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| Section | Weight | Objectives |
|---|---|---|
| App Design and Best Practices | 15% | - User experience principles
|
| Swift Programming Fundamentals | 30% | - Structures and classes
|
| Development Environment | 15% | - Debugging techniques
|
| SwiftUI Basics | 25% | - State management
|
| Data Handling and Logic | 15% | - Basic data processing
|
>> App-Development-with-Swift-Certified-User Latest Exam Discount <<
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NEW QUESTION # 20
Review the code.
var capitalCities = [ " USA " : " Washington D.C. " , " Spain " : " Madrid " , " Peru " : " Lima " ] Which two statements add the capital city of " Italy " to the dictionary? (Choose 2.)
Answer: A,E
Explanation:
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 .
NEW QUESTION # 21
Review the code snippet and identify what happens when the program is executed.
Answer: D
Explanation:
This question belongs to Swift Programming Language , specifically the objectives covering arrays , loops
, and range operators . The array has 6 elements, so menuItems.count is 6. The loop uses the half-open range operator:
for index in 1.. < (menuItems.count - 1)
That becomes:
for index in 1.. < 5
In Swift, the half-open range operator a.. < b includes the lower bound but does not include the upper bound.
So this loop runs with index values 1, 2, 3, 4, not 0 and not 5.
Array subscripting uses those indexes to print:
* menuItems[1] # " Burger "
* menuItems[2] # " Chicken "
* menuItems[3] # " Pasta "
* menuItems[4] # " Salad "
That means " Pizza " at index 0 is skipped, and " Steak " at index 5 is also skipped. Swift arrays use zero- based indexing, and count returns the number of elements in the array.
Therefore, the program prints only " Burger " , " Chicken " , " Pasta " , and " Salad " , so the correct answer is A .
NEW QUESTION # 22
Review the code snippet and then predict the output.
Answer: B
Explanation:
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 .
NEW QUESTION # 23
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 # 24 
Which two assignments of a value to direction are allowed? (Choose 2.)
Answer: C,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 # 25
......
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