Rust 0.10
46867cc3
Rust's pointers are one of its more unique and compelling features. Pointers are also one of the more confusing topics for newcomers to Rust. They can also be confusing for people coming from other languages that support pointers, such as C++. This guide will help you understand this important topic.
I have good news for you: you probably don't need to care about pointers, especially as you're getting started. Think of it this way: Rust is a language that emphasizes safety. Pointers, as the joke goes, are very pointy: it's easy to accidentally stab yourself. Therefore, Rust is made in a way such that you don't need them very often.
"But guide!" you may cry. "My co-worker wrote a function that looks like this:
fn succ(x: &int) -> int { *x + 1 }
So I wrote this code to try it out:
fn main() { let number = 5; let succ_number = succ(number); println!("{}", succ_number); }
And now I get an error:
error: mismatched types: expected `&int` but found `<VI0>` (expected &-ptr but found integral variable)What gives? It needs a pointer! Therefore I have to use pointers!"
Turns out, you don't. All you need is a reference. Try this on for size:
fn main() { let number = 5; let succ_number = succ(&number); println!("{}", succ_number); }
It's that easy! One extra little & there. This code will run, and print 6.
That's all you need to know. Your co-worker could have written the function like this:
fn succ(x: int) -> int { x + 1 } fn main() { let number = 5; let succ_number = succ(number); println!("{}", succ_number); }
No pointers even needed. Then again, this is a simple example. I assume that
your real-world succ function is more complicated, and maybe your co-worker
had a good reason for x to be a pointer of some kind. In that case, references
are your best friend. Don't worry about it, life is too short.
However.
Here are the use-cases for pointers. I've prefixed them with the name of the pointer that satisfies that use-case:
Five exceptions. That's it. Otherwise, you shouldn't need them. Be sceptical of pointers in Rust: use them for a deliberate purpose, not just to make the compiler happy.
If you're coming to Rust from a language like C or C++, you may be used to passing things by reference, or passing things by pointer. In some languages, like Java, you can't even have objects without a pointer to them. Therefore, if you were writing this Rust code:
struct Point { x: int, y: int, } fn main() { let p0 = Point { x: 5, y: 10}; let p1 = transform(p0); println!("{:?}", p1); }
I think you'd implement transform like this:
fn transform(p: &Point) -> Point { Point { x: p.x + 1, y: p.y + 1} } // and change this: let p1 = transform(&p0);
This does work, but you don't need to create those references! The better way to write this is simply:
struct Point { x: int, y: int, } fn transform(p: Point) -> Point { Point { x: p.x + 1, y: p.y + 1} } fn main() { let p0 = Point { x: 5, y: 10}; let p1 = transform(p0); println!("{:?}", p1); }
But won't this be inefficient? Well, that's a complicated question, but it's important to know that Rust, like C and C++, store aggregate data types 'unboxed,' whereas languages like Java and Ruby store these types as 'boxed.' For smaller structs, this way will be more efficient. For larger ones, it may be less so. But don't reach for that pointer until you must! Make sure that the struct is large enough by performing some tests before you add in the complexity of pointers.
Owned pointers are the conceptually simplest kind of pointer in Rust. A rough approximation of owned pointers follows:
These two properties make for three use cases.
Traits must be referenced through a pointer, because the struct that implements the trait may be a different size than a different struct that implements the trait. Therefore, unboxed traits don't make any sense, and aren't allowed.
Sometimes, you need a recursive data structure. The simplest is known as a 'cons list':
enum List<T> { Nil, Cons(T, ~List<T>), } fn main() { let list: List<int> = Cons(1, ~Cons(2, ~Cons(3, ~Nil))); println!("{:?}", list); }
This prints:
Cons(1, ~Cons(2, ~Cons(3, ~Nil)))The inner lists must be an owned pointer, because we can't know how many elements are in the list. Without knowing the length, we don't know the size, and therefore require the indirection that pointers offer.
This should almost never be a concern, but because creating an owned pointer boxes its value, it therefore makes referring to the value the size of the box. This may make passing an owned pointer to a function less expensive than passing the value itself. Don't worry yourself with this case until you've proved that it's an issue through benchmarks.
For example, this will work:
struct Point { x: int, y: int, } fn main() { let a = Point { x: 10, y: 20 }; spawn(proc() { println!("{}", a.x); }); }
This struct is tiny, so it's fine. If Point were large, this would be more
efficient:
struct Point { x: int, y: int, } fn main() { let a = ~Point { x: 10, y: 20 }; spawn(proc() { println!("{}", a.x); }); }
Now it'll be copying a pointer-sized chunk of memory rather than the whole struct.
Note: the
@form of managed pointers is deprecated and behind a feature gate (it requires a#[feature(managed_pointers)];attribute on the crate root; remember the semicolon!). There are replacements, currently there isstd::rc::Rcandstd::gc::Gcfor shared ownership via reference counting and garbage collection respectively.
