19 KiB
Lifetimes
Valid References with Lifetimes
Lifetimes are another type of generic generic that has been used in the past.
Rather than ensuring that a type has the behavior we want lifetimes ensure that references are valid for as long as we need them to be.
Every reference in Rust has a lifetime, which is the scope that a reference is valid
A majority of the time references are implicit and inferred, like how types are inferred for the most part
We must only annotate types when multiple types are possible
The same concept can happen with lifetimes, where references could be related in a few different ways
Rust requires us to annotate the relationships using generic lifetime parameters to ensure the actual references used at runtime are valid
Annotating lifetimes is not a concept or feature that is in other programming languages
Preventing Dangling References with Lifetimes
The main aim of lifetimes is to prevent dangling references, this causes a program to reference data other than the data it's intended to reference
Here is an example that will not compile due to the value being referred to going out of scope, then the reference being invalid and being used elsewhere
fn main() {
let r;
{
let x = 5;
r = &x;
}
println!("r: {r}");
}
This is initialization is allowed because if yo try to use a variable with no value associated then Rust will throw a compile-time error.
This proves that rust does not allow or have a null type
in this code r has a reference to x but x goes out of scope, this makes r not be valid reference, this invalid reference is attempted to be used.
The error message that would occur afterwards would say that x "does not live long enough". This refers to x going out of scope but r is still in scope
Hence we say that r "lives longer" because it has a larger scope
The Borrow Checker
The Rust compiler has a borrow checker that compares scopes to determine whether all borrows are valid
Here is an example with annotations showing the lifetimes of variables
fn main() {
let r; // ---------+-- 'a
// |
{ // |
let x = 5; // -+-- 'b |
r = &x; // | |
} // -+ |
// |
println!("r: {r}"); // |
} // ---------+
In this annotations r is represented by 'a and x is represented by 'b
'a is the lifetime of r and 'b is the lifetime of x
As you can see 'a encompasses more than 'b
At compile time Rust will compare the size of the two lifetimes and sees that r has a lifetime of 'a but that it refers to memory with a lifetime of 'b. This results in a rejection of the program because the reference becomes invalid before a use of it: the subject reference doesn't live as long as the reference.
The lifetime refers to how long it "lives" in memory or is available in memory
Here is the fix to the program above
fn main() {
let x = 5; // ----------+-- 'b
// |
let r = &x; // --+-- 'a |
// | |
println!("r: {r}"); // | |
// --+ |
} // ----------+
As you can now see the the lifetime of 'b is now larger than the 'a reference
Know you know that the reference r will always be valid because x is always valid
Generic Lifetimes in Functions
To illustrate this we will write a function that returns the longer of two string slices
This function takes in two string slices and returns a single string slice
Here is an example of the use of the function longest that does this functionality
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {result}");
}
Note that longest takes in a string slice, which are references, we don't want the function to take ownership of the strings
If we try to implement longest function like below, it won't compile
fn longest(x: &str, y: &str) -> &str {
if x.len() > y.len() {
x
} else {
y
}
}
Here is the compiler error
$ cargo run
Compiling chapter10 v0.1.0 (file:///projects/chapter10)
error[E0106]: missing lifetime specifier
--> src/main.rs:9:33
|
9 | fn longest(x: &str, y: &str) -> &str {
| ---- ---- ^ expected named lifetime parameter
|
= help: this function's return type contains a borrowed value, but the signature does not say whether it is borrowed from `x` or `y`
help: consider introducing a named lifetime parameter
|
9 | fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
| ++++ ++ ++ ++
For more information about this error, try `rustc --explain E0106`.
error: could not compile `chapter10` (bin "chapter10") due to 1 previous error
This error highlights that the return type needs a generic lifetime parameter because rust cannot determine whether the reference is referring to x or y
We don't know either because of the if and else in the body of the function refer to two different things depending on what the inputs are
We also don't know the concrete lifetimes of the references , so we cannot even analyze the lifetimes like we did before, so we cant ensure that the reference and lifetime will always be valid.
The borrow checker cannot check for this either so it throws an error.
This is because it doesn't know the lifetimes of x and y relate to the lifetime of the return value
To fix this error we should add a generic lifetime parameter that defines the relationship between the references so the borrow checker can perform its analysis
Lifetime Annotation Syntax
Lifetime annotations don't change based on how long any of the references live, instead they describe the relationship of the lifetimes of multiple references to each other without affecting the lifetimes
Like how functions can accept any type when the signature specifies a generic, functions can accept any lifetime by specifying a generic lifetime parameter
Lifetime annotations have a unusual syntax: the naming of a lifetime parameter must start with an apostrophe ' and are usually all lowercase and very short, just like generics
Most people use 'a as the first lifetime annotation
The placement of the lifetime goes after the & reference using a space to separate the lifetime from the reference's type
Here is an example to an i32 reference
&i32 // a reference
&'a i32 // a reference with an explicit lifetime
&'a mut i32 // a mutable reference with an explicit lifetime
One annotation by itself has no meaning because cannot be a relationship to another (or multiple) reference(s)
Lifetime Annotations in Function Signatures
To use lifetime annotations in a function's signature you must first declare the generic lifetime parameter inside <> between the function name and the parameter list (just like generics)
In this case the constraint we have is that we want the returned reference to be valid as long as both parameters are valid
This will be the relationship between the lifetime of the parameters and the return reference.
