One detail we didn’t discuss in the “References and Borrowing” section in Chapter 4 is that every reference in Rust has a lifetime, which is the scope for which that reference is valid. Most of the time, lifetimes are implicit and inferred, just like most of the time, types are inferred. We must annotate types when multiple types are possible. In a similar way, we must annotate lifetimes when the lifetimes of 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 will definitely be valid.
The concept of lifetimes is somewhat different from tools in other programming languages, arguably making lifetimes Rust’s most distinctive feature. Although we won’t cover lifetimes in their entirety in this chapter, we’ll discuss common ways you might encounter lifetime syntax so you can become familiar with the concepts.
The main aim of lifetimes is to prevent dangling references, which cause a program to reference data other than the data it’s intended to reference. Consider the program in Listing 10-17, which has an outer scope and an inner scope.
fn main() {
{
let r;
{
let x = 5;
r = &x;
}
println!("r: {}", r);
}
}
Listing 10-17: An attempt to use a reference whose value has gone out of scope
Note: The examples in Listings 10-17, 10-18, and 10-24 declare variables without giving them an initial value, so the variable name exists in the outer scope. At first glance, this might appear to be in conflict with Rust’s having no null values. However, if we try to use a variable before giving it a value, we’ll get a compile-time error, which shows that Rust indeed does not allow null values.
The outer scope declares a variable named r with no initial value, and the inner scope declares a variable named x with the initial value of 5. Inside the inner scope, we attempt to set the value of r as a reference to x. Then the inner scope ends, and we attempt to print the value in r. This code won’t compile because the value r is referring to has gone out of scope before we try to use it. Here is the error message:
$ cargo run
Compiling chapter10 v0.1.0 (file:///projects/chapter10)
error[E0597]: `x` does not live long enough
--> src/main.rs:7:17
|
7 | r = &x;
| ^^ borrowed value does not live long enough
8 | }
| - `x` dropped here while still borrowed
9 |
10 | println!("r: {}", r);
| - borrow later used here
error: aborting due to previous error
For more information about this error, try `rustc --explain E0597`.
error: could not compile `chapter10`.
To learn more, run the command again with --verbose.
The variable x doesn’t “live long enough.” The reason is that x will be out of scope when the inner scope ends on line 7. But r is still valid for the outer scope; because its scope is larger, we say that it “lives longer.” If Rust allowed this code to work, r would be referencing memory that was deallocated when x went out of scope, and anything we tried to do with r wouldn’t work correctly. So how does Rust determine that this code is invalid? It uses a borrow checker.
The Rust compiler has a borrow checker that compares scopes to determine whether all borrows are valid. Listing 10-18 shows the same code as Listing 10-17 but with annotations showing the lifetimes of the variables.
fn main() {
{
let r; // ---------+-- 'a
// |
{ // |
let x = 5; // -+-- 'b |
r = &x; // | |
} // -+ |
// |
println!("r: {}", r); // |
} // ---------+
}
Listing 10-18: Annotations of the lifetimes of r and x, named 'a and 'b, respectively
Here, we’ve annotated the lifetime of r with 'a and the lifetime of x with 'b. As you can see, the inner 'b block is much smaller than the outer 'a lifetime block. At compile time, Rust compares 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. The program is rejected because 'b is shorter than 'a: the subject of the reference doesn’t live as long as the reference.
Listing 10-19 fixes the code so it doesn’t have a dangling reference and compiles without any errors.
fn main() {
{
let x = 5; // ----------+-- 'b
// |
let r = &x; // --+-- 'a |
// | |
println!("r: {}", r); // | |
// --+ |
} // ----------+
}
Listing 10-19: A valid reference because the data has a longer lifetime than the reference
Here, x has the lifetime 'b, which in this case is larger than 'a. This means r can reference x because Rust knows that the reference in r will always be valid while x is valid.
Now that you know where the lifetimes of references are and how Rust analyzes lifetimes to ensure references will always be valid, let’s explore generic lifetimes of parameters and return values in the context of functions.
