is mutable, but not the `y`.

# Interior vs. Exterior Mutability

However, when we say something is ‘immutable’ in Rust, that doesn’t mean that it’s not able to be changed: we mean something has ‘exterior mutability’. Consider, for example, `Arc<T>`:

fn main() { use std::sync::Arc; let x = Arc::new(5); let y = x.clone(); }

    use std::sync::Arc;

    let x = Arc::new(5);
    let y = x.clone();

When we call `clone()`, the `Arc<T>` needs to update the reference count. Yet we’ve not used any `mut`s here, `x` is an immutable binding, and we didn’t take `&mut 5` or anything. So what gives?

To understand this, we have to go back to the core of Rust’s guiding philosophy, memory safety, and the mechanism by which Rust guarantees it, the ownership system, and more specifically, borrowing:

> 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`).

So, that’s the real definition of ‘immutability’: is this safe to have two pointers to? In `Arc<T>`’s case, yes: the mutation is entirely contained inside the structure itself. It’s not user facing. For this reason, it hands out `&T` with `clone()`. If it handed out `&mut T`s, though, that would be a problem.

Other types, like the ones in the `std::cell` module, have the opposite: interior mutability. For example:

fn main() { use std::cell::RefCell; let x = RefCell::new(42); let y = x.borrow_mut(); }

    use std::cell::RefCell;

    let x = RefCell::new(42);

    let y = x.borrow_mut();

RefCell hands out `&mut` references to what’s inside of it with the `borrow_mut()` method. Isn’t that dangerous? What if we do:

fn main() { use std::cell::RefCell; let x = RefCell::new(42); let y = x.borrow_mut(); let z = x.borrow_mut(); (y, z); }

    use std::cell::RefCell;

    let x = RefCell::new(42);

    let y = x.borrow_mut();
    let z = x.borrow_mut();

This will in fact panic, at runtime. This is what `RefCell` does: it enforces Rust’s borrowing rules at runtime, and `panic!`s if they’re violated. This allows us to get around another aspect of Rust’s mutability rules. Let’s talk about it first.

## Field-level mutability

Mutability is a property of either a borrow (`&mut`) or a binding (`let mut`). This means that, for example, you cannot have a `struct` with some fields mutable and some immutable:

fn main() { struct Point { x: i32, mut y: i32, // nope } }

    struct Point {
        x: i32,
        mut y: i32, // nope
    }

The mutability of a struct is in its binding:

fn main() { struct Point { x: i32, y: i32, } let mut a = Point { x: 5, y: 6 }; a.x = 10; let b = Point { x: 5, y: 6}; b.x = 10; // error: cannot assign to immutable field `b.x` }

    struct Point {
        x: i32,
        y: i32,
    }

    let mut a = Point { x: 5, y: 6 };

    a.x = 10;

    let b = Point { x: 5, y: 6};

    b.x = 10; // error: cannot assign to immutable field `b.x`

However, by using `Cell<T>`, you can emulate field-level mutability:

fn main() { use std::cell::Cell; struct Point { x: i32, y: Cell<i32>, } let point = Point { x: 5, y: Cell::new(6) }; point.y.set(7); println!("y: {:?}", point.y); }

    use std::cell::Cell;

    struct Point {
        x: i32,
        y: Cell<i32>,
    }

    let point = Point { x: 5, y: Cell::new(6) };

    point.y.set(7);

    println!("y: {:?}", point.y);

This will print `y: Cell { value: 7 }`. We’ve successfully updated `y`.