Ownership and functions in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to pass ownership of the string to the function.

Rust

fn take_ownership(s: String) {
    println!("{}", s);
}

fn main() {
    let s = String::from("hello");
    take_ownership([1]);
}

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

As

B&s

Cs.clone()

Dref s

Attempts:

3 left

💡 Hint

Common Mistakes

Using &s instead of s causes borrowing, not ownership transfer.

Using s.clone() copies the data, not moves ownership.

Passing s moves ownership to the function. Using s.clone() copies, &s borrows, and ref s is invalid here.

2fill in blank

medium

Complete the code to borrow the string without taking ownership.

Rust

fn calculate_length(s: &String) -> usize {
    s.len()
}

fn main() {
    let s1 = String::from("hello");
    let len = calculate_length([1]);
    println!("Length: {}", len);
}

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

A&s1

Bs1.clone()

Cs1

D*s1

Attempts:

3 left

💡 Hint

Common Mistakes

Passing s1 moves ownership, causing errors after the call.

Using s1.clone() copies data unnecessarily.

Using &s1 borrows the string, allowing the function to read it without taking ownership.

3fill in blank

hard

Fix the error in the function signature to allow mutable borrowing.

Rust

fn change(s: [1] String) {
    s.push_str(", world");
}

fn main() {
    let mut s = String::from("hello");
    change(&mut s);
    println!("{}", s);
}

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

Aref

B&

Cmut

D&mut

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of &mut prevents modification.

Using mut without & is invalid in function parameters.

The function must accept a mutable reference &mut String to modify the string without taking ownership.

4fill in blank

hard

Fill both blanks to correctly return ownership and length from the function.

Rust

fn calculate_length_and_return(s: [1]) -> (String, usize) {
    let length = s.len();
    ([2], length)
}

fn main() {
    let s1 = String::from("hello");
    let (s2, len) = calculate_length_and_return(s1);
    println!("{} has length {}", s2, len);
}

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

AString

Bs

C&String

Ds.clone()

Attempts:

3 left

💡 Hint

Common Mistakes

Using &String as parameter borrows, not owns.

Returning s.clone() copies unnecessarily.

The function takes ownership of String and returns it back with the length. Returning s gives ownership back.

5fill in blank

hard

Fill all three blanks to correctly use mutable references and modify the string.

Rust

fn append_world([2] s: [1] String) {
    s.[3](", world");
}

fn main() {
    let mut s = String::from("hello");
    append_world(&mut s);
    println!("{}", s);
}

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

A&mut

Bmut

Cpush_str

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of &mut prevents modification.

Using push instead of push_str requires a char, not a string slice.

The function takes a mutable reference &mut String and uses push_str to append text.

Practice

1. What happens when a function in Rust takes ownership of a variable passed to it?

easy

A. The variable is shared between the function and the caller simultaneously.

B. The variable is copied and both original and function have ownership.

C. The function borrows the variable temporarily without taking ownership.

D. The original variable can no longer be used after the function call.

Solution

1. Step 1: Understand ownership transfer in Rust functions

   When a function takes ownership, it means the variable's ownership moves to the function.

2. Step 2: Effect on original variable after function call

   Since ownership moved, the original variable cannot be used anymore after the call.

3. Final Answer:

   The original variable can no longer be used after the function call. -> Option D

4. Quick Check:

   Ownership transfer = original variable unusable [OK]

Hint: Ownership moves to function, original variable becomes invalid [OK]

Common Mistakes:

• Thinking ownership is shared, not moved
• Assuming variable is copied automatically
• Confusing borrowing with ownership transfer

2. Which of the following function signatures correctly borrows a string slice without taking ownership?

easy

A. fn print_name(name: String) { println!("{}", name); }

B. fn print_name(name: &String) { println!("{}", name); }

C. fn print_name(name: &str) { println!("{}", name); }

D. fn print_name(name: String) -> String { name }

Solution

1. Step 1: Identify borrowing syntax for string slices

   Borrowing a string slice uses &str type, which is a reference to a string slice.

