Lifetime annotations in Rust - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust code with lifetime annotations?

Consider the following Rust code snippet. What will it print when run?

Rust

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("apple");
    let string2 = "banana";
    let result = longest(string1.as_str(), string2);
    println!("{}", result);
}

A apple

B Compilation error due to lifetime mismatch

C applebanana

D banana

💡 Hint

Look at which string slice is longer and how lifetimes are used to ensure safety.

The function longest returns the string slice with the longer length. "banana" is longer than "apple", so it prints "banana". The lifetime annotations ensure both inputs and output live long enough.

❓ Predict Output

intermediate

What error does this Rust code produce?

Analyze the following Rust code. What error will the compiler produce?

Rust

fn invalid_ref<'a>() -> &'a str {
    let s = String::from("hello");
    &s
}

fn main() {
    let r = invalid_ref();
    println!("{}", r);
}

A error: `s` does not live long enough

B error: missing lifetime specifier

C error: cannot return value referencing local variable `s`

D No error, prints "hello"

💡 Hint

Think about the lifetime of local variables and returned references.

The function tries to return a reference to a local variable `s` which is dropped at the end of the function. This causes a compilation error: cannot return value referencing local variable `s`.

🔧 Debug

advanced

Does this Rust code compile and what does it print?

Examine the code below. What is the correct outcome?

Rust

fn choose_first<'a>(x: &'a str, y: &str) -> &'a str {
    x
}

fn main() {
    let s1 = String::from("hello");
    let result;
    {
        let s2 = String::from("world");
        result = choose_first(s1.as_str(), s2.as_str());
    }
    println!("{}", result);
}

A Compilation error because `choose_first` has mismatched lifetimes

B No compilation error, prints "hello"

C Compilation error because `result` has an invalid lifetime

D Compilation error because `s2` does not live long enough

💡 Hint

The function always returns `x`, and `y` is unused in the return value. Check if the borrow checker allows `s2` to be dropped early.

No compilation error occurs because `choose_first` always returns the reference `x` (`s1.as_str()`), and `s1` lives until the end of `main`. The reference `y` (`s2.as_str()`) is borrowed only during the function call and not stored or returned, so dropping `s2` after the block is fine. The code compiles and prints "hello".

📝 Syntax

advanced

Which option correctly declares a function with lifetime annotations?

Choose the correct Rust function signature that returns a reference with a lifetime tied to the input reference.

A fn get_str<'a>(s: &'a str) -> &'a str

B fn get_str(s: &str) -> &str<'a>

C fn get_str(s: &'a str) -> &'a str

D fn get_str<'a>(s: &str) -> &'a str

💡 Hint

Remember the syntax for lifetime parameters in function signatures.

Option A correctly declares a lifetime parameter `'a` and uses it for both input and output references. Option A has invalid syntax placing lifetime after &str. Option A uses `'a` without declaring it. Option A declares `'a` but input reference lacks it, causing mismatch.

🚀 Application

expert

How many items are in the resulting vector after this Rust code runs?

Given the following Rust code using lifetime annotations, how many elements does the vector `results` contain after execution?

Rust

fn collect_refs<'a>(inputs: Vec<&'a str>) -> Vec<&'a str> {
    let mut results = Vec::new();
    for s in inputs {
        results.push(s);
    }
    results
}

fn main() {
    let s1 = String::from("one");
    let s2 = String::from("two");
    let s3 = String::from("three");
    let inputs = vec![s1.as_str(), s2.as_str(), s3.as_str()];
    let results = collect_refs(inputs);
    println!("{}", results.len());
}

A 3

B 1

C Compilation error due to lifetime mismatch

D 0

💡 Hint

Consider the lifetimes of the strings and the references collected.

All three strings live long enough for their references to be collected into the vector. The function collects all references without issue, so the resulting vector has 3 elements.