Rust's Memory Management

Note: I accidentally published this. I won't take it down, but it is incomplete and I do not guarantee its correctness.

Note 2: I don't think this post is necessary anymore, as the tutorial has been updated significantly, and is quite understandable. As such I don't plan on finishing this post. Email me if you'd like to see it finished.

Rust seems to intimidate newcomers, as its memory model is fairly complex. I think part of the problem is that the language tutorial introduces the memory model by feature. Rather, I'll introduce it by concept, showing examples of code that breaks memory safety in C, and how Rust's memory model prevents the error. Hopefully I can convince you that Rust isn't as complex as it looks, and that the extra syntax is well-worth the zero-cost memory safety. I'll be comparing Rust code to equivalent C idioms using the zeromq library because it has a very clean API in both languages. As such, I assume basic familiarity with C. Rust is not a very suitable language for new programmers, and neither is this tutorial.

The core concepts driving Rust's memory management are, in reverse order of simplicity, ownership, mutability, and lifetime. It's not that other languages don't have them, it's just that they're implicit or convention. I won't be using many fancy features of Rust here. They aren't needed to explain the memory management, variables and functions suffice.


Ownership is simply who is responsibile for freeing an object. If you own an object, you are ensured it is valid.

An example in C would be:

void *context = zmq_ctx_new(); // zeromq owns the context, you just get to use it
// ...
zmq_ctx_destroy(context); // you need to tell zeromq to free it, you aren't allowed to do so

In Rust:

let context: ~Context = zmq_ctx_new();
// ....
// context implicitly freed when no one owns it anymore

Notice that the C uses an opaque void* and that Rust uses the explicitly-named ~Context. In C, you often use opaque types because the language offers no control over visibility of struct members. In Rust, you can mark the fields of a struct priv, which will disallow users of the library from using those fields. This is a bit nuanced, and I'll explain it in a later tutorial on privacy and the module system.

The tilde means "owned pointer", in that if you have an owned pointer to an object, you are the owner. Creating an owned pointer involves a heap allocation (malloc). I use an owned pointer in these examples to mirror the C, but idiomatic Rust uses values on the stack much more frequently than values on the heap.

Taking ownership

Rust goes one step further with ownership. The compiler asserts that an object cannot be owned multiple times. An object can be owned by a stack frame (locals and function arguments), a scope, or a struct. For lack of a better term, I'll call these "owners." You transfer ownership by "moving" into an owner. Once an owner loses ownership, the compiler will error if you try to use the object again through that owner. This disallows dangling pointers and prevents an entire class of error (and potentially vulnerability), "use after free."

In C, it's perfectly valid to do:

void *context = zmq_ctx_new();
use_context(context); // compiles fine, but clearly incorrect

Whereas in Rust:

let context: ~Context = zmq_ctx_new();
{ // introduce a scope
    let context2: ~Context = context; // context moved into the variable in this scope
} // context2 freed at scope close
use_context(context); // error: use of moved value: `context`

Not taking ownership

Most functions that take arguments don't need to take ownership. To express this, Rust has the "borrowed pointer." For example, in C you would write this:

int length(some_large_struct_t *datum) {
    return datum->x - datum->padding;

The function does not care who owns the object, it doesn't need to free it, as there's no reason the object can't be used again after this computation. In Rust, it's:

fn length(datum: &some_large_struct) -> int {
    datum.x - datum.padding

The first thing you'll notice is that there is no dereference to be found in the source. In C, -> is the dereferencing struct member extraction operator. In Rust, . will dereference if it needs to. You'll also notice that Rust uses & rather than * to indicate this type of a pointer. It's the same operator used to take a pointer (address-of), as in C. C, however, doesn't enforce the non-transfer of ownership. It's perfectly valid to do:

int length(some_large_struct_t *datum) {
    int l = datum->x - datum->padding;
    return l;

length has now incorrectly taken ownership of datum, by freeing it. Any code which had a pointer to datum has now been rendered incorrect. In Rust, there is no way to safely free memory. It is automatically free'd.

Multiple ownership

There are situations when ownership isn't so black and white, and for that Rust has the "managed pointer." The type of a managed pointer is @'static T. The 'static means "this type cannot contain any borrowed pointers." The shorthand for @'static T is just @T: it's impossible to have a managed pointer to a type that doesn't fulfill 'static. Using only owned and borrowed pointers, ownership forms a DAG. Managed pointers allow cycles. They are GC-managed, and allow multiple pointers to the same object. The way this is enabled is that the pointer's contents (the pointee) are immutable. If they were allowed to be mutable, data races would occur. In the next section I'll show the mutable version.

These have no direct comparison in C. The closest comparison is probably GObject's memory management API. Managed pointers aren't explicit memory management. As the name suggests, the runtime manages it for you. Managed pointers are usually considered as a "last resort". Even though GC in Rust can be fast because all managed pointers are task-local, no GC is always faster than some GC.


Mutability is whether you are allowed to mutate, or modify, some data. There are two pieces to mutability: mutability of things an object owns (inherited mutability), and mutability of data through a pointer.

Inherited Mutability

Say you have a point on the stack:

struct Point {
    x: int,
    y: int
let value: Point = Point { x: 24, y: 42 };

If you try to mutate the point by, say, changing the x field, the compiler will complain:

value.x = 42; //~ ERROR: cannot assign to immutable field

The error message is actualy a bit misleading: there is no such thing as a mutable field! "Point" is said to have inherited mutability; its mutability is determined by the mutability of the thing that holds it. So, to fix this example, we'd use:

let mut value: Point = Point { x: 24, y: 42 };
value.x = 42;

Inherited mutability inherits throughout the ownership tree: the value, and everything the value owns, recursively.