Let’s write a function that returns the longer of two string slices. This function will take two string slices and return a string slice. After we’ve implemented the longest function, the code in Listing 10-20 should print The longest string is abcd.
Filename: src/main.rs
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {}", result);
}
Listing 10-20: A main function that calls the longest function to find the longer of two string slices
Note that we want the function to take string slices, which are references, because we don’t want the longest function to take ownership of its parameters. Refer to the “String Slices as Parameters” section in Chapter 4 for more discussion about why the parameters we use in Listing 10-20 are the ones we want.
If we try to implement the longest function as shown in Listing 10-21, it won’t compile.
Filename: src/main.rs
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {}", result);
}
fn longest(x: &str, y: &str) -> &str {
if x.len() > y.len() {
x
} else {
y
}
}
Listing 10-21: An implementation of the longest function that returns the longer of two string slices but does not yet compile
Instead, we get the following error that talks about lifetimes:
$ 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 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`
error: aborting due to previous error
For more information about this error, try `rustc --explain E0106`.
error: could not compile `chapter10`.
To learn more, run the command again with --verbose.
The help text reveals that the return type needs a generic lifetime parameter on it because Rust can’t tell whether the reference being returned refers to x or y. Actually, we don’t know either, because the if block in the body of this function returns a reference to x and the else block returns a reference to y!
When we’re defining this function, we don’t know the concrete values that will be passed into this function, so we don’t know whether the if case or the else case will execute. We also don’t know the concrete lifetimes of the references that will be passed in, so we can’t look at the scopes as we did in Listings 10-18 and 10-19 to determine whether the reference we return will always be valid. The borrow checker can’t determine this either, because it doesn’t know how the lifetimes of x and y relate to the lifetime of the return value. To fix this error, we’ll add generic lifetime parameters that define the relationship between the references so the borrow checker can perform its analysis.
Lifetime annotations don’t change how long any of the references live. Just as functions can accept any type when the signature specifies a generic type parameter, functions can accept references with any lifetime by specifying a generic lifetime parameter. Lifetime annotations describe the relationships of the lifetimes of multiple references to each other without affecting the lifetimes.
Lifetime annotations have a slightly unusual syntax: the names of lifetime parameters must start with an apostrophe (') and are usually all lowercase and very short, like generic types. Most people use the name 'a. We place lifetime parameter annotations after the & of a reference, using a space to separate the annotation from the reference’s type.
Here are some examples: a reference to an i32 without a lifetime parameter, a reference to an i32 that has a lifetime parameter named 'a, and a mutable reference to an i32 that also has the lifetime 'a.
&i32 // a reference &'a i32 // a reference with an explicit lifetime &'a mut i32 // a mutable reference with an explicit lifetime
One lifetime annotation by itself doesn’t have much meaning, because the annotations are meant to tell Rust how generic lifetime parameters of multiple references relate to each other. For example, let’s say we have a function with the parameter first that is a reference to an i32 with lifetime 'a. The function also has another parameter named second that is another reference to an i32 that also has the lifetime 'a. The lifetime annotations indicate that the references first and second must both live as long as that generic lifetime.
Now let’s examine lifetime annotations in the context of the longest function. As with generic type parameters, we need to declare generic lifetime parameters inside angle brackets between the function name and the parameter list. The constraint we want to express in this signature is that all the references in the parameters and the return value must have the same lifetime. We’ll name the lifetime 'a and then add it to each reference, as shown in Listing 10-22.
Filename: src/main.rs
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {}", result);
}
fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
if x.len() > y.len() {
x
} else {
y
}
}
Listing 10-22: The longest function definition specifying that all the references in the signature must have the same lifetime 'a
This code should compile and produce the result we want when we use it with the main function in Listing 10-20.
