Rust doesn't have a garbage collector. At first, this was a bit shocking to me. Garbage collection is usually one of those things that most programming languages implement for memory efficiency. But Rust is memory efficient and Rust code is considered safe. So how does Rust manage memory?
Most languages run a garbage collector program to reclaim memory that is no longer used. It allows the programmer to not worry about managing unused memory (like in C). Instead, it periodically goes through the allocated memory, checks which parts are not referenced, and reclaims them. When a program runs the garbage collector is unique to the language and where the code is running. For example, JavaScript implements multiple strategies on when to run garbage collection.
Rust, however, deals with this problem in an entirely different way. Rust has a feature called the Ownership model. It's one of Rust's most unique features and the reason that Rust eliminates a whole class of memory-related issues. In this article, we'll dive deep into the ownership model.
What is ownership?
Memory in Rust is managed through a system that keeps track of who owns which piece of the memory. It comes with a set of rules that Rust can check at compile time rather than at run-time. This means that any potential memory management mistakes we're likely to make, are caught at compile time. While Rust does this on a general level and not just for memory management, combining ownership and compile-time checks saves the programmer a lot of headaches by simply eliminating mistakes.
The ownership rules are simple:
- Each value in Rust has an owner
- There can only be one owner at a time
- When the owner goes out of scope, the value is dropped and Rust reclaims that memory
This might seem restrictive. For example, what happens when you pass a variable to another function as a parameter? What if you pass by reference? What happens to return values? We'll answer these questions and go into detail on how Rust manages ownership of memory.
Stacks and Heaps
Rust uses stacks and heaps to store values. Here's a quick summary of stacks and heaps:
Stacks store values in the order it gets them and is sequential. It removes it in the opposite order. This is called Last In, First Out (LIFO). Adding data is called pushing onto the stack and removing data is called popping off the stack. Each value stored in a stack should have a known and fixed size. In Rust, the size must be known at compile time. Data types that use the stack are numbers, boolean, characters, and string literals (In Rust, string literals are hardcoded strings, which is why their size is known. Since variables in Rust are immutable by default, the string literal's value cannot be changed. This is often inconvenient but it's there when needed.). Adding and removing data from stacks is fast and easy.
Heaps store values whose size is unknown at compile time or can change at runtime. Heaps are less organized than stacks. In Rust, when using a data type that uses heaps, we request a certain amount of memory. Rust finds a place in the heap that can accommodate our request and returns a pointer (which is the address of the location). The pointer itself has a known and fixed size so it can be stored on the stack. When we want the data stored at the location we follow the pointer. Adding data from heaps is slower because we have to search for a suitable place first and then keep a reference to that piece of memory. Data types that use the heap are the String type, arrays, objects.
Why do we need to know stacks and heaps? Solving the problems with storing data in heaps is the reason ownership exists. Where a value is stored helps us understand the ownership rules and the compile-time errors that Rust will eventually throw at us when we make mistakes.
Now that we know how data is stored, let's look at a few code samples and see how ownership works.
Variable Scope
Let's take the following snippet of code:
let a = 1;
{
let b = 2;
println!("{}", b); // Prints value of b
}
println!("{}", b); // Throws an error because the scope of b ends with the closing curly brace
The result is fairly intuitive. Since the scope of b ends with the closing curly brace, we can no longer access it. When a variable does out of scope, Rust will call an in-built drop function that will destroy the variable and reclaim the memory.
Shallow Copy
Let's try something a bit different from the above snippet.
let a = "I love Rust";
let b = a;
println!("{}", a);
println!("{}", b);
What do you think the output will be? If you guessed that both will print "I love Rust" you are correct. But let's take a closer look at what's happening here. When we say let b = a, we're telling Rust to assign the value of a to b. Internally, Rust is making a shallow copy of a and assigning the copy to b. It's also worth noting that "I love Rust" is a string literal. Since variables in Rust are immutable by default, this value cannot be changed at run-time and their size is known at compile-time. That is why the values here are stored in a stack. This copying behavior only applies to values stored in the stack.
Move
Consider this code:
let a = String::from("I love Rust");
let b = a;
println!("{}", a);
println!("{}", b);
This looks very similar to the previous snippet but behaves in a very different way. The key here is the String::from("I love Rust") part. Instead of the stack, this makes Rust store the value in the heap. It's still immutable but how the value is stored is different. a instead of holding the value "I love Rust" now holds a pointer to the heap of memory holding the string.
Another interesting this to note here is that this snippet will throw an error. When we say let b = a, the ownership of the value "moves" to b and we can no longer access a. So the line where we print the value of a will throw an error. This seems weird and is different from what other programming languages do. Let's say we're copying the pointer to b instead of moving the ownership. Both variables will hold the pointer to the value. But when they both go out of scope, they will try to free the same piece of memory. This is called a double free error and is one of those memory safety bugs that we talked about earlier. Bugs like this can cause security vulnerabilities. That is why Rust won't let us access a and moves its ownership of the value to b instead of letting us copy it.
As a side note, let's take a look at the error Rust throws:
$ cargo run
Compiling ownership v0.1.0 (file:///projects/ownership)
error[E0382]: borrow of moved value: `a`
--> src/main.rs:5:28
|
2 | let a = String::from("I love Rust");
| -- move occurs because `a` has type `String`, which does not implement the `Copy` trait
3 | let b = a;
| -- value moved here
4 |
5 | println!("{}, world!", a);
| ^^ value borrowed here after move
error: aborting due to previous error
For more information about this error, try `rustc --explain E0382`.
error: could not compile `ownership`
To learn more, run the command again with --verbose.
