Using if as expression in Rust - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of if expression with integer values

What is the output of this Rust code snippet?

Rust

fn main() {
    let x = 10;
    let y = if x > 5 { 100 } else { 50 };
    println!("{}", y);
}

A 100

B 50

C Compilation error

D 10

💡 Hint

Remember that if in Rust can return a value and be assigned to a variable.

Since x is 10, which is greater than 5, the if expression returns 100, which is assigned to y and printed.

❓ Predict Output

intermediate

Using if expression with different types

What error or output does this Rust code produce?

Rust

fn main() {
    let condition = true;
    let value = if condition { 5 } else { "five" };
    println!("{:?}", value);
}

A Compilation error due to mismatched types

B 5

C "five"

D Runtime panic

💡 Hint

Both branches of an if expression must have the same type in Rust.

The if expression branches return different types: integer and string slice, causing a compilation error.

🔧 Debug

advanced

Identify the error in if expression usage

What is the error in this Rust code snippet?

Rust

fn main() {
    let number = 7;
    let result = if number > 5 {
        "big"
    } else if number < 5 {
        "small"
    };
    println!("{}", result);
}

A Prints "small"

B Prints "big"

C Compilation error: missing else branch for if expression

D Runtime panic due to uninitialized variable

💡 Hint

If expressions must cover all cases to produce a value.

The if expression lacks an else branch for the case when number == 5, so result is not assigned a value, causing a compilation error.

🧠 Conceptual

advanced

Why use if as an expression in Rust?

Which of the following best explains why Rust uses if as an expression rather than a statement?

A It prevents the use of else blocks in the code.

B It forces all conditions to be true before executing code.

C It disables the use of loops inside if blocks.

D It allows assigning the result of the if directly to a variable, making code concise and expressive.

💡 Hint

Think about how expressions can produce values.

Using if as an expression lets you assign its result to variables, which makes code shorter and clearer.

❓ Predict Output

expert

Output of nested if expressions with shadowing

What is the output of this Rust program?

Rust

fn main() {
    let x = 3;
    let y = if x > 2 {
        let x = x * 2;
        if x > 5 { x } else { x + 1 }
    } else {
        0
    };
    println!("{}", y);
}

A 7

B 6

C 0

D Compilation error due to shadowing

💡 Hint

Remember that inner let shadows outer variables and if expressions return values.

x is 3 > 2 true, so let x = 3 * 2 = 6 (shadowing outer x). Inner if 6 > 5 true returns 6, so y = 6 printed.