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 function signature also tells Rust that the string slice returned from the function will live at least as long as lifetime 'a. In practice, it means that the lifetime of the reference returned by the longest function is the same as the smaller of the lifetimes of the references passed in. These constraints are what we want Rust to enforce. Remember, when we specify the lifetime parameters in this function signature, we’re not changing the lifetimes of any values passed in or returned. Rather, we’re specifying that the borrow checker should reject any values that don’t adhere to these constraints. Note that the longest function doesn’t need to know exactly how long x and y will live, only that some scope can be substituted for 'a that will satisfy this signature.
When annotating lifetimes in functions, the annotations go in the function signature, not in the function body. Rust can analyze the code within the function without any help. However, when a function has references to or from code outside that function, it becomes almost impossible for Rust to figure out the lifetimes of the parameters or return values on its own. The lifetimes might be different each time the function is called. This is why we need to annotate the lifetimes manually.
When we pass concrete references to longest, the concrete lifetime that is substituted for 'a is the part of the scope of x that overlaps with the scope of y. In other words, the generic lifetime 'a will get the concrete lifetime that is equal to the smaller of the lifetimes of x and y. Because we’ve annotated the returned reference with the same lifetime parameter 'a, the returned reference will also be valid for the length of the smaller of the lifetimes of x and y.
Let’s look at how the lifetime annotations restrict the longest function by passing in references that have different concrete lifetimes. Listing 10-23 is a straightforward example.
Filename: src/main.rs
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);
}
}
fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
if x.len() > y.len() {
x
} else {
y
}
}
Listing 10-23: Using the longest function with references to String values that have different concrete lifetimes
In this example, string1 is valid until the end of the outer scope, string2 is valid until the end of the inner scope, and result references something that is valid until the end of the inner scope. Run this code, and you’ll see that the borrow checker approves of this code; it will compile and print The longest string is long string is long.
Next, let’s try an example that shows that the lifetime of the reference in result must be the smaller lifetime of the two arguments. We’ll move the declaration of the result variable outside the inner scope but leave the assignment of the value to the result variable inside the scope with string2. Then we’ll move the println! that uses result outside the inner scope, after the inner scope has ended. The code in Listing 10-24 will not compile.
Filename: src/main.rs
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);
}
fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
if x.len() > y.len() {
x
} else {
y
}
}
Listing 10-24: Attempting to use result after string2 has gone out of scope
When we try to compile this code, we’ll get this error:
$ cargo run
Compiling chapter10 v0.1.0 (file:///projects/chapter10)
error[E0597]: `string2` does not live long enough
--> src/main.rs:6:44
|
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
error: aborting due to previous error
For more information about this error, try `rustc --explain E0597`.
error: could not compile `chapter10`.
To learn more, run the command again with --verbose.
The error shows that for result to be valid for the println! statement, string2 would need to be valid until the end of the outer scope. Rust knows this because we annotated the lifetimes of the function parameters and return values using the same lifetime parameter 'a.
As humans, we can look at this code and see that string1 is longer than string2 and therefore result will contain a reference to string1. Because string1 has not gone out of scope yet, a reference to string1 will still be valid for the println! statement. However, the compiler can’t see that the reference is valid in this case. We’ve told Rust that the lifetime of the reference returned by the longest function is the same as the smaller of the lifetimes of the references passed in. Therefore, the borrow checker disallows the code in Listing 10-24 as possibly having an invalid reference.
Try designing more experiments that vary the values and lifetimes of the references passed in to the longest function and how the returned reference is used. Make hypotheses about whether or not your experiments will pass the borrow checker before you compile; then check to see if you’re right!
The way in which you need to specify lifetime parameters depends on what your function is doing. For example, if we changed the implementation of the longest function to always return the first parameter rather than the longest string slice, we wouldn’t need to specify a lifetime on the y parameter. The following code will compile:
Filename: src/main.rs
fn main() {
let string1 = String::from("abcd");
let string2 = "efghijklmnopqrstuvwxyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {}", result);
}
fn longest<'a>(x: &'a str, y: &str) -> &'a str {
x
}
In this example, we’ve specified a lifetime parameter 'a for the parameter x and the return type, but not for the parameter y, because the lifetime of y does not have any relationship with the lifetime of x or the return value.
