Function ownership interaction 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 print_string(s: String) {
    println!("{}", s);
}

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

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

Amy_string

B&my_string

Cmy_string.clone()

D"hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a reference instead of the owned value.

Passing my_string moves ownership to the function.

2fill in blank

medium

Complete the code to borrow the string without taking ownership.

Rust

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

fn main() {
    let my_string = String::from("world");
    print_string([1]);
    println!("{}", my_string);
}

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

Amy_string

B"world"

Cmy_string.clone()

D&my_string

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the owned value, which moves ownership.

Passing &my_string borrows the string, so ownership stays in main.

3fill in blank

hard

Fix the error by completing the code to return ownership back to the caller.

Rust

fn take_and_return(s: String) -> String {
    println!("{}", s);
    [1]
}

fn main() {
    let my_string = String::from("rust");
    let my_string = take_and_return(my_string);
    println!("{}", my_string);
}

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

As

B&s

Cs.clone()

Dreturn &s

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a reference instead of the owned value.

Returning s moves ownership back to the caller.

4fill in blank

hard

Fill both blanks to create a function that borrows a string slice and prints it.

Rust

fn print_slice(s: [1]) {
    println!("{}", s);
}

fn main() {
    let my_string = String::from("slice");
    print_slice([2]);
}

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

A&str

B&String

Cmy_string.as_str()

Dmy_string

Attempts:

3 left

💡 Hint

Common Mistakes

Using &String instead of &str in the function parameter.

The function takes a string slice &str, and we pass a slice using my_string.as_str().

5fill in blank

hard

Fill all three blanks to create a function that takes ownership, modifies, and returns the string.

Rust

fn add_exclamation(mut s: [1]) -> [2] {
    s.push_str([3]);
    s
}

fn main() {
    let my_string = String::from("hello");
    let my_string = add_exclamation(my_string);
    println!("{}", my_string);
}

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

AString

C"!"

D&str

Attempts:

3 left

💡 Hint

Common Mistakes

Using &str as parameter or return type instead of String.

The function takes ownership of a String, appends an exclamation mark, and returns the modified String.

Practice

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

easy

A. The function borrows the variable temporarily.

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

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

D. The variable is automatically cloned inside the function.

Solution

1. Step 1: Understand ownership transfer

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

2. Step 2: Effect on original variable

   After the function call, the original variable is no longer valid and cannot be used.

3. Final Answer:

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

4. Quick Check:

   Ownership moves = original invalid [OK]

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

Common Mistakes:

• Thinking ownership is copied, not moved
• Confusing borrowing with ownership transfer
• Assuming automatic cloning happens

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

easy

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

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

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

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

Solution

1. Step 1: Identify borrowing syntax

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

2. Step 2: Match function signature

   fn print_name(name: &str) { println!("{}", name); } uses &str, correctly borrowing without taking ownership.

3. Final Answer:

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

4. Quick Check:

   Borrow string slice = &str [OK]

Hint: Use &str to borrow string slices, not String type [OK]

Common Mistakes:

• Using String type takes ownership
• Using &String borrows but is less flexible than &str
• Returning String does not borrow

3. What is the output of the following Rust code?

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

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

medium

A. Inside: hello
Outside: hello

B. Inside: hello
Outside:

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

D. Runtime error: use after free

Solution

1. Step 1: Analyze ownership transfer in function call

   The function take_ownership takes ownership of s, moving it inside the function.

2. Step 2: Check usage after ownership moved

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

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Assuming variable is still valid after move
• Expecting runtime error instead of compile error
• Thinking variable is copied automatically

4. Identify the error in this Rust code and choose the fix:

fn add_one(num: &i32) -> i32 {
    *num + 1
}

fn main() {
    let x = 5;
    let y = add_one(x);
    println!("{}", y);
}

medium

A. No error, code runs correctly.

B. Change function parameter to num: i32 to take ownership.

C. Change return type to &i32 to return a reference.

D. Change call to add_one(&x) to pass a reference.

Solution

1. Step 1: Check function parameter type

   Function expects a reference &i32, but main passes x (i32) directly.

2. Step 2: Fix function call

   Pass a reference &x to match parameter type and avoid type mismatch error.

3. Final Answer:

   Change call to add_one(&x) to pass a reference. -> Option D

4. Quick Check:

   Pass reference when function expects &i32 [OK]

Hint: Match argument type with parameter type exactly [OK]

Common Mistakes:

• Passing value instead of reference
• Changing return type unnecessarily
• Ignoring compiler type mismatch errors

5. Given this Rust code, what is the correct way to write a function that takes ownership of a vector, adds an element, and returns ownership back?

fn add_element(vec: Vec<i32>) -> Vec<i32> {
    // Add 10 to the vector
    vec.push(10);
    vec
}

hard

A. fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec }

B. fn add_element(vec: &Vec<i32>) -> Vec<i32> { vec.push(10); vec.clone() }

C. fn add_element(vec: Vec<i32>) { vec.push(10); }

D. fn add_element(vec: &mut Vec<i32>) { vec.push(10); }

Solution

1. Step 1: Understand ownership and mutability

   To modify the vector, the function must take ownership and have mutable access, so parameter must be mutable.

2. Step 2: Return ownership after modification

   Return the modified vector to give ownership back to the caller.

3. Final Answer:

   fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec } -> Option A

4. Quick Check:

   Mutable ownership + return ownership = fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec } [OK]

Hint: Use mut parameter and return vector to keep ownership [OK]

Common Mistakes:

• Forgetting mut keyword for parameter
• Trying to modify borrowed vector without mut
• Not returning the vector after modification