For-in with where clause in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of a Swift for-in loop with where clause filtering even numbers

What will be printed by this Swift code?

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

Swift

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

A

2
4
6

B

1
3
5

C

1
2
3
4
5
6

D No output, syntax error

💡 Hint

The where clause filters elements before the loop body runs.

The where clause filters the numbers array to only even numbers (num % 2 == 0). So only 2, 4, and 6 are printed.

📝 Syntax

intermediate

Identify the syntax error in this Swift for-in loop with where clause

Which option correctly fixes the syntax error in this code?

let fruits = ["apple", "banana", "cherry"]
for fruit in fruits where fruit.hasPrefix("a")
    print(fruit)

Swift

let fruits = ["apple", "banana", "cherry"]
for fruit in fruits where fruit.hasPrefix("a")
    print(fruit)

A

Remove the where clause:
for fruit in fruits {
    print(fruit)
}

B

Change 'where' to 'if':
for fruit in fruits if fruit.hasPrefix("a") {
    print(fruit)
}

C

Add a colon after the where clause:
for fruit in fruits where fruit.hasPrefix("a"): {
    print(fruit)
}

D

Add curly braces around the loop body:
for fruit in fruits where fruit.hasPrefix("a") {
    print(fruit)
}

💡 Hint

Swift requires braces for loop bodies with multiple lines or where clauses.

The for-in loop with a where clause must have its body enclosed in curly braces. Option D adds the braces correctly.

❓ optimization

advanced

Optimize filtering in a Swift for-in loop with where clause

Given this code:

let numbers = Array(1...1000)
var sum = 0
for num in numbers where num % 5 == 0 {
    sum += num
}

Which option is the most efficient way to calculate the sum of multiples of 5?

Swift

let numbers = Array(1...1000)
var sum = 0
for num in numbers where num % 5 == 0 {
    sum += num
}

A

Use stride to iterate only multiples of 5:
for num in stride(from: 5, through: 1000, by: 5) {
    sum += num
}

B

Filter first then loop:
for num in numbers.filter({ $0 % 5 == 0 }) {
    sum += num
}

C Keep the original code as it is the most efficient.

D Use a while loop to check each number and add if divisible by 5.

💡 Hint

Avoid checking every number when you can jump directly to multiples.

Using stride lets you loop only over multiples of 5, avoiding unnecessary checks and improving performance.

🔧 Debug

advanced

Debug why this Swift for-in loop with where clause prints no output

Consider this code:

let words = ["Swift", "is", "fun"]
for word in words where word.count > 10 {
    print(word)
}

Why does it print nothing?

Swift

let words = ["Swift", "is", "fun"]
for word in words where word.count > 10 {
    print(word)
}

A The print statement is outside the loop body due to missing braces.

B The syntax of the where clause is incorrect, causing the loop to skip printing.

C No word in the array has more than 10 characters, so the where clause filters out all elements.

D The array is empty, so the loop never runs.

💡 Hint

Check the condition in the where clause against the array elements.

All words have fewer than or equal to 10 characters, so none satisfy the condition word.count > 10, resulting in no output.

🧠 Conceptual

expert

Understanding the behavior of multiple where clauses in nested for-in loops

What will be the output of this Swift code?

let numbers = [1, 2, 3]
let letters = ["a", "b", "c"]
for num in numbers where num % 2 != 0 {
    for letter in letters where letter != "b" {
        print("\(num)\(letter)")
    }
}

Swift

let numbers = [1, 2, 3]
let letters = ["a", "b", "c"]
for num in numbers where num % 2 != 0 {
    for letter in letters where letter != "b" {
        print("\(num)\(letter)")
    }
}

A

1a
1b
1c
3a
3b
3c

B

1a
1c
3a
3c

C

2a
2c

D No output due to syntax error

💡 Hint

Both loops filter elements with their own where clauses.

The outer loop runs only for odd numbers (1 and 3). The inner loop skips letter "b". So combinations printed are 1a, 1c, 3a, 3c.