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| Section | Weight | Objectives |
|---|---|---|
| SwiftUI Basics | 25% | - Layout and navigation
|
| App Design and Best Practices | 15% | - User experience principles
|
| Data Handling and Logic | 15% | - Basic data processing
|
| Development Environment | 15% | - Debugging techniques
|
| Swift Programming Fundamentals | 30% | - Basic types and operators
|
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NEW QUESTION # 18 
Which two assignments of a value to direction are allowed? (Choose 2.)
Answer: A,E
Explanation:
This question belongs to Swift Programming Language , specifically the domain covering basic Swift types and how Swift handles enumerations . The code defines an enum named CompassPoint with the cases north, south, east, and west, and then declares direction as type CompassPoint. In Swift, an enum case can be assigned using the fully qualified form CompassPoint.north, so B is valid. Swift also allows the shorthand form .north when the compiler already knows the expected type is CompassPoint, so D is also valid. Apple's Swift language documentation explains that once a variable is known to be of a specific enumeration type, you can set its value using the shorter dot syntax.
The other options are not allowed. A is invalid because enum cases are not assigned using constructor-style syntax like CompassPoint(north). C is invalid because north by itself is not enough unless it is written with dot shorthand in a context with inferred enum type. E is invalid because north is a case of the enum type, not a member accessed from the variable instance as direction.north. Swift enum cases are referenced from the enum type or by shorthand dot syntax, not as instance properties.
NEW QUESTION # 19
Complete the code that will add the BlueView to the NavigationStack and present the RedView modally.
|Complete the code by typing in the boxes.
Answer:
Explanation:
NavigationLink, .sheet
Explanation:
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.
NEW QUESTION # 20
You have created a view which includes some formatted text:
You decide to extract this formatted text into a subview so that you can reuse the formatting for other texts.
You highlight the Text and choose Extract subview.
You refactor ExtractedView to BigGreenTextView.
You add the line let text: String to the extracted view above the body.
You replace " That ' s all folks " with a reference to text
How should you call this extracted View from the original View?
Answer: B
Explanation:
This question belongs to View Building with SwiftUI , specifically the domain on extracting subviews to simplify the structure of an overlarge View .
When you create a custom SwiftUI view and add a stored property like:
let text: String
that property becomes part of the view's initializer. Since the property name is text, Swift expects you to pass the value using the parameter label text: when creating the view. So the correct call is:
BigGreenTextView(text: " That ' s all folks " )
That makes B the correct answer.
Why the others are wrong:
* A is incorrect because this is not using the expected parameter label.
* C would only be correct if there were already another variable named text in the calling scope and you wanted to pass that variable instead of the literal string shown in the question.
* D is incorrect because it omits the parameter label.
This is a standard SwiftUI pattern: extract reusable formatting into a separate custom View, give it an input property, and then pass the needed value through the initializer using the property label.
NEW QUESTION # 21
Complete the code by selecting the correct option from each drop-down list to create the following screen.
Note: You will receive partial credit for each correct answer.
Answer:
Explanation:

NEW QUESTION # 22
Review the code:
Given a struct called Animal, what line of code should be added on line 5 in order to produce the output shown?
Answer: C
Explanation:
This question belongs to View Building with SwiftUI , especially the objective domain on using List views to iterate through collections and displaying model data in SwiftUI. In the code, animals is the data source passed into List(animals) { animal in ... }. That closure iterates through each element of the collection one at a time, and the parameter animal represents the current Animal instance for that row. To show the name of the current animal in the UI, the correct statement is Text(animal.name). SwiftUI's list documentation explains that each row inside a List must be a SwiftUI View, and a Text view is a standard way to display a string value for each item in the collection.
Option D is therefore correct because it accesses the name property of the current animal object being processed by the List closure. This is the standard SwiftUI pattern when iterating over identifiable model objects: pass the collection into List, receive each item in the closure, and build a row view from that item's properties. Apple's SwiftUI tutorials repeatedly use this collection-driven row-building pattern for lists and navigation.
The other options are incorrect for clear reasons. A incorrectly refers to the type name Animal instead of the current instance. B uses Animals with the wrong identifier and an invalid indexing approach. C also treats animal as an index rather than the current element object. Since the closure already gives you the current Animal, you directly access its property with animal.name.
NEW QUESTION # 23
......
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