When returning a reference from a function, the lifetime parameter for the return type needs to match the lifetime parameter for one of the parameters. If the reference returned does not refer to one of the parameters, it must refer to a value created within this function, which would be a dangling reference because the value will go out of scope at the end of the function. Consider this attempted implementation of the longest function that won’t compile:
Filename: src/main.rs
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest(string1.as_str(), string2);
println!("The longest string is {}", result);
}
fn longest<'a>(x: &str, y: &str) -> &'a str {
let result = String::from("really long string");
result.as_str()
}
Here, even though we’ve specified a lifetime parameter 'a for the return type, this implementation will fail to compile because the return value lifetime is not related to the lifetime of the parameters at all. Here is the error message we get:
$ 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 error: aborting due to previous error For more information about this error, try `rustc --explain E0515`. error: could not compile `chapter10`. To learn more, run the command again with --verbose.
The problem is that result goes out of scope and gets cleaned up at the end of the longest function. We’re also trying to return a reference to result from the function. There is no way we can specify lifetime parameters that would change the dangling reference, and Rust won’t let us create a dangling reference. In this case, the best fix would be to return an owned data type rather than a reference so the calling function is then responsible for cleaning up the value.
Ultimately, lifetime syntax is about connecting the lifetimes of various parameters and return values of functions. Once they’re connected, Rust has enough information to allow memory-safe operations and disallow operations that would create dangling pointers or otherwise violate memory safety.
So far, we’ve only defined structs to hold owned types. It’s possible for structs to hold references, but in that case we would need to add a lifetime annotation on every reference in the struct’s definition. Listing 10-25 has a struct named ImportantExcerpt that holds a string slice.
Filename: src/main.rs
struct ImportantExcerpt<'a> {
part: &'a str,
}
fn main() {
let novel = String::from("Call me Ishmael. Some years ago...");
let first_sentence = novel.split('.').next().expect("Could not find a '.'");
let i = ImportantExcerpt {
part: first_sentence,
};
}
Listing 10-25: A struct that holds a reference, so its definition needs a lifetime annotation
This struct has one field, part, that holds a string slice, which is a reference. As with generic data types, we declare the name of the generic lifetime parameter inside angle brackets after the name of the struct so we can use the lifetime parameter in the body of the struct definition. This annotation means an instance of ImportantExcerpt can’t outlive the reference it holds in its part field.
The main function here creates an instance of the ImportantExcerpt struct that holds a reference to the first sentence of the String owned by the variable novel. The data in novel exists before the ImportantExcerpt instance is created. In addition, novel doesn’t go out of scope until after the ImportantExcerpt goes out of scope, so the reference in the ImportantExcerpt instance is valid.
You’ve learned that every reference has a lifetime and that you need to specify lifetime parameters for functions or structs that use references. However, in Chapter 4 we had a function in Listing 4-9, which is shown again in Listing 10-26, that compiled without lifetime annotations.
Filename: src/lib.rs
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[..]
}
fn main() {
let my_string = String::from("hello world");
// first_word works on slices of `String`s
let word = first_word(&my_string[..]);
let my_string_literal = "hello world";
// first_word works on slices of string literals
let word = first_word(&my_string_literal[..]);
// Because string literals *are* string slices already,
// this works too, without the slice syntax!
let word = first_word(my_string_literal);
}
Listing 10-26: A function we defined in Listing 4-9 that compiled without lifetime annotations, even though the parameter and return type are references
The reason this function compiles without lifetime annotations is historical: in early versions (pre-1.0) of Rust, this code wouldn’t have compiled because every reference needed an explicit lifetime. At that time, the 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 team found that Rust programmers were entering the same lifetime annotations over and over in particular situations. These situations were predictable and followed a few deterministic patterns. The developers programmed these patterns into the compiler’s code so the borrow checker could infer the lifetimes in these situations and wouldn’t need explicit annotations.
