Rust 0.13.0-nightly
56d544f7a

A 30-minute Introduction to Rust

Rust is a modern systems programming language focusing on safety and speed. It accomplishes these goals by being memory safe without using garbage collection.

This introduction will give you a rough idea of what Rust is like, eliding many details. It does not require prior experience with systems programming, but you may find the syntax easier if you've used a 'curly brace' programming language before, like C or JavaScript. The concepts are more important than the syntax, so don't worry if you don't get every last detail: you can read the Guide to get a more complete explanation.

Because this is about high-level concepts, you don't need to actually install Rust to follow along. If you'd like to anyway, check out the homepage for explanation.

To show off Rust, let's talk about how easy it is to get started with Rust. Then, we'll talk about Rust's most interesting feature, ownership, and then discuss how it makes concurrency easier to reason about. Finally, we'll talk about how Rust breaks down the perceived dichotomy between speed and safety.

1 Tools

Getting started on a new Rust project is incredibly easy, thanks to Rust's package manager, Cargo.

To start a new project with Cargo, use cargo new:

$ cargo new hello_world --bin

We're passing --bin because we're making a binary program: if we were making a library, we'd leave it off.

Let's check out what Cargo has generated for us:

$ cd hello_world
$ tree .
.
├── Cargo.toml
└── src
    └── main.rs

1 directory, 2 files

This is all we need to get started. First, let's check out Cargo.toml:

[package]

name = "hello_world"
version = "0.0.1"
authors = ["Your Name <you@example.com>"]

This is called a manifest, and it contains all of the metadata that Cargo needs to compile your project.

Here's what's in src/main.rs:

fn main() { println!("Hello, world!") }
fn main() {
    println!("Hello, world!")
}

Cargo generated a 'hello world' for us. We'll talk more about the syntax here later, but that's what Rust code looks like! Let's compile and run it:

$ cargo run
   Compiling hello_world v0.0.1 (file:///Users/you/src/hello_world)
     Running `target/hello_world`
Hello, world!

Using an external dependency in Rust is incredibly easy. You add a line to your Cargo.toml:

[package]

name = "hello_world"
version = "0.0.1"
authors = ["Your Name <someone@example.com>"]

[dependencies.semver]

git = "https://github.com/rust-lang/semver.git"

You added the semver library, which parses version numbers and compares them according to the SemVer specification.

Now, you can pull in that library using extern crate in main.rs.

extern crate semver; use semver::Version; fn main() { assert!(Version::parse("1.2.3") == Ok(Version { major: 1u, minor: 2u, patch: 3u, pre: vec!(), build: vec!(), })); println!("Versions compared successfully!"); }
extern crate semver;

use semver::Version;

fn main() {
    assert!(Version::parse("1.2.3") == Ok(Version {
        major: 1u,
        minor: 2u,
        patch: 3u,
        pre: vec!(),
        build: vec!(),
    }));

    println!("Versions compared successfully!");
}

Again, we'll discuss the exact details of all of this syntax soon. For now, let's compile and run it:

$ cargo run
    Updating git repository `https://github.com/rust-lang/semver.git`
   Compiling semver v0.0.1 (https://github.com/rust-lang/semver.git#bf739419)
   Compiling hello_world v0.0.1 (file:///home/you/projects/hello_world)
     Running `target/hello_world`
Versions compared successfully!

Because we only specified a repository without a version, if someone else were to try out our project at a later date, when semver was updated, they would get a different, possibly incompatible version. To solve this problem, Cargo produces a file, Cargo.lock, which records the versions of any dependencies. This gives us repeatable builds.

There is a lot more here, and this is a whirlwind tour, but you should feel right at home if you've used tools like Bundler, npm, or pip. There's no Makefiles or endless autotools output here. (Rust's tooling does play nice with external libraries written in those tools, if you need to.)

Enough about tools, let's talk code!

2 Ownership

Rust's defining feature is 'memory safety without garbage collection.' Let's take a moment to talk about what that means. Memory safety means that the programming language eliminates certain kinds of bugs, such as buffer overflows and dangling pointers. These problems occur when you have unrestricted access to memory. As an example, here's some Ruby code:

v = [];

v.push("Hello");

x = v[0];

v.push("world");

puts x

We make an array, v, and then call push on it. push is a method which adds an element to the end of an array.

