If and if-else statements in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested if-else statements

What is the output of this Swift code?

Swift

let number = 15
if number % 3 == 0 {
    if number % 5 == 0 {
        print("FizzBuzz")
    } else {
        print("Fizz")
    }
} else {
    print("No match")
}

A FizzBuzz

B Fizz

C Buzz

D No match

💡 Hint

Check if the number is divisible by 3 and then by 5 inside the nested if.

The number 15 is divisible by 3 (15 % 3 == 0), so the outer if is true. Inside, 15 % 5 == 0 is also true, so it prints "FizzBuzz".

❓ Predict Output

intermediate

Output of if-else with multiple conditions

What will be printed by this Swift code?

Swift

let score = 75
if score >= 90 {
    print("Grade A")
} else if score >= 80 {
    print("Grade B")
} else if score >= 70 {
    print("Grade C")
} else {
    print("Grade F")
}

A Grade A

B Grade C

C Grade B

D Grade F

💡 Hint

Check the conditions from top to bottom and see where 75 fits.

The score 75 is less than 90 and 80 but greater or equal to 70, so it prints "Grade C".

🔧 Debug

advanced

Identify the error in if-else syntax

What error does this Swift code produce?

Swift

let temperature = 30
if temperature > 25
    print("Hot day")
else
    print("Cool day")

A TypeError: Cannot compare Int with Int

B RuntimeError: Unexpected token

C No error, prints "Hot day"

D SyntaxError: Missing braces for if and else blocks

💡 Hint

Swift requires braces {} for if and else blocks.

In Swift, if and else blocks must be enclosed in braces {}. Missing braces cause a syntax error.

🧠 Conceptual

advanced

Understanding if-else statement flow

Consider this Swift code snippet: let x = 10 if x > 5 { print("A") } else if x > 8 { print("B") } else { print("C") } What will be printed?

A B

B C

C A

D No output

💡 Hint

The first true condition stops the rest from running.

The first condition x > 5 is true (10 > 5), so it prints "A" and skips the rest.

❓ Predict Output

expert

Output of complex if-else with logical operators

What is the output of this Swift code?

Swift

let a = 4
let b = 7
if a > 5 && b < 10 {
    print("X")
} else if a <= 5 || b > 10 {
    print("Y")
} else {
    print("Z")
}

A Y

B No output

C Z

D X

💡 Hint

Evaluate each condition carefully with logical AND and OR.

a > 5 is false (4 > 5 is false), so first if is false. Second condition a <= 5 is true (4 <= 5), so second if is true and prints "Y".

Practice

1. What does an if statement do in Swift?

easy

A. It runs code only if a condition is true.

B. It repeats code multiple times.

C. It stores data in a variable.

D. It defines a new function.

Solution

1. Step 1: Understand the purpose of if statement

   An if statement checks a condition and runs code only when that condition is true.

2. Step 2: Compare with other options

   Repeating code is done by loops, storing data is done by variables, and defining functions uses func.

3. Final Answer:

   It runs code only if a condition is true. -> Option A

4. Quick Check:

   If statement = runs code if true [OK]

Hint: If checks true condition to run code, else skips it [OK]

Common Mistakes:

• Confusing if with loops
• Thinking if stores data
• Mixing if with function definitions

2. Which of the following is the correct syntax for an if-else statement in Swift?

easy

A. if condition { } else { }

B. if (condition) then { } else { }

C. if condition then { } else { }

D. if condition: { } else: { }

Solution

1. Step 1: Recall Swift if-else syntax

   Swift uses if condition { } and else { } blocks without extra keywords like 'then' or colons.

2. Step 2: Check each option

   if condition { } else { } matches Swift syntax exactly. Options A, B, and C use invalid keywords or punctuation.

3. Final Answer:

   if condition { } else { } -> Option A

4. Quick Check:

   Swift if-else = if condition { } else { } [OK]

Hint: Swift uses braces {} without 'then' or colons in if-else [OK]

Common Mistakes:

• Adding 'then' keyword
• Using colons after if or else
• Putting condition in parentheses unnecessarily

3. What will be printed by this Swift code?

let number = 10
if number > 5 {
    print("Big")
} else {
    print("Small")
}

medium

A. Small

B. No output

C. 10

D. Big

Solution

1. Step 1: Evaluate the condition

   The variable number is 10. The condition number > 5 is true because 10 is greater than 5.

2. Step 2: Determine which block runs

   Since the condition is true, the code inside the if block runs, printing "Big".

3. Final Answer:

   Big -> Option D

4. Quick Check:

   10 > 5 is true, so prints "Big" [OK]

Hint: Check condition true? Run if block, else run else block [OK]

Common Mistakes:

• Confusing which block runs
• Printing variable instead of string
• Ignoring condition result

4. Find the error in this Swift code:

let age = 18
if age >= 18
    print("Adult")
else
    print("Minor")

medium

A. Variable age should be a string

B. Wrong comparison operator

C. Missing braces { } around if and else blocks

D. No error, code is correct

Solution

1. Step 1: Check Swift if-else syntax

   Swift requires braces { } around the code blocks after if and else.

2. Step 2: Identify missing braces

   The code lacks braces after if age >= 18 and else, causing a syntax error.

3. Final Answer:

   Missing braces { } around if and else blocks -> Option C

4. Quick Check:

   Swift needs braces for if-else blocks [OK]

Hint: Always use { } after if and else in Swift [OK]

Common Mistakes:

• Omitting braces in if-else
• Using wrong operators
• Wrong variable types

5. You want to write Swift code that prints "Positive", "Negative", or "Zero" depending on the value of an integer variable num. Which code correctly uses if-else statements to do this?

hard

A. if num >= 0 {
print("Positive")
} else if num <= 0 {
print("Negative")
} else {
print("Zero")
}

B. if num > 0 {
print("Positive")
} else if num < 0 {
print("Negative")
} else {
print("Zero")
}

C. if num > 0 {
print("Positive")
} else {
print("Negative")
}
print("Zero")

D. if num == 0 {
print("Zero")
} else if num > 0 {
print("Negative")
} else if num < 0 {
print("Positive")
}

Solution

1. Step 1: Understand the conditions needed

   We need to check if num is greater than 0, less than 0, or exactly 0.

2. Step 2: Analyze each option

   if num > 0 { print("Positive") } else if num < 0 { print("Negative") } else { print("Zero") } correctly checks >0, then <0, else zero. if num >= 0 { print("Positive") } else if num <= 0 { print("Negative") } else { print("Zero") } wrongly overlaps conditions. if num > 0 { print("Positive") } else { print("Negative") } print("Zero") prints "Zero" always. if num == 0 { print("Zero") } else if num > 0 { print("Negative") } else if num < 0 { print("Positive") } incorrectly prints "Negative" for positive numbers and "Positive" for negative numbers.

3. Final Answer:

   if num > 0 { print("Positive") } else if num < 0 { print("Negative") } else { print("Zero") } -> Option B

4. Quick Check:

   Check >0, else if <0, else zero [OK]

Hint: Check positive first, then negative, else zero last [OK]

Common Mistakes:

• Overlapping conditions causing wrong output
• Printing zero outside else block
• Wrong order of conditions