This piece of Rust history is relevant because it’s 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 analysis of references are called the lifetime elision rules. These aren’t rules for programmers to follow; they’re a set of particular cases that the compiler will consider, and if your code fits these cases, you don’t need to write the lifetimes explicitly.
The elision rules don’t provide full inference. If Rust deterministically applies the rules but there is still ambiguity as to what lifetimes the references have, the compiler won’t guess what the lifetime of the remaining references should be. In this case, instead of guessing, the compiler will give you an error that you can resolve by adding the lifetime annotations that specify how the references relate to each other.
Lifetimes on 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 what lifetimes references have 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 compiler gets to the end of the three rules and there are still references for which it can’t figure out lifetimes, the compiler will stop with an error. These rules apply to fn definitions as well as impl blocks.
The first rule is that each parameter that is a reference gets its own lifetime parameter. In other words, a function with one parameter gets one lifetime parameter: fn foo<'a>(x: &'a i32); a function with two parameters gets two separate lifetime parameters: fn foo<'a, 'b>(x: &'a i32, y: &'b i32); and so on.
The second rule is if there is exactly one input lifetime parameter, that lifetime is assigned to all output lifetime parameters: fn foo<'a>(x: &'a i32) -> &'a i32.
The third rule is if there are multiple input lifetime parameters, but one of them is &self or &mut self because this is a method, the lifetime of self is assigned to all output lifetime parameters. This third rule makes methods much nicer to read and write because fewer symbols are necessary.
Let’s pretend we’re the compiler. We’ll apply these rules to figure out what the lifetimes of the references in the signature of the first_word function in Listing 10-26 are. The signature starts without any lifetimes associated with the references:
fn first_word(s: &str) -> &str {
Then the compiler applies the first rule, which specifies that each parameter gets its own lifetime. We’ll call it 'a as usual, so now the signature is this:
fn first_word<'a>(s: &'a str) -> &str {
The second rule applies because there is exactly one input lifetime. The second rule specifies that the lifetime of the one input parameter gets assigned to the output lifetime, so the signature is now this:
fn first_word<'a>(s: &'a str) -> &'a str {
Now all the references in this function signature have lifetimes, and the compiler can continue its analysis without needing the programmer to annotate the lifetimes in this function signature.
Let’s look at another example, this time using the longest function that had no lifetime parameters when we started working with it in Listing 10-21:
fn longest(x: &str, y: &str) -> &str {
Let’s apply the first rule: each parameter gets its own lifetime. This time we have two parameters instead of one, so we have two lifetimes:
fn longest<'a, 'b>(x: &'a str, y: &'b str) -> &str {
You can see that the second rule doesn’t apply because there is more than one input lifetime. The third rule doesn’t apply either, because longest is a function rather than a method, so none of the parameters are self. After working through all three rules, we still haven’t figured out what the return type’s lifetime is. This is why we got an error trying to compile the code in Listing 10-21: the compiler worked through the lifetime elision rules but still couldn’t figure out all the lifetimes of the references in the signature.
Because the third rule really only applies in method signatures, we’ll look at lifetimes in that context next to see why the third rule means we don’t have to annotate lifetimes in method signatures very often.
When we implement methods on a struct with lifetimes, we use the same syntax as that of generic type parameters shown in Listing 10-11. Where we declare and use the lifetime parameters depends on whether they’re related to the struct fields or the method parameters and return values.
Lifetime names for struct fields always need to be declared after the impl keyword and then used after the struct’s name, because those lifetimes are part of the struct’s type.
In method signatures inside the impl block, references might be tied to the lifetime of references in the struct’s fields, or they might be independent. In addition, the lifetime elision rules often make it so that lifetime annotations aren’t necessary in method signatures. Let’s look at some examples using the struct named ImportantExcerpt that we defined in Listing 10-25.
