Function ownership interaction in Rust - Mini Project: Build & Apply

Function ownership interaction

📖 Scenario: Imagine you have a small program that manages a book title. You want to see how ownership works when you pass the book title to a function and get it back.

🎯 Goal: You will create a string variable for a book title, write a function that takes ownership of this string and returns it, and then print the returned title.

📋 What You'll Learn

Create a variable called book_title with the value "Rust Programming"

Write a function called take_and_return that takes ownership of a String parameter called title and returns it

Call the function take_and_return with book_title and store the result in a variable called returned_title

Print the value of returned_title

💡 Why This Matters

🌍 Real World

Understanding ownership is key in Rust to manage memory safely without a garbage collector.

💼 Career

Rust developers must master ownership rules to write efficient and safe code, especially when working with functions and data passing.

1

Create the initial book title

Create a variable called book_title and set it to the string "Rust Programming" using String::from.

Rust

# Create a variable called book_title with value "Rust Programming"
# Your code here

Hint

Use let book_title = String::from("Rust Programming"); to create the string.

2

Write the function to take and return ownership

Write a function called take_and_return that takes ownership of a String parameter named title and returns it.

Rust

let book_title = String::from("Rust Programming");
// Write the function take_and_return below
// Your code here

Hint

The function should accept title: String and return it directly.

3

Call the function and store the returned value

Call the function take_and_return with book_title and store the result in a variable called returned_title.

Rust

let book_title = String::from("Rust Programming");

fn take_and_return(title: String) -> String {
    title
}

// Call take_and_return with book_title and store in returned_title
// Your code here

Hint

Use let returned_title = take_and_return(book_title); to call the function and save the result.

4

Print the returned book title

Print the value of returned_title using println!.

Rust

let book_title = String::from("Rust Programming");

fn take_and_return(title: String) -> String {
    title
}

let returned_title = take_and_return(book_title);

// Print returned_title
// Your code here

Hint

Use println!("{}", returned_title); to display the 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