Free Practice Questions for the Apple App Development with Swift App-Development-with-Swift-Certified-User Exam (2026 Updated)

Comprehensive and Detailed Explanation From App Development with Swift domains:

This question falls under Swift Programming Language , specifically the domain for managing data using collection types , with emphasis on dictionaries . In Swift, a dictionary stores data as key-value pairs , so in this example the country name is the key and the capital city is the value. To add a new entry, Swift supports two standard approaches. The first is subscript assignment , which is shown in option C : capitalCities[ " Italy " ] = " Rome " . Apple’s documentation explains that you can add a key-value pair to a dictionary by assigning a value for a new key through the dictionary subscript.

The second correct approach is option E : capitalCities.updateValue( " Rome " , forKey: " Italy " ). Apple documents that updateValue(_:forKey:) updates the value for an existing key, or adds a new key-value pair if the key does not already exist . That makes it equally valid for inserting " Italy " : " Rome " into the dictionary.

The incorrect options fail for different reasons. A reverses the key and value, making " Rome " the key and " Italy " the value. B is wrong because append is used with arrays, not dictionaries. D is not the standard valid insertion syntax for Swift dictionaries in this context; Swift’s documented mutation approaches here are subscript assignment and updateValue. Therefore, the two correct answers are C and E .

This question is from Swift Programming Language , especially the domains for Optional types , safe and unsafe unwrapping , and control flow . The code uses the ternary conditional operator :

nextApplePhone.releaseDate != nil ? _____

In Swift, the ternary operator follows this structure:

condition ? valueIfTrue : valueIfFalse

So if nextApplePhone.releaseDate != nil is true, the expression must return the new release date. If it is false, it must keep lastReleaseDate unchanged. That means the missing part must be:

nextApplePhone.releaseDate! : lastReleaseDate

which is option D .

This works because nextApplePhone.releaseDate is declared as String?, so it is an optional. Once the condition confirms it is not nil, the code force-unwraps it with ! to access the underlying String value. If the optional is nil, the expression returns lastReleaseDate instead. Apple’s Swift documentation describes the ternary conditional operator as a shortcut for choosing one of two expressions based on a condition, and it explains that force unwrapping with ! accesses an optional’s wrapped value when you know it is not nil.

The other options are incorrect because they reverse the true/false logic, omit the needed unwrap, or contain invalid identifiers. Therefore, the correct completion is D .

This question belongs to View Building with SwiftUI , specifically the domain on creating a multi-view app with navigation stacks, links, and sheets .

To present a modal view in SwiftUI when a Boolean state changes, the correct modifier is .sheet . The matching sheet API for a Boolean binding is:

sheet(isPresented: $showInfo) {

// modal content

}

So the first blank must be .sheet , and the second blank must be (isPresented: .

The logic works like this:

@State stores the local Boolean that controls presentation.

Pressing the button calls showInfo.toggle(), changing the value from false to true.

When that Boolean becomes true, the .sheet(isPresented:) modifier presents the modal view.

When the modal is dismissed, SwiftUI updates the Boolean back as needed.

There is also a typing issue in the screenshot: the state variable appears as ShowInfo, while the button and binding use showInfo. Swift is case-sensitive, so those names must match. The corrected code should use the same identifier consistently, such as:

@State var showInfo = false

Therefore, the correct dropdown selections are:

sheet

(isPresented:

This question belongs to Swift Programming Language , specifically the domains covering control flow , loops , and range operators .

The code starts with:

var count = 0

var answer = 0

So both variables begin with the value 0.

In the first loop:

for index in 1...5 {

count = index

}

the closed range 1...5 includes 1, 2, 3, 4, and 5 . During each iteration, count is updated to the current value of index. After the loop finishes, the final value assigned to count is 5 .

Then the second loop runs:

for index in 1.. < count {

answer = index

}

Here, the half-open range 1.. < count means values starting at 1 up to, but not including , count. Since count is 5, this loop runs with index equal to 1, 2, 3, and 4 . Each time through the loop, answer is updated to the current index. After the last iteration, answer becomes 4 .

So the final value of answer is 4 . This question tests understanding of the difference between the closed range operator ... and the half-open range operator .. < , which is a key Swift control-flow concept.

