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.

Practice

1. What does using if as an expression in Rust allow you to do?

easy

A. Define functions inside an if statement

B. Only execute code blocks without returning values

C. Assign a value based on a condition directly to a variable

D. Create loops that run while a condition is true

Solution

1. Step 1: Understand the role of if as expression

   In Rust, if can return a value, unlike some languages where it only controls flow.

2. Step 2: Recognize assignment usage

   This means you can assign a variable directly using let x = if condition { value1 } else { value2 };.

3. Final Answer:

   Assign a value based on a condition directly to a variable -> Option C

4. Quick Check:

   if as expression = assign value [OK]

Hint: Think of if as a value picker, not just a decision maker [OK]

Common Mistakes:

• Thinking if can't return values
• Confusing if expression with loops
• Assuming if can define functions

2. Which of the following is the correct syntax for using if as an expression to assign a value in Rust?

easy

A. let x = if (condition) then 10 else 20;

B. let x = if condition { 10 } else { 20 };

C. let x = if condition: 10 else: 20;

D. let x = if condition { 10 } else 20;

Solution

1. Step 1: Recall Rust if expression syntax

   Rust uses curly braces for blocks and requires both if and else blocks to be enclosed in braces when returning values.

2. Step 2: Check each option

   let x = if condition { 10 } else { 20 }; correctly uses if condition { 10 } else { 20 };. Others use invalid syntax like missing braces or keywords.

3. Final Answer:

   let x = if condition { 10 } else { 20 }; -> Option B

4. Quick Check:

   Correct Rust if expression syntax = let x = if condition { 10 } else { 20 }; [OK]

Hint: Remember: Rust uses braces {} for both if and else blocks [OK]

Common Mistakes:

• Omitting braces around else block
• Using 'then' keyword (not in Rust)
• Using colons instead of braces

3. What is the output of this Rust code?

let number = 7;
let result = if number % 2 == 0 { "even" } else { "odd" };
println!("{}", result);

medium

A. even

B. 7

C. Compilation error

D. odd

Solution

1. Step 1: Evaluate the condition

   The condition checks if 7 modulo 2 equals 0. Since 7 % 2 is 1, the condition is false.

2. Step 2: Determine which branch runs

   Because the condition is false, the else branch runs, returning "odd".

3. Final Answer:

   odd -> Option D

4. Quick Check:

   7 is odd = odd [OK]

Hint: Check the condition result carefully before picking output [OK]

Common Mistakes:

• Assuming 7 % 2 == 0 is true
• Confusing output with the number itself
• Expecting a syntax error

4. Identify the error in this Rust code using if as expression:

let x = if true { 5 } else { "five" };

medium

A. Mismatched types in if and else branches

B. Missing semicolon after else block

C. Cannot use if as expression in let binding

D. Condition must be a variable, not a literal

Solution

1. Step 1: Check types returned by if and else

   The if branch returns an integer 5, the else branch returns a string "five".

2. Step 2: Understand Rust type rules for if expressions

   Both branches must return the same type because the if expression must have a single consistent type.

3. Final Answer:

   Mismatched types in if and else branches -> Option A

4. Quick Check:

   if branches must match types = Mismatched types in if and else branches [OK]

Hint: Ensure both if and else return same type [OK]

Common Mistakes:

• Ignoring type mismatch errors
• Thinking semicolon is missing
• Believing literals can't be conditions

5. You want to assign a variable status based on a score: if score >= 90, status is "Excellent"; if score >= 60, status is "Pass"; otherwise, "Fail". Which Rust code correctly uses nested if expressions to do this?

hard

A. let status = if score >= 90 { "Excellent" } else if score >= 60 { "Pass" } else { "Fail" };

B. let status = if score >= 90 { "Excellent" } else { if score >= 60 { "Pass" } else "Fail" };

C. let status = if score >= 90 { "Excellent" } else if score >= 60 { "Pass" }; else { "Fail" };

D. let status = if score >= 90 { "Excellent" } else { "Pass" } else { "Fail" };

Solution

1. Step 1: Understand nested if expressions

   Rust supports chaining if and else if expressions to select among multiple conditions.

2. Step 2: Check syntax correctness

   let status = if score >= 90 { "Excellent" } else if score >= 60 { "Pass" } else { "Fail" }; correctly uses else if and ends with else block. let status = if score >= 90 { "Excellent" } else { if score >= 60 { "Pass" } else { "Fail" } }; nests if inside else but is more verbose. Options C and D have syntax errors.

3. Final Answer:

   let status = if score >= 90 { "Excellent" } else if score >= 60 { "Pass" } else { "Fail" }; -> Option A

4. Quick Check:

   Use else if for clean multi-condition if expressions [OK]

Hint: Use else if for multiple conditions, not nested if inside else [OK]

Common Mistakes:

• Missing else block causing syntax error
• Incorrect semicolon placement
• Using multiple else blocks