Here is this implemented with the name 'a as the lifetime, it is added to each reference in the function signature
fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
if x.len() > y.len() {
x
} else {
y
}
}
This code should compile now with the use of lifetime annotations
The function signature now tells Rust that for some lifetime 'a, the function takes two parameters, both of which are string slices that live at least as long as lifetime 'a
The signature also tells Rust that the string slice returned from the function will live at least as long as life 'a.
What this means is that the smaller of the lifetime returned by the longest function is the same as the smaller of the lifetime values referred to by the function arguments
These relationships are what we want Rust to use when analyzing this code
Lifetime annotations go in the signature NOT the body of the function
Lifetime annotations are part of the contract of the function, like types in the signature
This makes the lifetime analysis easier on the Rust compiler
If there is a problem in the signature it makes it easier to identify and express in an error as well, and give clear solutions
Here is another example of the use of the longest function
fn main() {
let string1 = String::from("long string is long");
{
let string2 = String::from("xyz");
let result = longest(string1.as_str(), string2.as_str());
println!("The longest string is {result}");
}
}
In this example the lifetime of the return value of the longest function is the same as string2 which means that after the inner scope or {} the reference is no longer valid
This is both due to string2, result (which stores the reference) and the function signature states that it only lives as long as the shortest function
The value reference in result is long string is long and the program will print The longest string is long string is long
In this example the program will not compile
fn main() {
let string1 = String::from("long string is long");
let result;
{
let string2 = String::from("xyz");
result = longest(string1.as_str(), string2.as_str());
}
println!("The longest string is {result}");
}
This is due to string2 not living long enough for the reference in result to be used while in a valid state
Here is the error it would provide
$ cargo run
Compiling chapter10 v0.1.0 (file:///projects/chapter10)
error[E0597]: `string2` does not live long enough
--> src/main.rs:6:44
|
5 | let string2 = String::from("xyz");
| ------- binding `string2` declared here
6 | result = longest(string1.as_str(), string2.as_str());
| ^^^^^^^ borrowed value does not live long enough
7 | }
| - `string2` dropped here while still borrowed
8 | println!("The longest string is {result}");
| -------- borrow later used here
For more information about this error, try `rustc --explain E0597`.
error: could not compile `chapter10` (bin "chapter10") due to 1 previous error
This also states that string2 would need to be valid until the print! macro. Hence why it states that it doesn't live long enough
Even though the reference in this case is to string1 the compiler and the function signature states that the lifetime of the return value is the same as the shortest lifetime
Thinking in Terms of Lifetimes
The way that you need to specify lifetime parameters depends on what your function is doing
For example if this was your function
fn longest<'a>(x: &'a str, y: &str) -> &'a str {
x
}
This function's return reference's lifetime is the same as x so the returned reference only lives as long as what is passed into x
The return reference lifetime has no relations to the y lifetime
Any lifetime MUST have another relationship to another reference
Any lifetime in a function signature's return value MUST relate to AT LEAST one parameter
If it doesn't it would create a dangling reference, this is because the value would go out of scope at the end of the function
Here is an example of longest that creates both a dangling reference and an invalid lifetime
fn longest<'a>(x: &str, y: &str) -> &'a str {
let result = String::from("really long string");
result.as_str()
}
Here is the error
$ cargo run
Compiling chapter10 v0.1.0 (file:///projects/chapter10)
error[E0515]: cannot return value referencing local variable `result`
--> src/main.rs:11:5
|
11 | result.as_str()
| ------^^^^^^^^^
| |
| returns a value referencing data owned by the current function
| `result` is borrowed here
For more information about this error, try `rustc --explain E0515`.
error: could not compile `chapter10` (bin "chapter10") due to 1 previous error
The problem is that result goes out of scope and gets cleaned up at the end of the function.