2. Step 2: Check each signature for borrowing without ownership

   fn print_name(name: &str) { println!("{}", name); } uses &str, which borrows without taking ownership. fn print_name(name: &String) { println!("{}", name); } borrows a String reference but &str is more idiomatic for slices.

3. Final Answer:

   fn print_name(name: &str) { println!("{}", name); } -> Option C

4. Quick Check:

   Borrow string slice = &str parameter [OK]

Hint: Use &str to borrow string slices without ownership [OK]

Common Mistakes:

• Using String parameter takes ownership
• Confusing &String with &str
• Returning ownership when not needed

3. What will be the output of this Rust code?

fn take_ownership(s: String) {
    println!("Inside function: {}", s);
}

fn main() {
    let s = String::from("hello");
    take_ownership(s);
    println!("Outside function: {}", s);
}

medium

A. Inside function: hello
Compile error: use of moved value 's'

B. Inside function: hello
Outside function: hello

C. Inside function: hello
Outside function: <empty>

D. Runtime error: variable s not found

Solution

1. Step 1: Analyze ownership transfer in function call

   The function take_ownership takes ownership of s, so s is moved into the function.

2. Step 2: Check usage of s after function call

   After the call, s is no longer valid in main, so using it causes a compile error.

3. Final Answer:

   Inside function: hello Compile error: use of moved value 's' -> Option A

4. Quick Check:

   Ownership moved, use after move = compile error [OK]

Hint: Using moved variable after function call causes compile error [OK]

Common Mistakes:

• Assuming variable is copied automatically
• Expecting output after ownership moved
• Ignoring Rust's ownership rules

4. Identify the error in this Rust code and how to fix it:

fn calculate_length(s: String) -> usize {
    s.len()
}

fn main() {
    let s1 = String::from("hello");
    let len = calculate_length(s1);
    println!("The length of '{}' is {}.", s1, len);
}

medium

A. Error: s1 is moved into calculate_length; fix by passing &s1 instead.

B. Error: calculate_length returns wrong type; fix by returning &usize.

C. Error: println! macro syntax incorrect; fix by removing quotes.

D. No error; code runs and prints length correctly.

Solution

1. Step 1: Identify ownership issue in function call

   calculate_length takes ownership of s1, so s1 is moved and invalid after call.

2. Step 2: Fix by borrowing s1 instead of moving

   Change function parameter to &String and pass &s1 to avoid moving ownership.

3. Final Answer:

   Error: s1 is moved into calculate_length; fix by passing &s1 instead. -> Option A

4. Quick Check:

   Pass reference to avoid moving ownership [OK]

Hint: Pass references to keep ownership when calling functions [OK]

Common Mistakes:

• Trying to return reference to local variable
• Ignoring ownership move on function parameters
• Misunderstanding borrowing syntax

5. You want to write a function that takes ownership of a vector, doubles each element, and returns ownership back. Which function signature and body correctly implement this in Rust?

hard

A. fn double_vec(v: &Vec<i32>) -> Vec<i32> { v.iter().map(|x| x * 2).collect() }

B. fn double_vec(v: Vec<i32>) -> Vec<i32> { v.iter().map(|x| x * 2).collect() }

C. fn double_vec(v: Vec<i32>) { for x in &v { println!("{}", x * 2); } }

D. fn double_vec(v: &mut Vec<i32>) { for x in v { *x *= 2; } }

Solution

1. Step 1: Understand ownership and return requirements

   The function must take ownership of the vector and return ownership back after doubling elements.

2. Step 2: Analyze each option for ownership and functionality

   fn double_vec(v: Vec<i32>) -> Vec<i32> { v.iter().map(|x| x * 2).collect() } takes ownership (Vec<i32>), creates a new doubled vector, and returns it, satisfying requirements.

3. Final Answer:

   fn double_vec(v: Vec<i32>) -> Vec<i32> { v.iter().map(|x| x * 2).collect() } -> Option B

4. Quick Check:

   Take ownership, return new doubled vector [OK]

Hint: Take Vec ownership and return new Vec after processing [OK]

Common Mistakes:

• Borrowing vector but not returning ownership
• Not returning anything from function
• Modifying borrowed vector without ownership