In Chapter 3 of the Rust Book, Variables and Mutability, we go through a couple iterations on this theme in order to demonstrate the default, immutable behavior of variables in Rust:
fn main() {
let x = 5;
println!("The value of x is {}", x);
x = 6;
println!("The value of x is {}", x);
}
Which outputs:
error[E0384]: cannot assign twice to immutable variable `x`
--> src/main.rs:4:5
|
2 | let x = 5;
| -
| |
| first assignment to `x`
| help: make this binding mutable: `mut x`
3 | println!("The value of x is {}", x);
4 | x = 6;
| ^^^^^ cannot assign twice to immutable variable
However, because of Rust's take on shadowing variables, we can simply do this to change the value of the nonetheless "immutable" x
:
fn main() {
let x = 5;
println!("The value of x is {}", x);
let x = 6;
println!("The value of x is {}", x);
}
Which outputs (skipping the details):
The value of x is 5
The value of x is 6
Funnily enough, this code also produces the above pair of lines as output, despite the fact that we don't call let
but instead mut
the first time x
is bound to 5
:
fn main() {
let mut x = 5;
println!("The value of x is {}", x);
x = 6;
println!("The value of x is {}", x);
}
This ambiguity in how variables are (not really) protected from reassignment seems contrary to the stated goal of protecting the values bound to immutable - by Rust default - variables. From the same chapter (which also contains the section Shadowing):
It’s important that we get compile-time errors when we attempt to change a value that we previously designated as immutable because this very situation can lead to bugs. If one part of our code operates on the assumption that a value will never change and another part of our code changes that value, it’s possible that the first part of the code won’t do what it was designed to do. The cause of this kind of bug can be difficult to track down after the fact, especially when the second piece of code changes the value only sometimes.
In Rust, the compiler guarantees that when you state that a value won’t change, it really won’t change. That means that when you’re reading and writing code, you don’t have to keep track of how and where a value might change. Your code is thus easier to reason through.
If I can cause this important feature of my immutable x
to be side-stepped with an innocent enough call to let
, why do I need mut
? Is there some way to really, seriously-you-guys make x
immutable, such that no let x
can reassign its value?
I believe the confusion is because you're conflating names with storage.
In this example, there is one name (
x
), and two storage locations (x_0
andx_1
). The secondlet
is simply re-binding the namex
to refer to storage locationx_1
. Thex_0
storage location is entirely unaffected.In this example, there is one name (
x
), and one storage location (x_0
). Thex = 6
assignment is directly changing the bits of storage locationx_0
.You might argue that these do the same thing. If so, you would be wrong:
This outputs:
This is because changing which storage location
x
refers to has absolutely no effect on the storage locationx_0
, which is whaty_0
contains a pointer to. However,This fails to compile because you cannot mutate
x_0
while it is borrowed.Rust cares about protecting against unwanted mutation effects as observed through references. This doesn't conflict with allowing shadowing, because you're not changing values when you shadow, you're just changing what a particular name means in a way that cannot be observed anywhere else. Shadowing is a strictly local change.
So yes, you absolutely can keep the value of
x
from being changed. What you can't do is keep what the namex
refers to from being changed. At most, you can use something likeclippy
to deny shadowing as a lint.