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:

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

We can introduce mutability with the mut keyword:

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.

If you want to change what the binding points to, you’ll need a mutable reference:

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 you can mutate the thing that’s bound to y (*y = 5). A subtle distinction.

Of course, if you need both:

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:

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

fn foo(mut x: i32) {
# }

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>]arc:

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 muts 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 Ts, though, that would be a problem.

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

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:

use std::cell::RefCell;

let x = RefCell::new(42);

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

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]struct with some fields mutable and some immutable:

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

The mutability of a struct is in its binding:

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>]cell, you can emulate field-level mutability:

use std::cell::Cell;

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

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


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

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