Next, we make a new variable, x, that's equal to the first element of the array. Simple, but this is where the 'bug' will appear.

Let's keep going. We then call push again, pushing "world" onto the end of the array. v now is ["Hello", "world"].

Finally, we print x with the puts method. This prints "Hello."

All good? Let's go over a similar, but subtly different example, in C++:

#include<iostream>
#include<vector>
#include<string>

int main() {
    std::vector<std::string> v;

    v.push_back("Hello");

    std::string& x = v[0];

    v.push_back("world");

    std::cout << x;
}

It's a little more verbose due to the static typing, but it's almost the same thing. We make a std::vector of std::strings, we call push_back (same as push) on it, take a reference to the first element of the vector, call push_back again, and then print out the reference.

There's two big differences here: one, they're not exactly the same thing, and two...

$ g++ hello.cpp -Wall -Werror
$ ./a.out 
Segmentation fault (core dumped)

A crash! (Note that this is actually system-dependent. Because referring to an invalid reference is undefined behavior, the compiler can do anything, including the right thing!) Even though we compiled with flags to give us as many warnings as possible, and to treat those warnings as errors, we got no errors. When we ran the program, it crashed.

Why does this happen? When we prepend to an array, its length changes. Since its length changes, we may need to allocate more memory. In Ruby, this happens as well, we just don't think about it very often. So why does the C++ version segfault when we allocate more memory?

The answer is that in the C++ version, x is a reference to the memory location where the first element of the array is stored. But in Ruby, x is a standalone value, not connected to the underyling array at all. Let's dig into the details for a moment. Your program has access to memory, provided to it by the operating system. Each location in memory has an address. So when we make our vector, v, it's stored in a memory location somewhere:

location name value
0x30 v

(Address numbers made up, and in hexadecimal. Those of you with deep C++ knowledge, there are some simplifications going on here, like the lack of an allocated length for the vector. This is an introduction.)

When we push our first string onto the array, we allocate some memory, and v refers to it:

location name value
0x30 v 0x18
0x18 "Hello"

We then make a reference to that first element. A reference is a variable that points to a memory location, so its value is the memory location of the "Hello" string:

location name value
0x30 v 0x18
0x18 "Hello"
0x14 x 0x18

When we push "world" onto the vector with push_back, there's no room: we only allocated one element. So, we need to allocate two elements, copy the "Hello" string over, and update the reference. Like this:

location name value
0x30 v 0x08
0x18 GARBAGE
0x14 x 0x18
0x08 "Hello"
0x04 "world"

Note that v now refers to the new list, which has two elements. It's all good. But our x didn't get updated! It still points at the old location, which isn't valid anymore. In fact, the documentation for push_back mentions this:

If the new size() is greater than capacity() then all iterators and references (including the past-the-end iterator) are invalidated.

Finding where these iterators and references are is a difficult problem, and even in this simple case, g++ can't help us here. While the bug is obvious in this case, in real code, it can be difficult to track down the source of the error.

Before we talk about this solution, why didn't our Ruby code have this problem? The semantics are a little more complicated, and explaining Ruby's internals is out of the scope of a guide to Rust. But in a nutshell, Ruby's garbage collector keeps track of references, and makes sure that everything works as you might expect. This comes at an efficiency cost, and the internals are more complex. If you'd really like to dig into the details, this article can give you more information.

Garbage collection is a valid approach to memory safety, but Rust chooses a different path. Let's examine what the Rust version of this looks like:

fn main() { let mut v = vec![]; v.push("Hello"); let x = &v[0]; v.push("world"); println!("{}", x); }
fn main() {
    let mut v = vec![];

    v.push("Hello");

    let x = &v[0];

    v.push("world");

    println!("{}", x);
}

This looks like a bit of both: fewer type annotations, but we do create new variables with let. The method name is push, some other stuff is different, but it's pretty close. So what happens when we compile this code? Does Rust print "Hello", or does Rust crash?

