## Validating References with Lifetimes

When we talked about references in Chapter 4, we left out an important detail: 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. Similarly to when we have to annotate types because multiple types are possible, there are cases where the lifetimes of references could be related in a few different ways, so Rust needs us to annotate the relationships using generic lifetime parameters so that it can make sure the actual references used at runtime will definitely be valid.

Yes, it's a bit unusual, and will be different to tools you've used in other programming languages. Lifetimes are, in some ways, Rust's most distinctive feature.

Lifetimes are a big topic that can't be covered in entirety in this chapter, so we'll cover common ways you might encounter lifetime syntax in this chapter to get you familiar with the concepts. Chapter 19 will contain more advanced information about everything lifetimes can do.

### Lifetimes Prevent Dangling References

The main aim of lifetimes is to prevent dangling references, which will cause a program to reference data other than the data we're intending to reference. Consider the program in Listing 10-16, with an outer scope and an inner scope. 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 out the value in `r`:

    {
        let r;

        {
            let x = 5;
            r = &x;
        }

        println!("r: {}", r);
    }

Listing 10-16: An attempt to use a reference whose value has gone out of scope

> #### Uninitialized Variables Cannot Be Used

> The next few examples declare variables without giving them an initial value, so that the variable name exists in the outer scope. This might appear to be in conflict with Rust not having null. However, if we try to use a variable before giving it a value, we'll get a compile-time error. Try it out!

When we compile this code, we'll get an error:

    error: `x` does not live long enough
       |
    6  |         r = &x;
       |              - borrow occurs here
    7  |     }
       |     ^ `x` dropped here while still borrowed
    ...
    10 | }
       | - borrowed value needs to live until here

The variable `x` doesn't "live long enough." Why not? Well, `x` is going to go out of scope when we hit the closing curly brace on line 7, ending the inner scope. But `r` is valid for the outer scope; its scope is larger and 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 should not be allowed?

#### The Borrow Checker

The part of the compiler called the borrow checker compares scopes to determine that all borrows are valid. Listing 10-17 shows the same example from Listing 10-16 with annotations showing the lifetimes of the variables:

    {
        let r;         // -------+-- 'a
                       //        |
        {              //        |
            let x = 5; // -+-----+-- 'b
            r = &x;    //  |     |
        }              // -+     |
                       //        |
        println!("r: {}", r); // |
                       //        |
                       // -------+
    }

Listing 10-17: Annotations of the lifetimes of `r` and `x`, named `'a` and `'b` respectively

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 an object with a lifetime of `'b`. The program is rejected because the lifetime `'b` is shorter than the lifetime of `'a`: the subject of the reference does not live as long as the reference.

Let's look at an example in Listing 10-18 that doesn't try to make a dangling reference and compiles without any errors:

    # #![allow(unused_variables)]
    #fn main() {
    {
        let x = 5;            // -----+-- 'b
                              //      |
        let r = &x;           // --+--+-- 'a
                              //   |  |
        println!("r: {}", r); //   |  |
                              // --+  |
    }                         // -----+

    #}

Listing 10-18: 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`: Rust knows that the reference in `r` will always be valid while `x` is valid.

Now that we've shown where the lifetimes of references are in a concrete example and discussed how Rust analyzes lifetimes to ensure references will always be valid, let's talk about generic lifetimes of parameters and return values in the context of functions.

### Generic Lifetimes in Functions

Let's write a function that will return the longest of two string slices. We want to be able to call this function by passing it two string slices, and we want to get back a string slice. The code in Listing 10-19 should print `The longest string is abcd` once we've implemented the `longest` function:

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 10-19: A `main` function that calls the `longest` function to find the longest of two string slices

Note that we want the function to take string slices (which are references, as we talked about in Chapter 4) since we don't want the `longest` function to take ownership of its arguments. We want the function to be able to accept slices of a `String` (which is the type of the variable `string1`) as well as string literals (which is what variable `string2` contains).

Refer back to the "String Slices as Arguments" section of Chapter 4 for more discussion about why these are the arguments we want.

If we try to implement the `longest` function as shown in Listing 10-20, it won't compile:

Filename: src/main.rs

    fn longest(x: &str, y: &str) -> &str {
        if x.len() > y.len() {
            x
        } else {
            y
        }
    }

Listing 10-20: An implementation of the `longest` function that returns the longest of two string slices, but does not yet compile

Instead we get the following error that talks about lifetimes:

    error[E0106]: missing lifetime specifier
       |
    1  | 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`

The help text is telling us that the return type needs a generic lifetime parameter on it because Rust can't tell if the reference being returned refers to `x` or `y`. Actually, we don't know either, since the `if` block in the body of this function returns a reference to `x` and the `else` block returns a reference to `y`!

