For-in with where clause in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print only even numbers from the array.

Swift

let numbers = [1, 2, 3, 4, 5, 6]
for number in numbers where number [1] 2 == 0 {
    print(number)
}

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

A+

B%

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '%' causes a syntax error.

Using '-' or '*' does not correctly check for even numbers.

The where clause filters numbers where the remainder when divided by 2 is zero, meaning even numbers.

2fill in blank

medium

Complete the code to print names that start with the letter 'A'.

Swift

let names = ["Alice", "Bob", "Anna", "Mark"]
for name in names where name.hasPrefix([1]) {
    print(name)
}

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

A"B"

B"M"

C"Z"

D"A"

Attempts:

3 left

💡 Hint

Common Mistakes

Using a letter other than 'A' will filter wrong names.

Forgetting quotes causes a syntax error.

The hasPrefix method checks if the string starts with the given substring. Here, we want names starting with 'A'.

3fill in blank

hard

Fix the error in the code to print numbers greater than 10.

Swift

let values = [5, 12, 18, 7, 3]
for val in values where val [1] 10 {
    print(val)
}

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '<=' filters numbers less than or equal to 10.

Using '==' filters only numbers equal to 10.

The where clause should filter values greater than 10, so the operator must be '>'.

4fill in blank

hard

Fill both blanks to print words longer than 4 characters.

Swift

let words = ["apple", "dog", "banana", "cat"]
for word in words where word.[1] > [2] {
    print(word)
}

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

Acount

Blength

C4

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'length' is incorrect in Swift strings.

Comparing to 5 filters words longer than 5 characters, not 4.

The count property gives the length of the string. We compare it to 4 to find words longer than 4 characters.

5fill in blank

hard

Fill all three blanks to print numbers divisible by 3 and greater than 10.

Swift

let nums = [3, 9, 12, 15, 7, 20]
for num in nums where num [1] 3 == 0 && num [2] [3] {
    print(num)
}

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

A%

B>

C10

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' changes the filter condition.

Using incorrect operators causes syntax errors.

The modulo operator '%' checks divisibility by 3. The 'where' clause also checks if the number is greater than 10.

Practice

1. What does the where clause do in a for-in loop in Swift?

easy

A. It repeats the loop twice for each item.

B. It changes the type of the loop variable.

C. It filters items so the loop runs only on those that meet the condition.

D. It stops the loop after the first item.

Solution

1. Step 1: Understand the role of where in a loop

   The where clause adds a condition to filter items during iteration.

2. Step 2: Effect on loop execution

   Only items that satisfy the where condition are processed in the loop.

3. Final Answer:

   It filters items so the loop runs only on those that meet the condition. -> Option C

4. Quick Check:

   where filters loop items [OK]

Hint: Remember: where filters items in the loop [OK]

Common Mistakes:

• Thinking where changes variable type
• Assuming it repeats the loop multiple times per item
• Believing it stops the loop early

2. Which of the following is the correct syntax for using a where clause in a Swift for-in loop?

easy

A. for item where item.isActive in items { print(item) }

B. for item in items where item.isActive { print(item) }

C. for item in items if item.isActive { print(item) }

D. for item in items; where item.isActive { print(item) }

Solution

1. Step 1: Recall correct for-in where syntax

   The correct syntax places where after the sequence: for variable in sequence where condition.

2. Step 2: Check each option

   Only for item in items where item.isActive { print(item) } matches the correct syntax. The other options have misplaced or incorrect keywords: where before in, if instead of where, or a semicolon before where.

3. Final Answer:

   for item in items where item.isActive { print(item) } -> Option B

4. Quick Check:

   Correct where syntax [OK]

Hint: Put where right after in sequence [OK]

Common Mistakes:

• Placing where before in
• Using if instead of where
• Adding semicolon before where

3. Given the array let numbers = [1, 2, 3, 4, 5], what will this code print?

for num in numbers where num % 2 == 0 {
    print(num)
}

medium

A. 2 4

B. 1 2 3 4 5

C. 1 3 5

D. No output

Solution

1. Step 1: Understand the loop and condition

   The loop iterates over numbers and prints only those where num % 2 == 0, meaning even numbers.

2. Step 2: Identify even numbers in the array

   From [1, 2, 3, 4, 5], the even numbers are 2 and 4.

3. Final Answer:

   2 4 -> Option A

4. Quick Check:

   Print even numbers = 2 4 [OK]

Hint: Check condition filters even numbers only [OK]

Common Mistakes:

• Printing odd numbers instead
• Printing all numbers ignoring where
• Expecting no output

4. What is wrong with this Swift code?

let names = ["Anna", "Bob", "Cara"]
for name where name.hasPrefix("A") in names {
    print(name)
}

medium

A. The where clause is placed before in incorrectly.

B. The array names is not declared properly.

C. The hasPrefix method does not exist.

D. The loop variable name should be a constant.

Solution

1. Step 1: Check the for-in where syntax

   The where clause must come after in sequence, not before.

2. Step 2: Identify the syntax error

   The code places where before in names, which is invalid syntax.

3. Final Answer:

   The where clause is placed before in incorrectly. -> Option A

4. Quick Check:

   where must follow in [OK]

Hint: Always put where after in [OK]

Common Mistakes:

• Placing where before in
• Assuming hasPrefix is invalid
• Thinking loop variable must be var

5. You have a list of user records as dictionaries:

let users = [
  ["name": "Alice", "age": 25],
  ["name": "Bob", "age": 17],
  ["name": "Cara", "age": 30]
]

Using a for-in loop with a where clause, how do you print only the names of users who are 18 or older?

hard

A. for user in users { if user["age"] >= 18 { print(user["name"]!) } }

B. for user where (user["age"] as? Int ?? 0) >= 18 in users { print(user["name"]!) }

C. for user in users if (user["age"] as? Int ?? 0) >= 18 { print(user["name"]!) }

D. for user in users where (user["age"] as? Int ?? 0) >= 18 { print(user["name"]!) }

Solution

1. Step 1: Use correct for-in where syntax

   The where clause must come after in users to filter users by age.

2. Step 2: Check type casting and condition

   Use safe cast as? Int with default 0, then check if age >= 18.

3. Step 3: Print the name safely

   Access user["name"]! to print the name string.

4. Final Answer:

   for user in users where (user["age"] as? Int ?? 0) >= 18 { print(user["name"]!) } -> Option D

5. Quick Check:

   Correct for-in where with safe cast [OK]

Hint: Put where after in users with safe cast [OK]

Common Mistakes:

• Placing where before in
• Using if instead of where
• Not casting age safely
• Forgetting to unwrap name