C:\Users\Waqas Shahid\Desktop\Mudassir\Untitled.jpg

This question belongs to View Building with SwiftUI , especially the objectives for using List views to iterate through collections and structuring views with standard SwiftUI containers. The screenshot shows two grouped sets of rows: one headed MY FRIENDS and one headed MY PETS . In SwiftUI, the correct container for a scrollable table-style presentation of rows is List, and the correct way to divide that list into labeled groups is Section. Apple documents List as a container that presents data in a single-column row-based layout, and Section as a way to organize list content into grouped areas with headers and optional footers. That is exactly the structure shown in the image. ( developer.apple.com , developer.apple.com )

The ForEach(names, id: \.self) and ForEach(pets, id: \.self) lines are already iterating through the arrays, so each ForEach should be wrapped inside a Section. The section labels such as " My Friends " and " My Pets " are provided with the header: label. So the intended code structure is:

List {

Section {

ForEach(names, id: \.self) { name in Text(name) }

} header: {

Text( " My Friends " )

}

Section {

ForEach(pets, id: \.self) { pet in Text(pet) }

} header: {

Text( " My Pets " )

}

}

This matches the UI shown in the image and aligns directly with SwiftUI list and section composition patterns in App Development with Swift.

This question belongs to Swift Programming Language , specifically the objective on managing data using collection types , including dictionaries and arrays . In the dictionary shown, the emojis are the keys and the text descriptions are the values . Swift provides a keys property on dictionaries that returns a collection containing all dictionary keys. To convert that keys collection into an array, you use the Array(...) initializer. Therefore, the correct code is let emojis = Array(faces.keys). Apple documents both the Dictionary.keys property and the Array type used to store a sequence of values of the same type.

Option B is incorrect because faces.values would return the descriptions like " grinning " , " thinking " , and " happy " , not the emoji keys. Options A and C are incorrect because List is a SwiftUI view type, not the correct collection type for creating an array from dictionary contents. Also, the dictionary interface uses properties like .keys and .values, not method calls like .keys() or .values(). Apple’s dictionary documentation makes clear that keys is a property returning a collection of the dictionary’s keys.

This question belongs to View Building with SwiftUI , especially the domain on creating multiple views to implement app logic and sharing values between views. In FirstView, the value typed into the TextField is stored in number, which is a String. When the NavigationLink is tapped, the code passes Int(number) ?? 0 into SecondView. In Swift, converting a string to an integer with Int(...) uses an optional initializer, which means it returns an optional value and will produce nil if the string cannot be converted. The ?? 0 nil-coalescing operator then supplies 0 if conversion fails.

Inside SecondView, the body is:

Color(passedNumber == 3 ? .blue : .red)

This uses the ternary conditional operator. If passedNumber equals 3, the displayed color is blue; otherwise, it is red. Therefore, entering 3 into the TextField causes Int(number) to become 3, which makes the condition passedNumber == 3 true and displays a blue canvas. Any other listed number results in red.

So the correct answer is A. 3 . This matches the SwiftUI pattern of taking user input, converting it to the needed type, passing it into another view, and rendering UI conditionally based on that value.

This question belongs to Swift Programming Language , especially the domains covering control flow , loops , logical operators , and guard . The loop runs through 0.. < max, and since max = 101, the values of num are 0 through 100. Inside the loop, the guard statement keeps only values that satisfy both conditions:

num % 5 == 0 → the number must be divisible by 5

num % 2 != 0 → the number must be odd

So the code counts numbers from 0 to 100 that are odd multiples of 5 . Those values are:

5, 15, 25, 35, 45, 55, 65, 75, 85, 95

That gives a total of 10 numbers. Therefore count becomes 10, and the printed output is:

Total count: 10

The key Swift concept here is that guard ... else { continue } skips any loop iteration that does not meet the required condition. Only matching values reach count += 1. This is a standard use of guard for early exit and of the remainder operator % for divisibility checks. Therefore, the correct answer is B .

This question falls under View Building with SwiftUI , specifically the domain covering multi-view apps with navigation stacks, links, and sheets . The first blank must be NavigationLink because SwiftUI uses a navigation link inside a NavigationStack to push or present a destination view as part of the navigation hierarchy. Apple’s documentation states that people tap or click a NavigationLink to present a view inside a NavigationStack or NavigationSplitView. That matches the first code section, where tapping " Show Blue View " should navigate to BlueView().

The second blank must be .sheet because the code uses isPresented: $showRedView, which is the standard SwiftUI sheet modifier for modal presentation controlled by a Boolean binding. Apple documents sheet(isPresented:onDismiss:content:) as the modifier to use when you want to present a modal view when a Boolean becomes true. Since the button toggles showRedView, SwiftUI presents RedView() modally as a sheet.

So the completed structure is effectively:

NavigationLink( " Show Blue View " ) {

BlueView()

}

sheet(isPresented: $showRedView) {

RedView()

}

This directly aligns with SwiftUI navigation and modal presentation patterns in the App Development with Swift objective domains.