As 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 like we did in Listings 10-17 and 10-18 in order to determine that 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. We're going to add generic lifetime parameters that will define the relationship between the references so that the borrow checker can perform its analysis.

### Lifetime Annotation Syntax

Lifetime annotations don't change how long any of the references involved live. In the same way that functions can accept any type when the signature specifies a generic type parameter, functions can accept references with any lifetime when the signature specifies a generic lifetime parameter. What lifetime annotations do is relate the lifetimes of multiple references to each other.

Lifetime annotations have a slightly unusual syntax: the names of lifetime parameters must start with an apostrophe `'`. The names of lifetime parameters are usually all lowercase, and like generic types, their names are usually very short. `'a` is the name most people use as a default. Lifetime parameter annotations go after the `&` of a reference, and a space separates the lifetime annotation from the reference's type.

Here's some examples: we've got 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: lifetime annotations tell Rust how the generic lifetime parameters of multiple references relate to each other. If we have a function with the parameter `first` that is a reference to an `i32` that has the lifetime `'a`, and the function has another parameter named `second` that is another reference to an `i32` that also has the lifetime `'a`, these two lifetime annotations that have the same name indicate that the references `first` and `second` must both live as long as the same generic lifetime.

### Lifetime Annotations in Function Signatures

Let's look at lifetime annotations in the context of the `longest` function we're working on. Just like generic type parameters, generic lifetime parameters need to be declared within angle brackets between the function name and the parameter list. The constraint we want to tell Rust about for the references in the parameters and the return value is that they all must have the same lifetime, which we'll name `'a` and add to each reference as shown in Listing 10-21:

Filename: src/main.rs

    # #![allow(unused_variables)]
    #fn main() {
    fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
        if x.len() > y.len() {
            x
        } else {
            y
        }
    }

    #}

Listing 10-21: The `longest` function definition that specifies all the references in the signature must have the same lifetime, `'a`

This will compile and will produce the result we want when used with the `main` function in Listing 10-19.

The function signature now says that for some lifetime `'a`, the function will get two parameters, both of which are string slices that live at least as long as the lifetime `'a`. The function will return a string slice that also will last at least as long as the lifetime `'a`. This is the contract we are telling Rust we want it to enforce.

By specifying the lifetime parameters in this function signature, we are not changing the lifetimes of any values passed in or returned, but we are saying that any values that do not adhere to this contract should be rejected by the borrow checker. This function does not know (or need to know) exactly how long `x` and `y` will live, but only needs to know that there is some scope that can be substituted for `'a` that will satisfy this signature.

When annotating lifetimes in functions, the annotations go on the function signature, and not in any of the code in the function body. This is because Rust is able to analyze the code within the function without any help, but when a function has references to or from code outside that function, the lifetimes of the arguments or return values will potentially be different each time the function is called. This would be incredibly costly and often impossible for Rust to figure out. In this case, we need to annotate the lifetimes ourselves.

When concrete references are passed to `longest`, the concrete lifetime that gets substituted for `'a` is the part of the scope of `x` that overlaps with the scope of `y`. Since scopes always nest, another way to say this is that the generic lifetime `'a` will get the concrete lifetime 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 therefore be guaranteed to be valid as long as the shorter of the lifetimes of `x` and `y`.

Let's see how this restricts the usage of the `longest` function by passing in references that have different concrete lifetimes. Listing 10-22 is a straightforward example that should match your intuition from any language: `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. The borrow checker approves of this code; it will compile and print `The longest string is long string is long` when run:

Filename: src/main.rs

    # fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
    #     if x.len() > y.len() {
    #         x
    #     } else {
    #         y
    #     }
    # }
    #
    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);
        }
    }

Listing 10-22: Using the `longest` function with references to `String` values that have different concrete lifetimes

Next, let's try an example that will show 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`. Next, we'll move the `println!` that uses `result` outside of the inner scope, after it has ended. The code in Listing 10-23 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);
    }

Listing 10-23: Attempting to use `result` after `string2` has gone out of scope won't compile

If we try to compile this, we'll get this error:

    error: `string2` does not live long enough
       |
    6  |         result = longest(string1.as_str(), string2.as_str());
       |                                            ------- borrow occurs here
    7  |     }
       |     ^ `string2` dropped here while still borrowed
    8  |     println!("The longest string is {}", result);
    9  | }
       | - borrowed value needs to live until here

The error is saying that in order for `result` to be valid for the `println!`, `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 with the same lifetime parameter, `'a`.

