% Lifetimes
This is the last of three sections presenting Rust’s ownership system. This is one of Rust’s most distinct and compelling features, with which Rust developers should become quite acquainted. Ownership is how Rust achieves its largest goal, memory safety. There are a few distinct concepts, each with its own chapter:
- ownership, the key concept
- borrowing, and their associated feature ‘references’
- lifetimes, which you’re reading now
These three chapters are related, and in order. You’ll need all three to fully understand the ownership system.
Before we get to the details, two important notes about the ownership system.
Rust has a focus on safety and speed. It accomplishes these goals through many ‘zero-cost abstractions’, which means that in Rust, abstractions cost as little as possible in order to make them work. The ownership system is a prime example of a zero-cost abstraction. All of the analysis we’ll talk about in this guide is done at compile time. You do not pay any run-time cost for any of these features.
However, this system does have a certain cost: learning curve. Many new users to Rust experience something we like to call ‘fighting with the borrow checker’, where the Rust compiler refuses to compile a program that the author thinks is valid. This often happens because the programmer’s mental model of how ownership should work doesn’t match the actual rules that Rust implements. You probably will experience similar things at first. There is good news, however: more experienced Rust developers report that once they work with the rules of the ownership system for a period of time, they fight the borrow checker less and less.
With that in mind, let’s learn about lifetimes.
Lending out a reference to a resource that someone else owns can be complicated. For example, imagine this set of operations:
- I acquire a handle to some kind of resource.
- I lend you a reference to the resource.
- I decide I’m done with the resource, and deallocate it, while you still have your reference.
- You decide to use the resource.
Uh oh! Your reference is pointing to an invalid resource. This is called a dangling pointer or ‘use after free’, when the resource is memory. A small example of such a situation would be:
let r; // Introduce reference: `r`.
{
let i = 1; // Introduce scoped value: `i`.
r = &i; // Store reference of `i` in `r`.
} // `i` goes out of scope and is dropped.
println!("{}", r); // `r` still refers to `i`.
To fix this, we have to make sure that step four never happens after step three. In the small example above the Rust compiler is able to report the issue as it can see the lifetimes of the various values in the function.
When we have a function that takes arguments by reference the situation becomes more complex. Consider the following example:
fn skip_prefix(line: &str, prefix: &str) -> &str {
// ...
# line
}
let line = "lang:en=Hello World!";
let lang = "en";
let v;
{
let p = format!("lang:{}=", lang); // -+ `p` comes into scope.
v = skip_prefix(line, p.as_str()); // |
} // -+ `p` goes out of scope.
println!("{}", v);
Here we have a function skip_prefix
which takes two &str
references
as parameters and returns a single &str
reference. We call it
by passing in references to line
and p
: Two variables with different
lifetimes. Now the safety of the println!
-line depends on whether the
reference returned by skip_prefix
function references the still living
line
or the already dropped p
string.
Because of the above ambiguity, Rust will refuse to compile the example code. To get it to compile we need to tell the compiler more about the lifetimes of the references. This can be done by making the lifetimes explicit in the function declaration:
fn skip_prefix<'a, 'b>(line: &'a str, prefix: &'b str) -> &'a str {
// ...
# line
}
Let's examine the changes without going too deep into the syntax for now -
we'll get to that later. The first change was adding the <'a, 'b>
after the
method name. This introduces two lifetime parameters: 'a
and 'b
. Next each
reference in the function signature was associated with one of the lifetime
parameters by adding the lifetime name after the &
. This tells the compiler
how the lifetimes between different references are related.
As a result the compiler is now able to deduce that the return value of
skip_prefix
has the same lifetime as the line
parameter, which makes the v
reference safe to use even after the p
goes out of scope in the original
example.
In addition to the compiler being able to validate the usage of skip_prefix
return value, it can also ensure that the implementation follows the contract
established by the function declaration. This is useful especially when you are
implementing traits that are introduced later in the book.