Managed pointers, notated by an @, are used when having a single owner for
some data isn't convenient or possible. This generally happens when your
program is very large and complicated.
For example, let's say you're using an owned pointer, and you want to do this:
struct Point { x: int, y: int, } fn main() { let a = ~Point { x: 10, y: 20 }; let b = a; println!("{}", b.x); println!("{}", a.x); }
You'll get this error:
test.rs:10:20: 10:21 error: use of moved value: `a`
test.rs:10 println!("{}", a.x);
^
note: in expansion of format_args!
<std-macros>:158:27: 158:81 note: expansion site
<std-macros>:157:5: 159:6 note: in expansion of println!
test.rs:10:5: 10:25 note: expansion site
test.rs:8:9: 8:10 note: `a` moved here because it has type `~Point`, which is moved by default (use `ref` to override)
test.rs:8 let b = a;
^As the message says, owned pointers only allow for one owner at a time. When you assign a to b, a becomes invalid. Change your code to this, however:
struct Point { x: int, y: int, } fn main() { let a = @Point { x: 10, y: 20 }; let b = a; println!("{}", b.x); println!("{}", a.x); }
And it works:
10
10So why not just use managed pointers everywhere? There are two big drawbacks to managed pointers:
References are the third major kind of pointer Rust supports. They are simultaneously the simplest and the most complicated kind. Let me explain: references are considered 'borrowed' because they claim no ownership over the data they're pointing to. They're just borrowing it for a while. So in that sense, they're simple: just keep whatever ownership the data already has. For example:
use std::num::sqrt; struct Point { x: f32, y: f32, } fn compute_distance(p1: &Point, p2: &Point) -> f32 { let x_d = p1.x - p2.x; let y_d = p1.y - p2.y; sqrt(x_d * x_d + y_d * y_d) } fn main() { let origin = @Point { x: 0.0, y: 0.0 }; let p1 = ~Point { x: 5.0, y: 3.0 }; println!("{:?}", compute_distance(origin, p1)); }
This prints 5.83095189. You can see that the compute_distance function
takes in two references, but we give it a managed and unique pointer. Of
course, if this were a real program, we wouldn't have any of these pointers,
they're just there to demonstrate the concepts.
So how is this hard? Well, because we're ignoring ownership, the compiler needs to take great care to make sure that everything is safe. Despite their complete safety, a reference's representation at runtime is the same as that of an ordinary pointer in a C program. They introduce zero overhead. The compiler does all safety checks at compile time.
This theory is called 'region pointers,' and involve a concept called 'lifetimes'. Here's the simple explanation: would you expect this code to compile?
fn main() { println!("{}", x); let x = 5; }
Probably not. That's because you know that the name x is valid from where
it's declared to when it goes out of scope. In this case, that's the end of
the main function. So you know this code will cause an error. We call this
duration a 'lifetime'. Let's try a more complex example:
fn main() { let mut x = ~5; if *x < 10 { let y = &x; println!("Oh no: {:?}", y); return; } *x -= 1; println!("Oh no: {:?}", x); }
Here, we're borrowing a pointer to x inside of the if. The compiler, however,
is able to determine that that pointer will go out of scope without x being
mutated, and therefore, lets us pass. This wouldn't work:
fn main() { let mut x = ~5; if *x < 10 { let y = &x; *x -= 1; println!("Oh no: {:?}", y); return; } *x -= 1; println!("Oh no: {:?}", x); }
It gives this error:
test.rs:5:8: 5:10 error: cannot assign to `*x` because it is borrowed
test.rs:5 *x -= 1;
^~
test.rs:4:16: 4:18 note: borrow of `*x` occurs here
test.rs:4 let y = &x;
^~As you might guess, this kind of analysis is complex for a human, and therefore hard for a computer, too! There is an entire guide devoted to references and lifetimes that goes into lifetimes in great detail, so if you want the full details, check that out.
We've talked a lot about functions that accept various kinds of pointers, but what about returning them? Here's the rule of thumb: only return a unique or managed pointer if you were given one in the first place.
What does that mean? Don't do this:
fn foo(x: ~int) -> ~int { return ~*x; } fn main() { let x = ~5; let y = foo(x); }
Do this:
fn foo(x: ~int) -> int { return *x; } fn main() { let x = ~5; let y = ~foo(x); }
This gives you flexibility, without sacrificing performance. For example, this will also work:
fn foo(x: ~int) -> int { return *x; } fn main() { let x = ~5; let y = @foo(x); }
You may think that this gives us terrible performance: return a value and then
immediately box it up?!?! Isn't that the worst of both worlds? Rust is smarter
than that. There is no copy in this code. main allocates enough room for the
@int, passes a pointer to that memory into foo as x, and then foo writes
the value straight into that pointer. This writes the return value directly into
the allocated box.
This is important enough that it bears repeating: pointers are not for optimizing returning values from your code. Allow the caller to choose how they want to use your output.