If–else expression in Rust - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple if–else expression

What is the output of this Rust code snippet?

Rust

fn main() {
    let x = 10;
    let y = if x > 5 { "big" } else { "small" };
    println!("{}", y);
}

A small

B Compilation error

C 10

D big

💡 Hint

Check the condition x > 5 and what the if–else expression returns.

The variable y is assigned the string "big" because x is 10, which is greater than 5.

❓ Predict Output

intermediate

If–else expression with numeric result

What value does variable result hold after running this code?

Rust

fn main() {
    let a = 3;
    let result = if a % 2 == 0 { a * 2 } else { a + 1 };
    println!("{}", result);
}

A 4

B 3

C 6

D Compilation error

💡 Hint

Check if a is even or odd and what the if–else expression returns.

Since 3 is odd, the else branch runs, so result is 3 + 1 = 4.

❓ Predict Output

advanced

Nested if–else expression output

What is printed by this Rust program?

Rust

fn main() {
    let n = 7;
    let description = if n < 5 {
        "small"
    } else if n < 10 {
        "medium"
    } else {
        "large"
    };
    println!("{}", description);
}

A large

B small

C medium

D Compilation error

💡 Hint

Check the conditions in order and see which branch matches n = 7.

7 is not less than 5, but it is less than 10, so the second branch runs.

❓ Predict Output

advanced

If–else expression with mismatched types

What error does this Rust code produce?

Rust

fn main() {
    let flag = true;
    let value = if flag { 10 } else { "ten" };
    println!("{:?}", value);
}

A Runtime panic

B Type mismatch error

C Compilation succeeds and prints 10

D Compilation succeeds and prints "ten"

💡 Hint

Check the types returned by the if and else branches.

The if branch returns an integer, the else branch returns a string slice, causing a type mismatch error.

🧠 Conceptual

expert

Why must if–else expressions have matching types?

In Rust, why must the if and else branches of an if–else expression return the same type?

A Because Rust uses static typing and the expression must have a single, known type at compile time.

B Because Rust allows dynamic typing only inside if–else expressions.

C Because the else branch is optional and can be omitted.

D Because Rust automatically converts all types to strings in if–else expressions.

💡 Hint

Think about how Rust checks types before running the program.

Rust is statically typed, so every expression must have a single, known type at compile time. If branches return different types, the compiler cannot determine the expression's type.