1 % References and Borrowing
3 This guide is two of three presenting Rust’s ownership system. This is one of
4 Rust’s most unique and compelling features, with which Rust developers should
5 become quite acquainted. Ownership is how Rust achieves its largest goal,
6 memory safety. There are a few distinct concepts, each with its own
9 * [ownership][ownership], the key concept
10 * borrowing, which you’re reading now
11 * [lifetimes][lifetimes], an advanced concept of borrowing
13 These three chapters are related, and in order. You’ll need all three to fully
14 understand the ownership system.
16 [ownership]: ownership.html
17 [lifetimes]: lifetimes.html
21 Before we get to the details, two important notes about the ownership system.
23 Rust has a focus on safety and speed. It accomplishes these goals through many
24 ‘zero-cost abstractions’, which means that in Rust, abstractions cost as little
25 as possible in order to make them work. The ownership system is a prime example
26 of a zero cost abstraction. All of the analysis we’ll talk about in this guide
27 is _done at compile time_. You do not pay any run-time cost for any of these
30 However, this system does have a certain cost: learning curve. Many new users
31 to Rust experience something we like to call ‘fighting with the borrow
32 checker’, where the Rust compiler refuses to compile a program that the author
33 thinks is valid. This often happens because the programmer’s mental model of
34 how ownership should work doesn’t match the actual rules that Rust implements.
35 You probably will experience similar things at first. There is good news,
36 however: more experienced Rust developers report that once they work with the
37 rules of the ownership system for a period of time, they fight the borrow
38 checker less and less.
40 With that in mind, let’s learn about borrowing.
44 At the end of the [ownership][ownership] section, we had a nasty function that looked
48 fn foo(v1: Vec<i32>, v2: Vec<i32>) -> (Vec<i32>, Vec<i32>, i32) {
49 // do stuff with v1 and v2
51 // hand back ownership, and the result of our function
55 let v1 = vec![1, 2, 3];
56 let v2 = vec![1, 2, 3];
58 let (v1, v2, answer) = foo(v1, v2);
61 This is not idiomatic Rust, however, as it doesn’t take advantage of borrowing. Here’s
65 fn foo(v1: &Vec<i32>, v2: &Vec<i32>) -> i32 {
66 // do stuff with v1 and v2
72 let v1 = vec![1, 2, 3];
73 let v2 = vec![1, 2, 3];
75 let answer = foo(&v1, &v2);
77 // we can use v1 and v2 here!
80 Instead of taking `Vec<i32>`s as our arguments, we take a reference:
81 `&Vec<i32>`. And instead of passing `v1` and `v2` directly, we pass `&v1` and
82 `&v2`. We call the `&T` type a ‘reference’, and rather than owning the resource,
83 it borrows ownership. A binding that borrows something does not deallocate the
84 resource when it goes out of scope. This means that after the call to `foo()`,
85 we can use our original bindings again.
87 References are immutable, like bindings. This means that inside of `foo()`,
88 the vectors can’t be changed at all:
91 fn foo(v: &Vec<i32>) {
103 error: cannot borrow immutable borrowed content `*v` as mutable
108 Pushing a value mutates the vector, and so we aren’t allowed to do it.
112 There’s a second kind of reference: `&mut T`. A ‘mutable reference’ allows you
113 to mutate the resource you’re borrowing. For example:
124 This will print `6`. We make `y` a mutable reference to `x`, then add one to
125 the thing `y` points at. You’ll notice that `x` had to be marked `mut` as well.
126 If it wasn’t, we couldn’t take a mutable borrow to an immutable value.
128 You'll also notice we added an asterisk (`*`) in front of `y`, making it `*y`,
129 this is because `y` is a `&mut` reference. You'll also need to use them for
130 accessing the contents of a reference as well.