We also try to return a reference at the end of the function to result
There is no way we can specify lifetime parameters that would change the dangling reference in this case
Rust will not let you create a dangling reference
The fix for this would be to transfer ownership out of the function
Lifetime Annotations in Struct Definitions
Structs can also hold references
When it holds a reference it needs a lifetime annotation on every reference in the struct's def
Here is an example where the struct holds a single string
struct ImportantExcerpt<'a> {
part: &'a str,
}
fn main() {
let novel = String::from("Call me Ishmael. Some years ago...");
let first_sentence = novel.split('.').next().unwrap();
let i = ImportantExcerpt {
part: first_sentence,
};
}
Notice that the lifetime annotation is the same as a function's signature
The lifetime goes in a <> after the name of the structure
This annotation means that an instance of the struct can't outlive the reference
Lifetime Elision
In this case the function does not have a lifetime annotation and it compiles
fn first_word(s: &str) -> &str {
let bytes = s.as_bytes();
for (i, &item) in bytes.iter().enumerate() {
if item == b' ' {
return &s[0..i];
}
}
&s[..]
}
The reason why this function compiles without lifetime annotations is due to previous versions of Rust
In earlier versions of Rust (pre-1.0), this code wouldn't have compiled because every reference needed an explicit lifetime
At that time a function signature would have been written like this
fn first_word<'a>(s: &'a str) -> &'a str {
After writing a lot of Rust code, the Rust development team found that Rust programmers were writing a lot of the same code
These patterns were predictable and followed a few deterministic patterns. So the development team allowed the compiler and borrow checker to infer the lifetime annotation by using these common lifetime annotation patterns
Its important to know that it is possible that more deterministic patterns will emerge and be added to the compiler.
In the future even fewer lifetime annotations might be required
The patterns programmed into Rust's lifetime analysis of references are called the lifetime elision rules
These aren't rules for programmers, they are a set of particular cases that the compiler will consider. If your code fits these cases you don't have two explicitly define the lifetimes.
The elision rules don't provide full inference.
If there is still ambiguity as to what lifetimes are after the compiler applies the rules, the compiler will give an error instead of guessing
Instead of guessing the compiler will give you an error that you can resolve by adding the lifetime annotations
Lifetimes on a function or method parameters are called input lifetimes, and lifetimes on return values are called output lifetimes
The compiler uses three rules to figure out the lifetimes of the references when there aren't explicit annotations
The first rule applies to input lifetimes and the second and third rules apply to output lifetimes
If the complier gets to the end of these three rules and still cant figure it out then the compiler will stop with an error
These rules apply to fn definitions and impl blocks
The first rule is that the compiler assigns a lifetime parameter to each parameter that is a reference.
Each parameter gets its own separate lifetime annotation that has no relationship to any other lifetime (fn foo<'a, 'b>(x: &'a i32, y: &'b i32)
The second rule is that if there is exactly one input parameter, that lifetime is assigned to all output parameters (fn foo<'a>(x: &'a i32) -> &'a i32)
The third rule is that there are multiple input lifetime parameters, but one of them is &self or &mut self. Due to it being a method, the lifetime of self is assigned to all output lifetime parameters
This rule makes methods much cleaner to read and write because fewer symbols are necessary
Lifetime Annotations in Method Definitions
When we implement methods on struct with lifetimes, we use the same syntax as generic type parameters.
Where we declare and use the lifetime parameters depends on whether they are related to the struct fields or the method parameters and return values
In method signatures inside the impl block, references might be tied to the lifetime of references in the strut's fields or they might be independent
The lifetime elision rules often make it so that lifetime annotations aren't necessary in method signatures
Lets look at some examples using the struct ImportantExcerpt
First the method named level whose only parameter is a reference to self and whose return value is an i32, which is not a reference to anything
impl<'a> ImportantExcerpt<'a> {
fn level(&self) -> i32 {
3
}
}
The lifetime parameter declaration after impl and its use after the type name are required but we're not required to annotate the lifetime of the reference to self because of the first elision rule
Here is an example where the third elision rule is applicable
impl<'a> ImportantExcerpt<'a> {
fn announce_and_return_part(&self, announcement: &str) -> &str {
println!("Attention please: {announcement}");
self.part
}
}
There are two input lifetimes, so Rust applies the first lifetime elision rule and give both &self and announcement their own lifetimes.
Then because, because one of the parameters is &self the return type gets the lifetime of &self and all lifetimes have been dealt with
The Static Lifetime
One special lifetime is the 'static, which denotes that the affected reference can live for the entire duration of the program
All string literals have the 'static lifetime always
Here is how we can annotate it
let s: &'static str = "I have a static lifetime.";
The string literal is stored in the program's binary which is always available
You might see suggestions to use 'static lifetime in the error message.
Think about whether or not the value being referenced/the reference will always be valid before adding it
Most of the time the suggestion comes from attempting to create a dangling reference or a mismatch of the available lifetimes, instead of adding a 'static lifetime annotation
Instead the solution is to fix those problems