Note It's important to understand that lifetime annotations are descriptive, not prescriptive. This means that how long a reference is valid is determined by the code, not by the annotations. The annotations, however, give information about lifetimes to the compiler that uses them to check the validity of references. The compiler can do so without annotations in simple cases, but needs the programmers support in complex scenarios.
The 'a
reads ‘the lifetime a’. Technically, every reference has some lifetime
associated with it, but the compiler lets you elide (i.e. omit, see
"Lifetime Elision" below) them in common cases. Before we
get to that, though, let’s look at a short example with explicit lifetimes:
fn bar<'a>(...)
We previously talked a little about function syntax, but we didn’t
discuss the <>
s after a function’s name. A function can have ‘generic
parameters’ between the <>
s, of which lifetimes are one kind. We’ll discuss
other kinds of generics later in the book, but for now, let’s
focus on the lifetimes aspect.
We use <>
to declare our lifetimes. This says that bar
has one lifetime,
'a
. If we had two reference parameters with different lifetimes, it would
look like this:
fn bar<'a, 'b>(...)
Then in our parameter list, we use the lifetimes we’ve named:
...(x: &'a i32)
If we wanted a &mut
reference, we’d do this:
...(x: &'a mut i32)
If you compare &mut i32
to &'a mut i32
, they’re the same, it’s that
the lifetime 'a
has snuck in between the &
and the mut i32
. We read &mut i32
as ‘a mutable reference to an i32
’ and &'a mut i32
as ‘a mutable
reference to an i32
with the lifetime 'a
’.
You’ll also need explicit lifetimes when working with struct
s that
contain references:
struct Foo<'a> {
x: &'a i32,
}
fn main() {
let y = &5; // This is the same as `let _y = 5; let y = &_y;`.
let f = Foo { x: y };
println!("{}", f.x);
}
As you can see, struct
s can also have lifetimes. In a similar way to functions,
struct Foo<'a> {
# x: &'a i32,
# }
declares a lifetime, and
# struct Foo<'a> {
x: &'a i32,
# }
uses it. So why do we need a lifetime here? We need to ensure that any reference
to a Foo
cannot outlive the reference to an i32
it contains.
Let’s implement a method on Foo
:
struct Foo<'a> {
x: &'a i32,
}
impl<'a> Foo<'a> {
fn x(&self) -> &'a i32 { self.x }
}
fn main() {
let y = &5; // This is the same as `let _y = 5; let y = &_y;`.
let f = Foo { x: y };
println!("x is: {}", f.x());
}
As you can see, we need to declare a lifetime for Foo
in the impl
line. We repeat
'a
twice, like on functions: impl<'a>
defines a lifetime 'a
, and Foo<'a>
uses it.
If you have multiple references, you can use the same lifetime multiple times:
fn x_or_y<'a>(x: &'a str, y: &'a str) -> &'a str {
# x
# }
This says that x
and y
both are alive for the same scope, and that the
return value is also alive for that scope. If you wanted x
and y
to have
different lifetimes, you can use multiple lifetime parameters:
fn x_or_y<'a, 'b>(x: &'a str, y: &'b str) -> &'a str {
# x
# }
In this example, x
and y
have different valid scopes, but the return value
has the same lifetime as x
.
A way to think about lifetimes is to visualize the scope that a reference is valid for. For example:
fn main() {
let y = &5; // -+ `y` comes into scope.
// |
// Stuff... // |
// |
} // -+ `y` goes out of scope.
Adding in our Foo
:
struct Foo<'a> {
x: &'a i32,
}
fn main() {
let y = &5; // -+ `y` comes into scope.
let f = Foo { x: y }; // -+ `f` comes into scope.
// |
// Stuff... // |
// |
} // -+ `f` and `y` go out of scope.
Our f
lives within the scope of y
, so everything works. What if it didn’t?
