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