First, we’ll use a 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:
struct ImportantExcerpt<'a> {
part: &'a str,
}
impl<'a> ImportantExcerpt<'a> {
fn level(&self) -> i32 {
3
}
}
impl<'a> ImportantExcerpt<'a> {
fn announce_and_return_part(&self, announcement: &str) -> &str {
println!("Attention please: {}", announcement);
self.part
}
}
fn main() {
let novel = String::from("Call me Ishmael. Some years ago...");
let first_sentence = novel.split('.').next().expect("Could not find a '.'");
let i = ImportantExcerpt {
part: first_sentence,
};
}
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 lifetime elision rule applies:
struct ImportantExcerpt<'a> {
part: &'a str,
}
impl<'a> ImportantExcerpt<'a> {
fn level(&self) -> i32 {
3
}
}
impl<'a> ImportantExcerpt<'a> {
fn announce_and_return_part(&self, announcement: &str) -> &str {
println!("Attention please: {}", announcement);
self.part
}
}
fn main() {
let novel = String::from("Call me Ishmael. Some years ago...");
let first_sentence = novel.split('.').next().expect("Could not find a '.'");
let i = ImportantExcerpt {
part: first_sentence,
};
}
There are two input lifetimes, so Rust applies the first lifetime elision rule and gives both &self and announcement their own lifetimes. Then, because one of the parameters is &self, the return type gets the lifetime of &self, and all lifetimes have been accounted for.
One special lifetime we need to discuss is 'static, which means that this reference can live for the entire duration of the program. All string literals have the 'static lifetime, which we can annotate as follows:
#![allow(unused)]
fn main() {
let s: &'static str = "I have a static lifetime.";
}
The text of this string is stored directly in the program’s binary, which is always available. Therefore, the lifetime of all string literals is 'static.
You might see suggestions to use the 'static lifetime in error messages. But before specifying 'static as the lifetime for a reference, think about whether the reference you have actually lives the entire lifetime of your program or not. You might consider whether you want it to live that long, even if it could. Most of the time, the problem results from attempting to create a dangling reference or a mismatch of the available lifetimes. In such cases, the solution is fixing those problems, not specifying the 'static lifetime.
Let’s briefly look at the syntax of specifying generic type parameters, trait bounds, and lifetimes all in one function!
fn main() {
let string1 = String::from("abcd");
let string2 = "xyz";
let result = longest_with_an_announcement(
string1.as_str(),
string2,
"Today is someone's birthday!",
);
println!("The longest string is {}", result);
}
use std::fmt::Display;
fn longest_with_an_announcement<'a, T>(
x: &'a str,
y: &'a str,
ann: T,
) -> &'a str
where
T: Display,
{
println!("Announcement! {}", ann);
if x.len() > y.len() {
x
} else {
y
}
}
This is the longest function from Listing 10-22 that returns the longer of two string slices. But now it has an extra parameter named ann of the generic type T, which can be filled in by any type that implements the Display trait as specified by the where clause. This extra parameter will be printed before the function compares the lengths of the string slices, which is why the Display trait bound is necessary. Because lifetimes are a type of generic, the declarations of the lifetime parameter 'a and the generic type parameter T go in the same list inside the angle brackets after the function name.
We covered a lot in this chapter! Now that you know about generic type parameters, traits and trait bounds, and generic lifetime parameters, you’re ready to write code without repetition that works in many different situations. Generic type parameters let you apply the code to different types. Traits and trait bounds ensure that even though the types are generic, they’ll have the behavior the code needs. You learned how to use lifetime annotations to ensure that this flexible code won’t have any dangling references. And all of this analysis happens at compile time, which doesn’t affect runtime performance!
Believe it or not, there is much more to learn on the topics we discussed in this chapter: Chapter 17 discusses trait objects, which are another way to use traits. There are also more complex scenarios involving lifetime annotations that you will only need in very advanced scenarios; for those, you should read the Rust Reference. But next, you’ll learn how to write tests in Rust so you can make sure your code is working the way it should.
© 2010 The Rust Project Developers
Licensed under the Apache License, Version 2.0 or the MIT license, at your option.
https://doc.rust-lang.org/book/ch10-03-lifetime-syntax.html