Neither. It refuses to compile:

$ cargo run
   Compiling hello_world v0.0.1 (file:///Users/you/src/hello_world)
main.rs:8:5: 8:6 error: cannot borrow `v` as mutable because it is also borrowed as immutable
main.rs:8     v.push("world");
              ^
main.rs:6:14: 6:15 note: previous borrow of `v` occurs here; the immutable borrow prevents subsequent moves or mutable borrows of `v` until the borrow ends
main.rs:6     let x = &v[0];
                       ^
main.rs:11:2: 11:2 note: previous borrow ends here
main.rs:1 fn main() {
...
main.rs:11 }
           ^
error: aborting due to previous error

When we try to mutate the array by pushing it the second time, Rust throws an error. It says that we "cannot borrow v as mutable because it is also borrowed as immutable." What's up with "borrowed"?

In Rust, the type system encodes the notion of ownership. The variable v is an "owner" of the vector. When we make a reference to v, we let that variable (in this case, x) 'borrow' it for a while. Just like if you own a book, and you lend it to me, I'm borrowing the book.

So, when I try to modify the vector with the second call to push, I need to be owning it. But x is borrowing it. You can't modify something that you've lent to someone. And so Rust throws an error.

So how do we fix this problem? Well, we can make a copy of the element:

fn main() { let mut v = vec![]; v.push("Hello"); let x = v[0].clone(); v.push("world"); println!("{}", x); }
fn main() {
    let mut v = vec![];

    v.push("Hello");

    let x = v[0].clone();

    v.push("world");

    println!("{}", x);
}

Note the addition of clone(). This creates a copy of the element, leaving the original untouched. Now, we no longer have two references to the same memory, and so the compiler is happy. Let's give that a try:

$ cargo run
   Compiling hello_world v0.0.1 (file:///Users/you/src/hello_world)
     Running `target/hello_world`
Hello

Same result. Now, making a copy can be inefficient, so this solution may not be acceptable. There are other ways to get around this problem, but this is a toy example, and because we're in an introduction, we'll leave that for later.

The point is, the Rust compiler and its notion of ownership has saved us from a bug that would crash the program. We've achieved safety, at compile time, without needing to rely on a garbage collector to handle our memory.

3 Concurrency

Rust's ownership model can help in other ways, as well. For example, take concurrency. Concurrency is a big topic, and an important one for any modern programming language. Let's take a look at how ownership can help you write safe concurrent programs.

Here's an example of a concurrent Rust program:

fn main() { for _ in range(0u, 10u) { spawn(proc() { println!("Hello, world!"); }); } }
fn main() {
    for _ in range(0u, 10u) {
        spawn(proc() {
            println!("Hello, world!");
        });
    }
}

This program creates ten threads, who all print Hello, world!. The spawn function takes one argument, a proc. 'proc' is short for 'procedure,' and is a form of closure. This closure is executed in a new thread, created by spawn itself.

One common form of problem in concurrent programs is a 'data race.' This occurs when two different threads attempt to access the same location in memory in a non-synchronized way, where at least one of them is a write. If one thread is attempting to read, and one thread is attempting to write, you cannot be sure that your data will not be corrupted. Note the first half of that requirement: two threads that attempt to access the same location in memory. Rust's ownership model can track which pointers own which memory locations, which solves this problem.

Let's see an example. This Rust code will not compile:

fn main() { let mut numbers = vec![1i, 2i, 3i]; for i in range(0u, 3u) { spawn(proc() { for j in range(0, 3) { numbers[j] += 1 } }); } }
fn main() {
    let mut numbers = vec![1i, 2i, 3i];

    for i in range(0u, 3u) {
        spawn(proc() {
            for j in range(0, 3) { numbers[j] += 1 }
        });
    }
}

It gives us this error:

6:71 error: capture of moved value: `numbers`
    for j in range(0, 3) { numbers[j] += 1 }
               ^~~~~~~
7:50 note: `numbers` moved into closure environment here because it has type `proc():Send`, which is non-copyable (perhaps you meant to use clone()?)
    spawn(proc() {
        for j in range(0, 3) { numbers[j] += 1 }
    });
6:79 error: cannot assign to immutable dereference (dereference is implicit, due to indexing)
        for j in range(0, 3) { numbers[j] += 1 }
                           ^~~~~~~~~~~~~~~

It mentions that "numbers moved into closure environment". Because we referred to numbers inside of our proc, and we create three procs, we would have three references. Rust detects this and gives us the error: we claim that numbers has ownership, but our code tries to make three owners. This may cause a safety problem, so Rust disallows it.

