Labeled statements for nested loops in Swift - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to break out of the outer loop using the label.

Swift

outerLoop: for i in 1...3 {
    for j in 1...3 {
        if i == 2 && j == 2 {
            [1]
        }
        print("i = \(i), j = \(j)")
    }
}

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

Acontinue outerLoop

Bbreak

Ccontinue

Dbreak outerLoop

Attempts:

3 left

💡 Hint

Common Mistakes

Using break without the label only exits the inner loop.

Using continue instead of break does not exit the loop.

Using break outerLoop exits the outer loop immediately when the condition is met.

2fill in blank

medium

Complete the code to continue the outer loop when the inner loop reaches j == 2.

Swift

outer: for i in 1...3 {
    for j in 1...3 {
        if j == 2 {
            [1]
        }
        print("i = \(i), j = \(j)")
    }
}

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

Acontinue outer

Bbreak outer

Cbreak

Dcontinue

Attempts:

3 left

💡 Hint

Common Mistakes

Using break instead of continue changes the loop behavior.

Using continue without label only affects the inner loop.

continue outer skips the rest of the current iteration of the outer loop and moves to the next iteration.

3fill in blank

hard

Fix the error in the code to correctly break out of the outer loop.

Swift

outerLoop: for x in 1...3 {
    for y in 1...3 {
        if x == 3 && y == 1 {
            [1]
        }
        print("x = \(x), y = \(y)")
    }
}

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

Abreak outerloop

Bbreak OuterLoop

Cbreak outerLoop

Dbreak

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect case for the label name.

Using break without label only breaks inner loop.

The label name is case-sensitive. It must match exactly as outerLoop.

4fill in blank

hard

Fill the blank to continue the outer loop and skip the current inner loop iteration when j == 3.

Swift

outerLoop: for i in 1...3 {
    for j in 1...3 {
        if j == 3 {
            [1]
        }
        print("i = \(i), j = \(j)")
    }
}

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

Acontinue outerLoop

Bbreak

Ccontinue

Dbreak outerLoop

Attempts:

3 left

💡 Hint

Common Mistakes

Using break instead of continue changes the loop behavior.

Using continue without label only affects inner loop.

continue outerLoop skips the rest of the current iteration of the outer loop when j == 3.

5fill in blank

hard

Fill all three blanks to break out of the outer loop when i == 2 and j == 3, and print a message after the loops.

Swift

outer: for i in 1...3 {
    for j in 1...3 {
        if i == [1] && j == [2] {
            [3]
        }
        print("i = \(i), j = \(j)")
    }
}
print("Exited loops")

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

A2

B3

Cbreak outer

Dcontinue outer

Attempts:

3 left

💡 Hint

Common Mistakes

Using continue instead of break changes loop behavior.

Using wrong label name or case.

Incorrect condition values.

The code breaks out of the outer loop when i is 2 and j is 3 using break outer.