See how helpful the error messages are? It clearly states that the value moved here when we assign a to b. The message also tells us that the type String does not implement a Copy trait. This Copy trait is what allows values in stack memory to be copied. Data types which store values in the stack implement this trait but types using the heap don't. These helpful messages are one of the reasons I'm loving Rust so much.
Clone
Let's say in the previous code we did want to copy the value to b. Rust lets us make a copy of the value stored in the heap by using the clone method.
let a = String::from("I love Rust");
let b = a.clone();
println!("{}", a);
println!("{}", b);
This snippet won't throw an error. Instead, it'll create another piece on the heap, store the "I love Rust" string in it, and return the pointer. This way, Rust forces us to be conscious of our memory use by explicitly stating that we're making a deep copy.
The Copy Trait
Data types that implement the Copy trait will let us copy values that are stored on the stack. As a general rule, any scalar data types implement the Copy trait, which includes:
- All integer types
- Booleans
- Floating points
- Characters
- Tuples, only if they contain types that implement the
Copytrait.
Functions
Let's take a look at functions and how they handle ownership. For the most part, functions will behave like variables. Values in stacks will be copied when passed as arguments to functions.
fn main() {
let a = 10;
makes_a_copy(a);
println!("{}", a); // Will print the value of a
}
fn makes_a_copy(some_int: u32) {
// Will print the value of some_int which is a copy of the value passed to this function.
println!("{}", some_int);
}
Ownership of values in the heap will be moved to the called function:
fn main() {
let a = String::from("I am Ferris, your Rust spirit guide");
moves_here(a);
println!("{}", a);
// Throws an error because ownership moved to moves_here
// function and a is no longer accessible
}
fn moves_here(some_str: String) {
println!("{}", some_str); // Prints: I am Ferris, your Rust spirit guide.
} // Rust will call drop here and the memory will be reclaimed.
Return Values
You can return values from functions that works just like passing arguments to functions. Ownership of the return values follows the same rules as function arguments.
fn main() {
let a = String::from("I am Ferris, your Rust spirit guide");
let b = moves_here(a);
println!("{}", b);
// a is no longer accessible but the same value
// is now stored in b which can be accessed
}
fn moves_here(my_string: String) -> String {
println!("{}", my_string); // Prints: I am Ferris, your Rust spirit guide.
my_string
}
This will work for some cases but becomes tedious when we want to keep using the value even after passing it to a function? Passing and returning becomes tedious and leads to a lot of boilerplate code. What if we want to pass a value to a function but not transfer ownership? Rust lets us reference a piece of memory without transferring ownership. If you've learned C before, this is similar to passing arguments by reference. But it does come with a few rules.
References
Let's modify our previous code snippet so we can keep using our value after passing it to another function.
fn main() {
let a = String::from("I am Ferris, your Rust spirit guide");
borrows_here(&a);
println!("{}", a);
// a is still accessible because we didn't transfer ownership.
// We only passed a reference to a.
}
fn borrows_here(my_string: &String) {
println!("{}", my_string); // Prints: I am Ferris, your Rust spirit guide.
}
// Scope of my_string ends here but the value
// it points to is still in memory because
// my_string is only a reference to that piece of memory
When we pass a reference to a function, the pointer to the value is stored in the argument. Remember that pointers can be stored in stacks. In the above snippet, my_string stores the pointer to our original variable a. When borrows_here finishes execution, my_string will be dropped but the value it pointed to is still in memory. This is known as borrowing.
This leads to a question: Can we mutate/change a value that's been borrowed? The answer is no. Rust will throw and error when we try to mutate a borrowed value even when the original variable is mutable.
fn main() {
let mut a = String::from("I am Ferris");
borrows_here(&a);
println!("{}", a);
// a is still accessible because we didn't transfer ownership.
// We only passed a reference to a.
}
fn borrows_here(my_string: &String) {
my_string.push_str(", your Rust spirit guide.");
// This will throw an error
}
We can let Rust know that a reference is mutable by making a small change to our code.
fn main() {
let mut a = String::from("I am Ferris");
borrows_here(&mut a);
println!("{}", a);
// a is still accessible because we didn't transfer ownership.
// We only passed a reference to a.
}
fn borrows_here(my_string: &String) {
my_string.push_str(", your Rust spirit guide.");
// This will work
}
This is called a mutable reference and let's us change borrowed values.
Like I mentioned before, there are certain rules when referencing values:
- We can have only one mutable reference to a value in the same scope. We can have multiple mutable references by creating limited scopes using
{}but we cannot have simultaneous mutable references. - Mutable and Immutable references cannot coexist in the same scope
- Multiple immutable references are allowed
The reason for not allowing multiple mutable references is to eliminate data race conditions. It's a state where multiple references has access to a value and can independently change the values. This sometimes leads to bugs because you don't have the value you're expecting. Race conditions are difficult to debug. Rust eliminates this at compile time!
Similarly, when you're using an immutable reference along side a mutable one, we don't want a value to unexpectedly change while you're reading from it.
Another mistake Rust prevents us from making is to catch dangling pointers at compile-time. A dangling pointer is a pointer that references a value that doesn't exist anymore.
fn dangling_pointer() -> &String {
let a = String::from("I won't exist beyond the scope of this function");
&a
// We're trying to return a reference to a but after this line,
// a will be dropped. Rust will throw an error here.
}
Rust's Ownership model is a difficult concept to understand for newcomers. I initially had a hard time understanding the rules and facing some of the errors were annoying. But think about it this way. The rules and errors are there to prevent us from making mistakes, even though they may seem restrictive. Every error thrown by the ownership rules is a future bug prevented.
I hope I helped in your understanding of the ownership rules in Rust. I'm having fun learning Rust, however slow the process may be.