Rust lifetimes

<'a>

Syntax used to declare and use a lifetime parameter named 'a

Zero-Copy Parser Example

struct Highlight<'a> { // 1. We DECLARE a lifetime generic named 'a
    content: &'a str,  // 2. We USE 'a to tag this specific field
}

We can think of 'a kind of like a variable, similar to x=5. But instead of a value it holds the duration of time (scope).

<'a>: We are saying "This struct cannot exist on its own, it is dependent on some data that lives for a duration we call 'a."
&'a str: This tells use that this holds a reference to a string, that must live at least as long as 'a.

The Highlight struct is essentially signing a contract with the compiler stating

I, the struct Highlight, promise that I will never outlive the data inside content. If the string data goes out of scope, I must go out of scope too.

Parser Function

let raw_data = String::from("Header data... IMPORTANT: This is the payload ...footer data");
let highlight = find_important_part(&raw_data);

fn find_important_part<'a>(text:  &'a str) -> Highlight<'a> {
    let parts: Vec<&str> = text.split("IMPORTANT:").collect();
    if parts.len() > 1 {
        // We want the part after "IMPORTANT:"
        Highlight {
            content:  parts[1].trim(),
        }
    } else {
        Highlight { content: "" }
    }
}

My initial confusion here was here:

Highlight {
    content:  parts[1].trim(),
}

Setting content to parts[1].trim() looked like we were just "getting" a lifetime reference from nowhere, because I had stated the struct Highlight is defined needing content: &'a str,. Digging a little deeper I found Rust's Lifetime Elision rules. Which means although the trim() function in the std written as pub fn trim(&self) -> &str is expanded to fn trim<'b>(&'b self) -> &'b str. Which satisfies our definition.

When multiple variables share the same lifetime, Rust will also scope it to the smallest time frame.

Longest String Example

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

fn main() {
    let s1 = String::from("small");
    let s2 = String::from("longer string");
    println!("Longest: {}", longer(&s1, &s2));
}

Here we have an example of both x and y using the same lifetime 'a, this code works fine because both s1 and s2 live with the main() scope, so they both end at the same time. But what if we changed to scope of one of them?

let s1 = String::from("long lived");
let result;
{
    let s2 = String::from("short");
    // We pass &s1 (long) and &s2 (short) to 'longer'
    // Therefore, 'a becomes the lifespan of s2 (the shorter one)
    result = longer(&s1, &s2);
} // s2 dies here, so 'a ends here.

// even if 'result' actually points to 's1' at runtime!
println!("Result: {}", result);
}
// Error: `s2 does not live long enough
// Error: borrowed value does not live long enough

The solution here is to deal with the ownership of the string. Instead of &'a str we can change the return to String. The con here is we are taking a small performance hit, this is probably still an acceptable solution unless you are writing some high performance code or working with embeded systems.

The Embeded systems way which has higher performance would be create a type that can hold either type.

// Can hold EITHER a reference to 'a OR a reference to 'b
enum Longest<'a, 'b> {
    Left(&'a str),
    Right(&'b str),
}

fn longer<'a, 'b>(x: &'a str, y: &'b str) -> Longest<'a, 'b> {
    if x.len() > y.len() {
        Longest::Left(x)
    } else {
        Longest::Right(y)
    }
}

fn main() {
    let s1 = String::from("long lived");
    let result;
    {
        let s2 = String::from("short");
        
        // We get back an Enum, preserving the specific lifetime of what we picked
        let outcome = longer(&s1, &s2);
        
        match outcome {
            Longest::Left(val) => {
                // val is &'a str (Alive!)
                result = val; 
                println!("Result is safe: {}", result);
            },
            Longest::Right(val) => {
                // val is &'b str (About to die!)
                println!("Result is short-lived: {}", val);
            }
        }
    } // s2 dies here. 
      // If we had matched 'Left', 'result' would still be valid
}

<'a>

Zero-Copy Parser Example

Parser Function

Variables Sharing Lifetimes

Longest String Example