Move semantics in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move ownership of the String to variable b.

Rust

let a = String::from("hello");
let b = [1];

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

Aref a

B&a

Ca

Da.clone()

Attempts:

3 left

💡 Hint

Common Mistakes

Using &a borrows instead of moves ownership.

Using a.clone() creates a copy, not a move.

In Rust, assigning a to b moves ownership, so b owns the String now.

2fill in blank

medium

Complete the code to fix the error by cloning the String before moving.

Rust

let s1 = String::from("hello");
let s2 = [1];

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

As1

Bs1.clone()

C&s1

Dref s1

Attempts:

3 left

💡 Hint

Common Mistakes

Using s1 moves ownership, invalidating s1.

Using &s1 only borrows, not clones.

Cloning s1 creates a new String, so s2 owns a copy and s1 remains valid.

3fill in blank

hard

Fix the error by completing the code to borrow the String instead of moving it.

Rust

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

let s = String::from("hello");
print_string(s);
print_string(s);

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

A&String

BString

Cstr

D&str

Attempts:

3 left

💡 Hint

Common Mistakes

Using String moves ownership, so second call fails.

Using str is a string slice type, but not a reference.

Borrowing with &String allows multiple uses without moving ownership.

4fill in blank

hard

Complete the code to create a vector of Strings by moving ownership inside the loop.

Rust

let words = vec![String::from("apple"), String::from("banana")];
let mut new_words = Vec::new();
for word in words[1] {
    new_words.push(word);
}

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

A.into_iter()

Bclone()

C.clone()

Attempts:

3 left

💡 Hint

Common Mistakes

Using iter() borrows, causing ownership errors.

Cloning unnecessarily copies data.

Using into_iter() moves ownership of each String from words. No need to clone when pushing.

5fill in blank

hard

Fill all three blanks to implement a function that takes ownership of a String, modifies it, and returns it.

Rust

fn add_exclamation[1] {
    let mut s = [2];
    s.push('!');
    [3]
}

let greeting = String::from("Hello");
let excited = add_exclamation(greeting);

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

A(s: String) -> String

Bs

D(s: &String) -> String

Attempts:

3 left

💡 Hint

Common Mistakes

Using a reference in the signature causes ownership issues.

Not returning the String loses ownership.

The function takes ownership of s, modifies it, and returns it to transfer ownership back.

Practice

1. What happens to a variable in Rust after its value is moved to another variable?

easy

A. The original variable can no longer be used.

B. The original variable keeps a copy of the value.

C. The original variable is automatically cloned.

D. The original variable becomes a reference to the new variable.

Solution

1. Step 1: Understand move semantics

   In Rust, moving a value transfers ownership from one variable to another.

2. Step 2: Effect on original variable

   After the move, the original variable is invalid and cannot be used to prevent double free errors.

3. Final Answer:

   The original variable can no longer be used. -> Option A

4. Quick Check:

   Move means original variable unusable [OK]

Hint: Moving means original variable is invalid after transfer [OK]

Common Mistakes:

• Thinking the original variable still holds a copy
• Assuming Rust automatically clones on move
• Confusing move with borrowing

2. Which of the following Rust code snippets correctly moves ownership of a String from a to b?

easy

A. let b = a.clone();

B. let b = &a;

C. let b = a;

D. let b = *a;

Solution

1. Step 1: Identify move syntax

   Assigning let b = a; moves ownership of a to b.

2. Step 2: Check other options

   &a borrows, a.clone() copies, *a is invalid for String.

3. Final Answer:

   let b = a; -> Option C

4. Quick Check:

   Simple assignment moves ownership [OK]

Hint: Use simple assignment to move ownership, not references or clone [OK]

Common Mistakes:

• Using references instead of moving
• Using clone which copies instead of moves
• Dereferencing a String which is invalid

3. What is the output of this Rust code?

let s1 = String::from("hello");
let s2 = s1;
println!("{}", s1);

medium

A. Compilation error

B. hello

C. Runtime error

D. Empty string

Solution

1. Step 1: Analyze ownership transfer

   Assigning s2 = s1 moves ownership of the String from s1 to s2.

2. Step 2: Using moved variable

   Trying to print s1 after move causes a compile-time error because s1 is invalid.

3. Final Answer:

   Compilation error -> Option A

4. Quick Check:

   Using moved variable causes compile error [OK]

Hint: Using moved variable causes compile error, not runtime [OK]

Common Mistakes:

• Expecting output 'hello' despite move
• Thinking it causes runtime error
• Assuming Rust clones automatically

4. Identify the error in this Rust code related to move semantics:

fn main() {
    let v = vec![1, 2, 3];
    let v2 = v;
    println!("{}", v.len());
}

medium

A. Vector v cannot be moved to v2.

B. Cannot print vector length with println!.

C. Missing semicolon after let v2 = v.

D. Using v.len() after v was moved to v2.

Solution

1. Step 1: Understand move of vector

   Assigning v2 = v moves ownership of vector v to v2.

2. Step 2: Using moved variable causes error

   Calling v.len() after move is invalid and causes compile error.

3. Final Answer:

   Using v.len() after v was moved to v2. -> Option D

4. Quick Check:

   Using moved variable causes compile error [OK]

Hint: After move, original variable cannot be used [OK]

Common Mistakes:

• Thinking vector cannot be moved
• Ignoring ownership rules after move
• Confusing move with borrowing

5. Given this Rust function:

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

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

What is the best way to fix the code so main can print s after calling take_ownership?

hard

A. Clone s before passing: take_ownership(s.clone()).

B. All of the above.

C. Return s from take_ownership and assign it back.

D. Change take_ownership to take a reference: fn take_ownership(s: &String).

Solution

1. Step 1: Understand ownership transfer in function

   Passing s to take_ownership moves ownership, invalidating s in main.

2. Step 2: Ways to keep s usable

   Change take_ownership to take a reference: fn take_ownership(s: &String). borrows s so ownership is not moved.
   Clone s before passing: take_ownership(s.clone()). clones s so original remains.
   Return s from take_ownership and assign it back. returns ownership back to main.

3. Step 3: Best fix

   All three options correctly fix the problem.

4. Final Answer:

   All of the above. -> Option B

5. Quick Check:

   Borrow, clone, or return ownership fixes [OK]

Hint: Borrow, clone, or return ownership to keep variable usable [OK]

Common Mistakes:

• Ignoring ownership transfer in function calls
• Not cloning when needed
• Not returning ownership back