Why Swift loops are safe by default - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a safe loop that iterates over an array.

Swift

let numbers = [1, 2, 3, 4, 5]
for number in [1] {
    print(number)
}

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

Anumbers

B0..<5

C1...5

DArray(1...5)

Attempts:

3 left

💡 Hint

Common Mistakes

Using a range instead of the array variable causes errors or unexpected behavior.

Using the array name directly in the loop ensures safe iteration over its elements.

2fill in blank

medium

Complete the code to safely iterate over the indices of an array.

Swift

let fruits = ["Apple", "Banana", "Cherry"]
for index in [1] {
    print(fruits[index])
}

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

A1...fruits.count

B0..<fruits.count

C0...fruits.count

D1..<fruits.count

Attempts:

3 left

💡 Hint

Common Mistakes

Using a closed range causes an index out of range error.

Using 0..

3fill in blank

hard

Fix the error in the loop to prevent out-of-bounds access.

Swift

let letters = ["a", "b", "c"]
for i in 0...[1] {
    print(letters[i])
}

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

Aletters.count - 1

Bletters.count - 2

Cletters.count + 1

Dletters.count

Attempts:

3 left

💡 Hint

Common Mistakes

Using letters.count causes a runtime error due to invalid index.

The last valid index is letters.count - 1; using this prevents out-of-bounds errors.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that filters keys with values greater than 10.

Swift

let data = ["a": 5, "b": 15, "c": 20]
let filtered = data.filter { $0.[1] > [2] }

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

Avalue

B10

Ckey

D15

Attempts:

3 left

💡 Hint

Common Mistakes

Using $0.key instead of $0.value causes wrong filtering.

Accessing the value with $0.value and comparing it to 10 filters correctly.

5fill in blank

hard

Fill all three blanks to create a safe loop that prints uppercase keys with values less than 50.

Swift

let scores = ["math": 45, "science": 55, "english": 40]
for ([1], [2]) in scores {
    if [3] < 50 {
        print([1].uppercased())
    }
}

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

Asubject

Bscore

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names or using incorrect comparisons causes errors.

Using (subject, score) and checking score < 50 prints uppercase subject names safely.

Practice

1. Why are Swift loops considered safe by default?

easy

A. They allow infinite loops without any checks.

B. They check conditions before each iteration to avoid errors.

C. They require manual index management to prevent errors.

D. They do not support looping over collections.

Solution

1. Step 1: Understand Swift loop behavior

   Swift loops automatically check the loop condition before each iteration to ensure safety.

2. Step 2: Compare options with loop safety

   Only They check conditions before each iteration to avoid errors. describes this safety feature correctly; others describe unsafe or incorrect behaviors.

3. Final Answer:

   They check conditions before each iteration to avoid errors. -> Option B

4. Quick Check:

   Loop safety = condition checks before iteration [OK]

Hint: Remember: Swift loops always check conditions first [OK]

Common Mistakes:

• Thinking Swift loops allow infinite loops without checks
• Believing manual index management is needed
• Assuming loops don't support collections

2. Which of the following is the correct syntax for a safe Swift for-in loop over an array named numbers?

easy

A. loop number in numbers { print(number) }

B. for (int i = 0; i < numbers.count; i++) { print(numbers[i]) }

C. for number to numbers { print(number) }

D. for number in numbers { print(number) }

Solution

1. Step 1: Identify correct Swift loop syntax

   Swift uses for variable in collection { } syntax for safe iteration.

2. Step 2: Check each option

   Only for number in numbers { print(number) } matches correct syntax; A uses invalid 'loop', B is C-style (not Swift), C uses invalid 'to' keyword.

3. Final Answer:

   for number in numbers { print(number) } -> Option D

4. Quick Check:

   Swift for-in syntax = for number in numbers { print(number) } [OK]

Hint: Use 'for item in collection' for safe Swift loops [OK]

Common Mistakes:

• Using C-style for loops in Swift
• Using incorrect keywords like 'loop' or 'to'
• Forgetting 'in' keyword

3. What will be the output of this Swift code?

let numbers = [1, 2, 3]
for i in 0..

medium

A. 1 2 3

B. 0 1 2

C. Error: Index out of range

D. No output

Solution

1. Step 1: Understand the loop range

   The loop runs from 0 up to but not including numbers.count (3), so i = 0,1,2.

2. Step 2: Print values at each index

   numbers[0] = 1, numbers[1] = 2, numbers[2] = 3; all print in order.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Loop prints array elements safely [OK]

Hint: Ranges like 0..

Common Mistakes:

• Confusing range end as inclusive
• Expecting index out of range error
• Thinking loop prints indices instead of values

4. Identify the error in this Swift loop and how to fix it:

let fruits = ["apple", "banana", "cherry"]
for i in 0...fruits.count {
    print(fruits[i])
}

medium

A. Using closed range causes index out of range error.

B. Missing 'in' keyword in the loop.

C. Array 'fruits' is not declared properly.

D. Loop variable 'i' should be a String.

Solution

1. Step 1: Analyze the loop range

   The range 0...fruits.count includes fruits.count (3), but valid indices are 0 to 2.

2. Step 2: Identify the error and fix

   Accessing fruits[3] causes index out of range error; use 0..

3. Final Answer:

   Using closed range causes index out of range error. -> Option A

4. Quick Check:

   Closed range includes invalid index [OK]

Hint: Use half-open range 0..

Common Mistakes:

• Using closed range instead of half-open
• Ignoring array index limits
• Confusing loop variable type

5. You want to safely iterate over a dictionary scores to print keys and values. Which Swift loop ensures safety and prevents runtime errors?

hard

A. for i in 0...scores.count { print(scores[i]) }

B. for key in scores.keys { print(scores[key]) }

C. for (key, value) in scores { print(key, value) }

D. while true { print(scores) }

Solution

1. Step 1: Understand dictionary iteration in Swift

   Swift allows safe iteration over dictionaries using for (key, value) in dictionary.

2. Step 2: Evaluate options for safety

   for (key, value) in scores { print(key, value) } uses correct syntax and safely accesses keys and values. for i in 0...scores.count { print(scores[i]) } is invalid for dictionaries, C risks optional unwrapping, D causes infinite loop.

3. Final Answer:

   for (key, value) in scores { print(key, value) } -> Option C

4. Quick Check:

   Safe dictionary loop = for (key, value) in scores { print(key, value) } [OK]

Hint: Use for (key, value) in dictionary for safe iteration [OK]

Common Mistakes:

• Using index-based loops on dictionaries
• Forgetting optional unwrapping for dictionary values
• Creating infinite loops unintentionally