Why lifetimes exist in Rust - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a reference with a lifetime.

Rust

fn print_str<[1]>(s: &[1] str) { println!("{}", s); }

Drag options to blanks, or click blank then click option'

A'a

Bstatic

Cmut

Dref

Attempts:

3 left

💡 Hint

Common Mistakes

Using mut or ref instead of a lifetime parameter.

The 'a syntax declares a lifetime parameter for the reference &'a str.

2fill in blank

medium

Complete the function signature to specify that the returned reference has the same lifetime as the input.

Rust

fn first_char<[1]>(s: &[1] str) -> &[1] str { &s[0..1] }

Drag options to blanks, or click blank then click option'

A'static

B'b

C'a

D'z

Attempts:

3 left

💡 Hint

Common Mistakes

Using different lifetime names for input and output.

The lifetime 'a is used to link the input and output lifetimes, ensuring the returned reference is valid as long as the input.

3fill in blank

hard

Fix the error by adding the correct lifetime annotation to the struct.

Rust

struct Holder<[1]> { value: &[1] str }

Drag options to blanks, or click blank then click option'

A'a

B'static

C'b

D'z

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the lifetime parameter or using inconsistent names.

The struct needs a lifetime parameter 'a to specify how long the reference inside it is valid.

4fill in blank

hard

Fill both blanks to define a function that returns the longer of two string slices with correct lifetimes.

Rust

fn longer<'[1]>(x: &'[1] str, y: &'[1] str) -> &'[2] str { if x.len() > y.len() { x } else { y } }

Drag options to blanks, or click blank then click option'

Aa

Bb

Cc

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Using different lifetime names for inputs and output.

The lifetime 'a is used consistently to tie the input references and the output reference together.

5fill in blank

hard

Fill all three blanks to define a struct with two references having different lifetimes.

Rust

struct Pair<'[1], '[2]> { first: &'[1] str, second: &'[2] str }

Drag options to blanks, or click blank then click option'

Aa

Bb

Cc

Dd

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same lifetime for both references when different lifetimes are needed.

The struct uses two different lifetime parameters 'a and 'b for its two references.

Practice

1.

Why does Rust use lifetimes like 'a in its code?

easy

A. To automatically free memory like a garbage collector

B. To make the code run faster by skipping checks

C. To allow multiple mutable references at the same time

D. To ensure references do not outlive the data they point to

Solution

1. Step 1: Understand what lifetimes represent

   Lifetimes in Rust show how long a reference is valid to prevent dangling references.

2. Step 2: Connect lifetimes to safety

   They ensure references do not outlive the data they point to, avoiding bugs.

3. Final Answer:

   To ensure references do not outlive the data they point to -> Option D

4. Quick Check:

   Lifetimes = prevent dangling references [OK]

Hint: Lifetimes prevent references to missing data [OK]

Common Mistakes:

• Thinking lifetimes speed up code
• Confusing lifetimes with garbage collection
• Believing lifetimes allow multiple mutable references

2.

Which of the following is the correct way to declare a function with a lifetime parameter 'a for a reference argument in Rust?

fn example(____ s: &'a str) -> &'a str { s }

easy

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

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

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

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

Solution

1. Step 1: Recall lifetime syntax in function signatures

   Lifetimes are declared before the function name with 'a and used on references.

2. Step 2: Match correct syntax

   fn example<'a>(s: &'a str) -> &'a str correctly declares 'a and uses it on both input and output references.

3. Final Answer:

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

4. Quick Check:

   Lifetime syntax = fn example<'a>(s: &'a str) -> &'a str [OK]

Hint: Declare lifetime before fn name and use on references [OK]

Common Mistakes:

• Putting lifetime after & instead of before type
• Forgetting to declare lifetime before function name
• Using lifetime on non-reference types

3.

What will this Rust code print?

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

fn main() {
    let s1 = String::from("apple");
    let s2 = "banana";
    let result = longest(&s1, s2);
    println!("{}", result);
}

medium

A. apple

B. banana

C. Compilation error due to lifetimes

D. Runtime error due to invalid reference

Solution

1. Step 1: Understand function longest behavior

   It returns the longer string slice between x and y.

2. Step 2: Compare lengths of "apple" and "banana"

   "apple" has 5 letters, "banana" has 6, so it returns "banana".

3. Final Answer:

   banana -> Option B

4. Quick Check:

   Longest string = banana [OK]

Hint: Compare string lengths carefully to find longest [OK]

Common Mistakes:

• Assuming s1 is longer without counting
• Thinking lifetimes cause compile error here
• Confusing references with ownership causing runtime error

4.

Identify the error in this Rust code related to lifetimes:

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 word;
    {
        let s = String::from("hello world");
        word = first_word(&s);
    }
    println!("{}", word);
}

medium

A. Error: cannot borrow s as mutable

B. Error: missing lifetime annotation in function signature

C. Error: returned reference may outlive borrowed data

D. No error, code runs fine

Solution

1. Step 1: Analyze lifetime of variable s

   Variable s is created inside inner block and dropped after it ends.

2. Step 2: Check lifetime of returned reference word

   word holds reference to s, but s is dropped before println!, causing dangling reference.

3. Final Answer:

   Error: returned reference may outlive borrowed data -> Option C

4. Quick Check:

   Dangling reference error = Error: returned reference may outlive borrowed data [OK]

Hint: References must not outlive their data scope [OK]

Common Mistakes:

• Ignoring scope of s and thinking code is safe
• Assuming missing lifetime annotation causes error here
• Confusing mutable borrow errors with lifetime errors

5.

You want to write a function that returns a reference to the longer string slice between two inputs, but one input is owned String and the other is a string slice &str. How should you use lifetimes to make this safe and correct?

hard

A. Convert the owned String to a slice and use a single lifetime for both inputs and output

B. Return the owned String directly without lifetimes

C. Use different lifetimes for each input and output to avoid conflicts

D. Clone both inputs and return an owned String without references

Solution

1. Step 1: Understand input types and lifetimes

   Owned String must be converted to a string slice (&str) to use lifetimes uniformly.

2. Step 2: Use one lifetime parameter for both inputs and output

   This ensures the returned reference is valid as long as both inputs are valid.

3. Final Answer:

   Convert the owned String to a slice and use a single lifetime for both inputs and output -> Option A

4. Quick Check:

   Single lifetime with slices = Convert the owned String to a slice and use a single lifetime for both inputs and output [OK]

Hint: Use &str slices and one lifetime for safe references [OK]

Common Mistakes:

• Returning owned String without references when references expected
• Using multiple lifetimes causing borrow conflicts
• Forgetting to convert String to &str before comparison