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Rust Ownership, Move and Borrow - Part 2

13 Nov 21

The rules that govern borrowing with respect to mutability are straightforward: we can have multiple immutable borrows, or one mutable borrow. To gain a better understanding though, consider what happens when dealing with nested data. In this example, our User has a nested Config as well as a list of bookmarks:

// #[derive(Debug)] instructs the compiler to auto-generate the code needed to
// satisfy the Debug trait (think interface). We need this to be able to print
// out the values

#[derive(Debug)]
struct Config {
max_bookmarks: usize,
}

#[derive(Debug)]
struct User {
config: Config,
bookmarks: Vec<String>,
}

fn main() {
let mut user = User{
bookmarks: vec![],
config: Config{max_bookmarks: 1000},
};

let config = user.config;
println!("{:?} {:?}", user, config);
}

The above code moves user.config to a new owner, the config variable. When we try to print user and config, the compiler tells us that user has been "partially moved" and cannot be used. If we change the last line to print user.bookmarks instead of user, the code works:

println!("{:?} {:?}", user.bookmarks, config);

We can see from these examples that Rust allows nested fields to be moved (and borrowed) independently, but doing so invalidates the whole (user in this case). This makes sense: user is no longer valid as a whole, its config is no longer valid (it's been moved to a different owner).

While we're able to partially move and borrow, mutability is, by default, "inherited". We cannot control the mutability of individual fields. If we don't declare our user as mutable, then bookmarks isn't mutable:

fn main() {
let user = User{
bookmarks: vec![],
config: Config{max_bookmarks: 1000},
};

// will not work as user isn't mutable
user.bookmarks.push("https://www.openmymind.net".to_owned());
}

However, when we move, we can change mutability. The following will work:

fn add(mut user: User) {
user.bookmarks.push("https://www.openmymind.net".to_owned());
}

fn main() {
let user = User{
bookmarks: vec![],
config: Config{max_bookmarks: 1000},
};
add(user)
}

This is important: it isn't the data which is or isn't mutable, it's the binding.

As a final example, consider the following:

fn main() {
let name: Option<String> = Some("Leto".to_owned());
match name {
None => println!("no name"),
Some(name) => println!("we have a name: {}", name),
}
println!("{:?}", name)
}

The compiler will tell us that our last line is invalid because name has been partially moved. Note that name is an Option<String>, and we can think of the actual string value, "Leto", as a nested field. The second arm of our match takes ownership of the the string value. The result is that name, the Option<String>, is no longer valid since part of its data has been moved. We can solve this by matching against a borrow:

fn main() {
let name: Option<String> = Some("Leto".to_owned());
match &name {
None => println!("no name"),
Some(name) => println!("we have a name: {}", name),
}
println!("{:?}", name)
}

It's worth remembering that in addition to moving and borrowing, there's also copying. If we replace our Option<String> with an Option<T> where T implements copying, our initial borrow-free version works at the "cost" of copying the value:

fn main() {
let id: Option<i32> = Some(9001);
match id {
None => println!("no id"),
Some(id) => println!("we have a id: {}", id),
}
println!("{:?}", id)
}

(I say "'cost' of copying the value' because, the Copy trait is typically implemented on types where copying isn't only cheap, but also expected.)

Why?

Understanding the problems solved by this ownership model, even just conceptually, should help better understand the practical implications.

Why? - Safety

The main reason all of this is necessary is to have guaranteed compile-time safety without the overhead of a garbage collector. If, like me, you've spent most (or all) of your programming life with a garbage collector, you'll need to make a difficult adjustment to your perspective. We're used to the runtime keeping track of what is and isn't used and cleaning things up as needed. No more. This code, thankfully, won't compile:

#[derive(Debug)]
struct User {
id: i32
}

fn main() {
let mut users = vec![User{id: 1}, User{id: 2}];
let last1 = users.last();
users.remove(1);
print!("last1 {:?}\n", last1);
}

last borrows the value from our users vector which is then mutated via remove. Does last still point to valid data? In this specific case we're removing the last item, the same one returned by last(). If we had to manage our own memory remove probably would have been implement to free the removed memory; resulting in last1 pointing to freed memory. What if instead of remove(1) we'd had pushed a new value onto the vector? The vector might have needed to grow, moving all data to new memory. In that case last1 would be left pointing to the old freed location.

Even in this trivial case, without explicit ownership, we don't know who is responsible for freeing data. Start making this code more complex, with functions calling functions and third party libraries, and even the most diligent teams won't always know when data should be freed or when it has already been freed.

Rust's ownership model unambiguously answers the question of who is responsible for data. And, because the answer is based on a set of verifiable rules, we don't need to rely on fallible approaches, such as conventions and programmer diligence. Instead, the compiler can ensure correctness and insert code to free memory where needed.

Why? - Optimizations

The ownership model is both information we give to the compiler as well as a set of constraint about what is and isn't allowed. The compiler can leverage this to better optimize our code. Here's an example taken from The Rustonomicon:

fn compute(input: &u32, output: &mut u32) {
if *input > 10 {
*output = 1;
}
if *input > 5 {
*output *= 2;
}
}

Ideally, we'd like to store *input in a local variable so that we didn't have to dereference it twice. At a quick glance, you might think this should be safe, but what if input and output reference the same value?:

compute(&x, &mut x)

Languages that allow this cannot safely optimize the compute function. In Rust, this isn't valid, we can't immutably borrow while mutably borrowing, so the code can be optimized.

Why? - Simplicity

It might seem a bit silly to say the the ownership model is about simplicity, but hear me out. When you start, you'll probably look at basic examples or try to write simple programs. These often look and feel needlessly complicated. And you might think to yourself: what am I not getting? Like our very first example in Part 1 , couldn't the compiler just compile this and run:

fn main() {
let u1 = User{id: 9000};

let u2 = u1;
print!("{:?}", u2);

// this is an error
print!("{:?}", u1);
}

Maybe I'm wrong, but I believe the answer is: Yes, the compiler could compile this and run it with 100% safety. But that would turn the explicit and consistent nature of ownership into something fluid. Simple cases would become simpler, but complex cases would become more complicated. Sometimes assignments would move and invalidate the original, and sometimes it wouldn't. Either way, you'd still have to understand ownership, it's just that we'd introduce a second ownership mode. No thanks! (not to mention that it would make the compiler more complicated too).

Conclusion

Part 3 will conclude this introduction on ownership with a brief summary of everything we discovered and talked about so far. However, since this is such a fundamental aspect of Rust, we'll continue to revisit ownership while exploring other topics.

As a final word of encouragement, if you still feel overwhelmed, don't worry about it. Start small and try your best. Be persistent, but also consider asking the community for help. In particular, I've found people in the official rust forums incredibly friendly, helpful and fast.