132 Otherwise, `&mut` references are like references. There _is_ a large
133 difference between the two, and how they interact, though. You can tell
134 something is fishy in the above example, because we need that extra scope, with
135 the `{` and `}`. If we remove them, we get an error:
138 error: cannot borrow `x` as immutable because it is also borrowed as mutable
141 note: previous borrow of `x` occurs here; the mutable borrow prevents
142 subsequent moves, borrows, or modification of `x` until the borrow ends
145 note: previous borrow ends here
152 As it turns out, there are rules.
156 Here’s the rules about borrowing in Rust:
158 First, any borrow must last for a scope no greater than that of the owner.
159 Second, you may have one or the other of these two kinds of borrows, but not
160 both at the same time:
162 * one or more references (`&T`) to a resource,
163 * exactly one mutable reference (`&mut T`).
166 You may notice that this is very similar, though not exactly the same as,
167 to the definition of a data race:
169 > There is a ‘data race’ when two or more pointers access the same memory
170 > location at the same time, where at least one of them is writing, and the
171 > operations are not synchronized.
173 With references, you may have as many as you’d like, since none of them are
174 writing. However, as we can only have one `&mut` at a time, it is impossible to
175 have a data race. This is how Rust prevents data races at compile time: we’ll
176 get errors if we break the rules.
178 With this in mind, let’s consider our example again.
180 ## Thinking in scopes
193 This code gives us this error:
196 error: cannot borrow `x` as immutable because it is also borrowed as mutable
201 This is because we’ve violated the rules: we have a `&mut T` pointing to `x`,
202 and so we aren’t allowed to create any `&T`s. One or the other. The note
203 hints at how to think about this problem:
206 note: previous borrow ends here
213 In other words, the mutable borrow is held through the rest of our example. What
214 we want is for the mutable borrow to end _before_ we try to call `println!` and
215 make an immutable borrow. In Rust, borrowing is tied to the scope that the
216 borrow is valid for. And our scopes look like this:
221 let y = &mut x; // -+ &mut borrow of x starts here
225 println!("{}", x); // -+ - try to borrow x here
226 // -+ &mut borrow of x ends here
229 The scopes conflict: we can’t make an `&x` while `y` is in scope.
231 So when we add the curly braces:
237 let y = &mut x; // -+ &mut borrow starts here
239 } // -+ ... and ends here
241 println!("{}", x); // <- try to borrow x here
244 There’s no problem. Our mutable borrow goes out of scope before we create an
245 immutable one. But scope is the key to seeing how long a borrow lasts for.
247 ## Issues borrowing prevents
249 Why have these restrictive rules? Well, as we noted, these rules prevent data
250 races. What kinds of issues do data races cause? Here’s a few.
252 ### Iterator invalidation
254 One example is ‘iterator invalidation’, which happens when you try to mutate a
255 collection that you’re iterating over. Rust’s borrow checker prevents this from
259 let mut v = vec![1, 2, 3];
266 This prints out one through three. As we iterate through the vector, we’re
267 only given references to the elements. And `v` is itself borrowed as immutable,
268 which means we can’t change it while we’re iterating:
271 let mut v = vec![1, 2, 3];
282 error: cannot borrow `v` as mutable because it is also borrowed as immutable
285 note: previous borrow of `v` occurs here; the immutable borrow prevents
286 subsequent moves or mutable borrows of `v` until the borrow ends
289 note: previous borrow ends here
297 We can’t modify `v` because it’s borrowed by the loop.
301 References must not live longer than the resource they refer to. Rust will
302 check the scopes of your references to ensure that this is true.
304 If Rust didn’t check this property, we could accidentally use a reference
305 which was invalid. For example:
320 error: `x` does not live long enough
323 note: reference must be valid for the block suffix following statement 0 at
331 note: ...but borrowed value is only valid for the block suffix following
338 In other words, `y` is only valid for the scope where `x` exists. As soon as
339 `x` goes away, it becomes invalid to refer to it. As such, the error says that
340 the borrow ‘doesn’t live long enough’ because it’s not valid for the right
343 The same problem occurs when the reference is declared _before_ the variable it
344 refers to. This is because resources within the same scope are freed in the
345 opposite order they were declared:
358 error: `x` does not live long enough
361 note: reference must be valid for the block suffix following statement 0 at
370 note: ...but borrowed value is only valid for the block suffix following
379 In the above example, `y` is declared before `x`, meaning that `y` lives longer
380 than `x`, which is not allowed.