Function ownership interaction in Rust

Introduction

Functions in Rust can take ownership of values, borrow them, or return ownership. Understanding this helps you manage memory safely without errors.

When you want a function to use data without changing the original.

When you want a function to take full control of data and possibly modify or drop it.

When you want to return ownership of data from a function to use it later.

When you want to avoid copying large data by borrowing it temporarily.

Syntax

Rust

fn function_name(param: Type) -> ReturnType {
    // function body
}

Parameters can be passed by value (taking ownership) or by reference (borrowing).

Returning a value transfers ownership back to the caller.

Examples

This function takes ownership of a String. After calling it, the original String cannot be used.

Rust

fn take_ownership(s: String) {
    println!("{}", s);
} // s is dropped here

This function borrows a String reference. The original String can still be used after the call.

Rust

fn borrow_string(s: &String) {
    println!("{}", s);
} // s is borrowed, not owned

This function creates a String and returns it, transferring ownership to the caller.

Rust

fn give_ownership() -> String {
    let s = String::from("hello");
    s // ownership moved to caller
}

Sample Program

This program shows three ways functions interact with ownership: taking ownership, borrowing, and returning ownership.

Rust

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

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

fn give_ownership() -> String {
    let s = String::from("hello from function");
    s
}

fn main() {
    let s1 = String::from("hello");
    take_ownership(s1);
    // s1 can no longer be used here

    let s2 = String::from("world");
    borrow_string(&s2);
    println!("After borrow: {}", s2); // s2 still valid

    let s3 = give_ownership();
    println!("Received ownership: {}", s3);
}

Important Notes

When a function takes ownership, the original variable can no longer be used after the call.

Borrowing with references allows safe temporary access without losing ownership.

Returning ownership lets you create and send data back to the caller.

Summary

Functions can take ownership, borrow, or return ownership of values.

Ownership rules help prevent bugs and memory errors.

Use borrowing to let functions read data without taking ownership.

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