We can look at this code as humans and see that `string1` is longer, 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!`. However, what we've told Rust with the lifetime parameters is 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 10-23 as possibly having an invalid reference.

Try designing some 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 your experiments will pass the borrow checker or not before you compile, then check to see if you're right!

### Thinking in Terms of Lifetimes

The exact way 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 argument rather than the longest string slice, we wouldn't need to specify a lifetime on the `y` parameter. This code compiles:

Filename: src/main.rs

    # #![allow(unused_variables)]
    #fn main() {
    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`, since 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 of one of the arguments. If the reference returned does not refer to one of the arguments, the only other possibility is that it refers to a value created within this function, which would be a dangling reference since 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 longest<'a>(x: &str, y: &str) -> &'a str {
        let result = String::from("really long string");
        result.as_str()
    }

Even though we've specified a lifetime parameter `'a` for the return type, this implementation fails to compile because the return value lifetime is not related to the lifetime of the parameters at all. Here's the error message we get:

    error: `result` does not live long enough
      |
    3 |     result.as_str()
      |     ^^^^^^ does not live long enough
    4 | }
      | - borrowed value only lives until here
      |
    note: borrowed value must be valid for the lifetime 'a as defined on the block
    at 1:44...
      |
    1 | fn longest<'a>(x: &str, y: &str) -> &'a str {
      |                                             ^

The problem is that `result` will go out of scope and get cleaned up at the end of the `longest` function, and we're trying to return a reference to `result` from the function. There's 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 that the calling function is then responsible for cleaning up the value.

Ultimately, lifetime syntax is about connecting the lifetimes of various arguments 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.

### Lifetime Annotations in Struct Definitions

Up until now, we've only defined structs to hold owned types. It is possible for structs to hold references, but we need to add a lifetime annotation on every reference in the struct's definition. Listing 10-24 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 10-24: 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. Just like with generic data types, we have to declare the name of the generic lifetime parameter inside angle brackets after the name of the struct so that we can use the lifetime parameter in the body of the struct definition.

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`.

### Lifetime Elision

In this section, we've learned that every reference has a lifetime, and we need to specify lifetime parameters for functions or structs that use references. However, in Chapter 4 we had a function in the "String Slices" section, shown again in Listing 10-25, that compiled without lifetime annotations:

Filename: src/lib.rs

    # #![allow(unused_variables)]
    #fn main() {
    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[..]
    }

    #}

Listing 10-25: A function we defined in Chapter 4 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 of pre-1.0 Rust, this indeed wouldn't have compiled. 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 typing the same lifetime annotations over and over in particular situations. These situations were predictable and followed a few deterministic patterns. The Rust team then programmed these patterns into the Rust compiler's code so that the borrow checker can infer the lifetimes in these situations without forcing the programmer to explicitly add the annotations.

We mention this piece of Rust history because it's entirely 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; the rules are 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's still ambiguity as to what lifetimes the references have, it won't guess what the lifetime of the remaining references should be. In this case, the compiler will give you an error that can be resolved by adding the lifetime annotations that correspond to your intentions for how the references relate to each other.

First, some definitions: Lifetimes on function or method parameters are called input lifetimes, and lifetimes on return values are called output lifetimes.

Now, on to the rules that the compiler uses to figure out what lifetimes references have when there aren't explicit annotations. The first rule applies to input lifetimes, and the second two rules apply to output lifetimes. If the compiler gets to the end of the three rules and there are still references that it can't figure out lifetimes for, the compiler will stop with an error.

1. 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 arguments gets two separate lifetime parameters: `fn foo<'a, 'b>(x: &'a i32, y: &'b i32)`, and so on.

2. 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`.

3. If there are multiple input lifetime parameters, but one of them is `&self` or `&mut self` because this is a method, then the lifetime of `self` is assigned to all output lifetime parameters. This makes writing methods much nicer.