What to do here? Rust has two types that helps us: Arc<T> and Mutex<T>. "Arc" stands for "atomically reference counted." In other words, an Arc will keep track of the number of references to something, and not free the associated resource until the count is zero. The 'atomic' portion refers to an Arc's usage of concurrency primitives to atomically update the count, making it safe across threads. If we use an Arc, we can have our three references. But, an Arc does not allow mutable borrows of the data it holds, and we want to modify what we're sharing. In this case, we can use a Mutex<T> inside of our Arc. A Mutex will synchronize our accesses, so that we can ensure that our mutation doesn't cause a data race.

Here's what using an Arc with a Mutex looks like:

use std::sync::{Arc,Mutex}; fn main() { let numbers = Arc::new(Mutex::new(vec![1i, 2i, 3i])); for i in range(0u, 3u) { let number = numbers.clone(); spawn(proc() { let mut array = number.lock(); (*array)[i] += 1; println!("numbers[{}] is {}", i, (*array)[i]); }); } }
use std::sync::{Arc,Mutex};

fn main() {
    let numbers = Arc::new(Mutex::new(vec![1i, 2i, 3i]));

    for i in range(0u, 3u) {
        let number = numbers.clone();
        spawn(proc() {
            let mut array = number.lock();

            (*array)[i] += 1;

            println!("numbers[{}] is {}", i, (*array)[i]);
        });
    }
}

We first have to use the appropriate library, and then we wrap our vector in an Arc with the call to Arc::new(). Inside of the loop, we make a new reference to the Arc with the clone() method. This will increment the reference count. When each new numbers variable binding goes out of scope, it will decrement the count. The lock() call will return us a reference to the value inside the Mutex, and block any other calls to lock() until said reference goes out of scope.

We can compile and run this program without error, and in fact, see the non-deterministic aspect:

$ cargo run
   Compiling hello_world v0.0.1 (file:///Users/you/src/hello_world)
     Running `target/hello_world`
numbers[1] is 2
numbers[0] is 1
numbers[2] is 3
$ cargo run
     Running `target/hello_world`
numbers[2] is 3
numbers[1] is 2
numbers[0] is 1

Each time, we get a slightly different output, because each thread works in a different order. You may not get the same output as this sample, even.

The important part here is that the Rust compiler was able to use ownership to give us assurance at compile time that we weren't doing something incorrect with regards to concurrency. In order to share ownership, we were forced to be explicit and use a mechanism to ensure that it would be properly handled.

4 Safety and speed

Safety and speed are always presented as a continuum. On one hand, you have maximum speed, but no safety. On the other, you have absolute safety, with no speed. Rust seeks to break out of this mode by introducing safety at compile time, ensuring that you haven't done anything wrong, while compiling to the same low-level code you'd expect without the safety.

As an example, Rust's ownership system is entirely at compile time. The safety check that makes this an error about moved values:

fn main() { let vec = vec![1i, 2, 3]; for i in range(1u, 3) { spawn(proc() { println!("{}", vec[i]); }); } }
fn main() {
    let vec = vec![1i, 2, 3];

    for i in range(1u, 3) {
        spawn(proc() {
            println!("{}", vec[i]);
        });
    }
}

carries no runtime penalty. And while some of Rust's safety features do have a run-time cost, there's often a way to write your code in such a way that you can remove it. As an example, this is a poor way to iterate through a vector:

fn main() { let vec = vec![1i, 2, 3]; for i in range(1u, vec.len()) { println!("{}", vec[i]); } }
let vec = vec![1i, 2, 3];

for i in range(1u, vec.len()) {
     println!("{}", vec[i]);
}

The reason is that the access of vec[i] does bounds checking, to ensure that we don't try to access an invalid index. However, we can remove this while retaining safety. The answer is iterators:

fn main() { let vec = vec![1i, 2, 3]; for x in vec.iter() { println!("{}", x); } }
let vec = vec![1i, 2, 3];

for x in vec.iter() {
    println!("{}", x);
}

This version uses an iterator that yields each element of the vector in turn. Because we have a reference to the element, rather than the whole vector itself, there's no array access bounds to check.

5 Learning More

I hope that this taste of Rust has given you an idea if Rust is the right language for you. We talked about Rust's tooling, how encoding ownership into the type system helps you find bugs, how Rust can help you write correct concurrent code, and how you don't have to pay a speed cost for much of this safety.

To continue your Rustic education, read the guide for a more in-depth exploration of Rust's syntax and concepts.