This code won’t work:
struct Foo<'a> {
x: &'a i32,
}
fn main() {
let x; // -+ `x` comes into scope.
// |
{ // |
let y = &5; // ---+ `y` comes into scope.
let f = Foo { x: y }; // ---+ `f` comes into scope.
x = &f.x; // | | This causes an error.
} // ---+ `f` and y go out of scope.
// |
println!("{}", x); // |
} // -+ `x` goes out of scope.
Whew! As you can see here, the scopes of f
and y
are smaller than the scope
of x
. But when we do x = &f.x
, we make x
a reference to something that’s
about to go out of scope.
Named lifetimes are a way of giving these scopes a name. Giving something a name is the first step towards being able to talk about it.
The lifetime named ‘static’ is a special lifetime. It signals that something
has the lifetime of the entire program. Most Rust programmers first come across
'static
when dealing with strings:
let x: &'static str = "Hello, world.";
String literals have the type &'static str
because the reference is always
alive: they are baked into the data segment of the final binary. Another
example are globals:
static FOO: i32 = 5;
let x: &'static i32 = &FOO;
This adds an i32
to the data segment of the binary, and x
is a reference
to it.
Rust supports powerful local type inference in the bodies of functions but not in their item signatures. It's forbidden to allow reasoning about types based on the item signature alone. However, for ergonomic reasons, a very restricted secondary inference algorithm called “lifetime elision” does apply when judging lifetimes. Lifetime elision is concerned solely with inferring lifetime parameters using three easily memorizable and unambiguous rules. This means lifetime elision acts as a shorthand for writing an item signature, while not hiding away the actual types involved as full local inference would if applied to it.
When talking about lifetime elision, we use the terms input lifetime and output lifetime. An input lifetime is a lifetime associated with a parameter of a function, and an output lifetime is a lifetime associated with the return value of a function. For example, this function has an input lifetime:
fn foo<'a>(bar: &'a str)
This one has an output lifetime:
fn foo<'a>() -> &'a str
This one has a lifetime in both positions:
fn foo<'a>(bar: &'a str) -> &'a str
Here are the three rules:
-
Each elided lifetime in a function’s arguments becomes a distinct lifetime parameter.
-
If there is exactly one input lifetime, elided or not, that lifetime is assigned to all elided lifetimes in the return values of that function.
-
If there are multiple input lifetimes, but one of them is
&self
or&mut self
, the lifetime ofself
is assigned to all elided output lifetimes.
Otherwise, it is an error to elide an output lifetime.
Here are some examples of functions with elided lifetimes. We’ve paired each example of an elided lifetime with its expanded form.
fn print(s: &str); // elided
fn print<'a>(s: &'a str); // expanded
fn debug(lvl: u32, s: &str); // elided
fn debug<'a>(lvl: u32, s: &'a str); // expanded
In the preceding example, lvl
doesn’t need a lifetime because it’s not a
reference (&
). Only things relating to references (such as a struct
which contains a reference) need lifetimes.
fn substr(s: &str, until: u32) -> &str; // elided
fn substr<'a>(s: &'a str, until: u32) -> &'a str; // expanded
fn get_str() -> &str; // ILLEGAL, no inputs
fn frob(s: &str, t: &str) -> &str; // ILLEGAL, two inputs
fn frob<'a, 'b>(s: &'a str, t: &'b str) -> &str; // Expanded: Output lifetime is ambiguous
fn get_mut(&mut self) -> &mut T; // elided
fn get_mut<'a>(&'a mut self) -> &'a mut T; // expanded
fn args<T: ToCStr>(&mut self, args: &[T]) -> &mut Command; // elided
fn args<'a, 'b, T: ToCStr>(&'a mut self, args: &'b [T]) -> &'a mut Command; // expanded
fn new(buf: &mut [u8]) -> BufWriter; // elided
fn new<'a>(buf: &'a mut [u8]) -> BufWriter<'a>; // expanded