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How can I force a struct's field to always be immutable in Rust?

In Rust, you don't specify mutability inside a struct, but it is inherited from the variable binding. That's great, but is it possible to force a field to be always immutable, even when the root is mutable?

Something like this hypothetical syntax:

struct A {
    immut s: Shape, // immutable by design
    bla: Bla, // this field inheriting (im)mutability
}
let mut a = make_a();
a.s = x/*...*/; // illegal

This would help to maintain nice semantic restrictions in a program, just like Java's final does (in a very limited way).

Also, we could imagine this kind of struct having some non-owning references to internal immutable data, taking advantage of this immutability...

like image 472
LP_ Avatar asked May 19 '14 17:05

LP_


2 Answers

It's impossible to have immutability of a single field. That was an option in an ancient version of Rust (think before 0.8), but it was dropped because the rules confused a LOT of people. How was it confusing, you might ask? Think about it like this: if a field is declared mutable and struct is declared mutable and the reference used was an immutable reference (&) then the field is _______.

The best, as Lily Ballard noted, is that you can declare your Shape field as private and make a getter method using impl A {...}.

mod inner {
    pub struct A {
        s: i32, // can't be seen outside of module
        pub bla: i32,
    }

    impl A {
        pub fn new() -> Self {
            Self { s: 0, bla: 42 }
        }

        pub fn get_s(&self) -> i32 {
            self.s
        }
    }
}
let mut a = inner::A::new();
a.s = 42; // illegal
println!("{}", a.s); // also illegal
println!("{}", a.get_s()); // could be made to serve as a read-only method
error[E0616]: field `s` of struct `main::inner::A` is private
  --> src/main.rs:20:5
   |
20 |     a.s = 42; // illegal
   |     ^^^

error[E0616]: field `s` of struct `main::inner::A` is private
  --> src/main.rs:21:20
   |
21 |     println!("{}", a.s); // also illegal
   |                    ^^^

There is proposition that might drop notions of mutability and immutability completely (you can't say a struct never changes). See Niko's explanation for that change.

like image 120
Daniel Fath Avatar answered Nov 14 '22 07:11

Daniel Fath


You can create a struct and only implement the Deref trait for it. Without the DerefMut trait it won't be possible for contained values to be mutated.

https://doc.rust-lang.org/std/ops/trait.Deref.html

This way the compiler will make the member usable as if it's not wrapped in another struct, no need for any written getter method call.

use std::ops::Deref;

/// A container for values that can only be deref'd immutably.
struct Immutable<T> {
    value: T,
}

impl<T> Immutable<T> {
    pub fn new(value: T) -> Self {
        Immutable { value }
    }
}

impl<T> Deref for Immutable<T> {
    type Target = T;

    fn deref(&self) -> &Self::Target {
        &self.value
    }
}
struct Foo {
    bar: Immutable<Vec<u8>>,
    baz: usize,
}

impl Foo {
    pub fn new(vec: Vec<u8>) -> Self {
        Foo {
            bar: Immutable::new(vec),
            baz: 1337,
        }
    }

    pub fn mutate(&mut self) {
        self.bar.push(0); // This will cause a compiler error
    }
}
|
|         self.bar.push(0);
|         ^^^^^^^^ cannot borrow as mutable
|
= help: trait `DerefMut` is required to modify through a dereference, but it is not implemented for `runnable::immutable::Immutable<std::vec::Vec<u8>>`
like image 12
danthedaniel Avatar answered Nov 14 '22 06:11

danthedaniel