Let's pretend we're the compiler and apply these rules to figure out what the lifetimes of the references in the signature of the `first_word` function in Listing 10-25 are. The signatures starts without any lifetimes associated with the references:

    fn first_word(s: &str) -> &str {

Then we (as the compiler) apply the first rule, which says each parameter gets its own lifetime. We're going to call it `'a` as usual, so now the signature is:

    fn first_word<'a>(s: &'a str) -> &str {

On to the second rule, which applies because there is exactly one input lifetime. The second rule says the lifetime of the one input parameter gets assigned to the output lifetime, so now the signature is:

    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 do another example, this time with the `longest` function that had no lifetime parameters when we started working with in Listing 10-20:

    fn longest(x: &str, y: &str) -> &str {

Pretending we're the compiler again, let's apply the first rule: each parameter gets its own lifetime. This time we have two parameters, so we have two lifetimes:

    fn longest<'a, 'b>(x: &'a str, y: &'b str) -> &str {

Looking at the second rule, it doesn't apply since there is more than one input lifetime. Looking at the third rule, this also does not apply because this is a function rather than a method, so none of the parameters are `self`. So we're out of rules, but we haven't figured out what the return type's lifetime is. This is why we got an error trying to compile the code from Listing 10-20: the compiler worked through the lifetime elision rules it knows, but still can't figure out all the lifetimes of the references in the signature.

Because the third rule only really applies in method signatures, let's look at lifetimes in that context now, and see why the third rule means we don't have to annotate lifetimes in method signatures very often.

### Lifetime Annotations in Method Definitions

When we implement methods on a struct with lifetimes, the syntax is again the same as that of generic type parameters that we showed in Listing 10-10: the place that lifetime parameters are declared and used depends on whether the lifetime parameter is related to the struct fields or the method arguments 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, since 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 10-24.

First, here's a method named `level`. The only parameter is a reference to `self`, and the return value is just an `i32`, not a reference to anything:

    # #![allow(unused_variables)]
    #fn main() {
    # struct ImportantExcerpt<'a> {
    #     part: &'a str,
    # }
    #
    impl<'a> ImportantExcerpt<'a> {
        fn level(&self) -> i32 {
            3
        }
    }

    #}

The lifetime parameter declaration after `impl` and use after the type name is required, but we're not required to annotate the lifetime of the reference to `self` because of the first elision rule.

Here's an example where the third lifetime elision rule applies:

    # #![allow(unused_variables)]
    #fn main() {
    # struct ImportantExcerpt<'a> {
    #     part: &'a str,
    # }
    #
    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 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.

### The Static Lifetime

There is one special lifetime we need to discuss: `'static`. The `'static` lifetime is the entire duration of the program. All string literals have the `'static` lifetime, which we can choose to annotate as follows:

    # #![allow(unused_variables)]
    #fn main() {
    let s: &'static str = "I have a static lifetime.";

    #}

The text of this string is stored directly in the binary of your program and the binary of your program is always available. Therefore, the lifetime of all string literals is `'static`.

You may see suggestions to use the `'static` lifetime in error message help text, but before specifying `'static` as the lifetime for a reference, think about whether the reference you have is one that actually lives the entire lifetime of your program or not (or even if you want it to live that long, if it could). Most of the time, the problem in the code is an attempt to create a dangling reference or a mismatch of the available lifetimes, and the solution is fixing those problems, not specifying the `'static` lifetime.

### Generic Type Parameters, Trait Bounds, and Lifetimes Together

Let's briefly look at the syntax of specifying generic type parameters, trait bounds, and lifetimes all in one function!

    # #![allow(unused_variables)]
    #fn main() {
    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 10-21 that returns the longest of two string slices, but with an extra argument named `ann`. The type of `ann` is the generic type `T`, which may be filled in by any type that implements the `Display` trait as specified by the `where` clause. This extra argument will be printed out 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 both the lifetime parameter `'a` and the generic type parameter `T` go in the same list within the angle brackets after the function name.

## Summary

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 that isn't duplicated but can be used in many different situations. Generic type parameters mean the code can be applied to different types. Traits and trait bounds ensure that even though the types are generic, those types will have the behavior the code needs. Relationships between the lifetimes of references specified by lifetime annotations ensure that this flexible code won't have any dangling references. And all of this happens at compile time so that run-time performance isn't affected!

Believe it or not, there's even more to learn in these areas: Chapter 17 will discuss trait objects, which are another way to use traits. Chapter 19 will be covering more complex scenarios involving lifetime annotations. Chapter 20 will get to some advanced type system features. Up next, though, let's talk about how to write tests in Rust so that we can make sure our code using all these features is working the way we want it to!