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App-Development-with-Swift-Certified-User인증시험대비공부자료최신버전덤프
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최신 Apple App Development with Swift App-Development-with-Swift-Certified-User 무료샘플문제 (Q33-Q38):
질문 # 33
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.)
- A. capitalCities[ " Rome " ] = " Italy "
- B. capitalCities.append([ " Italy " : " Rome " ])
- C. capitalCities.updateValue( " Rome " , forKey: " Italy " )
- D. capitalCities[ " Italy " ] = " Rome "
- E. capitalCities = capitalCities + [ " Italy " : " Rome " ]
정답:C,D
설명:
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 .
질문 # 34
Review the code.
Note: You might need to scroll to see the entire block of code.

A breakpoint is set on line 3. When the application is run. it will stop at line 3. You need to debug the code.
Drag each debugging control from the left to the correct instruction on the right. You will receive partial credit for each correct answer

정답:
설명:

Explanation:
This question belongs to Xcode Developer Tools , especially the objective on using debugging techniques including breakpoints and stepping controls .
When execution stops at a breakpoint on line 3, Step Over runs that line without entering into another function call, so it is the correct action for moving past line 3 while staying in the current function. Step Into is used when execution reaches line 4 and you want to enter the display(numbers) function, which takes you into the function body starting at line 8. Once inside that function, Step Out continues execution until the current function returns, which is exactly what "step out from line 8" means.
Deactivate breakpoints turns breakpoint handling off so the debugger no longer stops on active breakpoints.
Continue program execution resumes the app until the next breakpoint or until the program finishes.
So the correct control order is:
1 = Continue
2 = Deactivate breakpoints
3 = Step Over
4 = Step Into
5 = Step Out

질문 # 35
Review the code.
struct ContentView: View {
let fruits = [ " Apple " , " Banana " , " Kiwi " ]
var body: some View {
List(fruits, id: .self) { fruit in
Text(fruit)
.font(.headline)
.padding()
}
}
}
Which of the following statements is true about the code?
- A. The List view is using the fruits array to display the contents as individual rows.
- B. The id: .self in the List view is not necessary and may be omitted.
- C. The id: .self in the List view should be rewritten as id: /self
- D. KeyPaths are used here to extract the font and padding properties dynamically.
정답:A
설명:
Comprehensive and Detailed Explanation From App Development with Swift domains:
This question belongs to View Building with SwiftUI , especially the domain covering List Views to iterate through collections . In the code, fruits is an array of strings, and the List initializer is being used to create one row for each item in that collection. Apple's SwiftUI documentation explains that List can present rows from a collection of data, and when the data elements are not supplied through a type that already provides identity, you can provide an id key path so SwiftUI can uniquely identify each row. Here, id: .self tells SwiftUI to use each string value itself as the identifier.
Option D is therefore the correct statement because the List is clearly rendering the contents of the fruits array as separate rows, and each row shows a Text(fruit) view. Apple's app development tutorials describe List as a container view that displays rows of data arranged in a single scrollable column, which matches exactly what this code is doing.
Option A is false because for an array of String values in this form, id: .self is used to identify each row.
Option B is false because the key path is not related to .font(.headline) or .padding(); those are standard view modifiers, not dynamic property extraction in this example. Option C is false because Swift key-path syntax uses a backslash, as in .self, not /self. Apple's KeyPath documentation shows that Swift key paths use the backslash form.
질문 # 36
For each statement about Navigation in SwiftUI. select True or False.

정답:
설명:

Explanation:
* You can treat a NavigationLink like a button, and run some Swift code when it is pressed. - False
* NavigationSplitView can be used to do navigation differently on different device sizes. - True
* You can put a header on your present View in a NavigationStack using .navigationTitle. - True
* If you have a NavigationLink that goes to another View with a NavigationLink, you need to declare a NavigationStack at each level. - False This question belongs to View Building with SwiftUI , specifically the objective on creating a multi-view app with navigation stacks, links, and sheets . Statement 1 is False because NavigationLink is primarily a navigation control that pushes or presents a destination in a navigation container; Apple documents it as creating a navigation link that presents a destination, not as a general-purpose action control like Button.
Statement 2 is True because NavigationSplitView is designed for adaptive navigation and can present navigation in different ways depending on platform and available space. Apple documents NavigationSplitView as a container for navigation across multiple columns, and this adaptive behavior is exactly why it is used differently across device sizes.
Statement 3 is True because .navigationTitle(...) sets the navigation title for a view shown inside a navigation container. Apple explicitly describes a view's navigation title as something used to visually display the current navigation state of an interface.
Statement 4 is False because you do not need a separate NavigationStack at every level. Apple describes NavigationStack as the container that manages a stack of views, and NavigationLink pushes additional destinations onto that stack. Nested destinations can keep navigating within the same stack.
질문 # 37
Review the code snippet.

What is the output from each print statement?
정답:
설명:
Answer the question by typing in the box.
10
Explanation:
This question belongs to Swift Programming Language , specifically the domain covering structs, classes, properties, methods, and the difference between structures and classes .
The key point is that Printer is declared as a class :
class Printer {
var copies: Int
init(copies: Int) {
self.copies = copies
}
}
In Swift, classes are reference types . That means when you assign one class instance to another variable, both variables refer to the same object in memory rather than creating a separate copy. Apple's Swift language guide explains that classes are passed by reference, while structures are value types. So in this code:
var printer1 = Printer(copies: 2)
var printer2 = printer1
both printer1 and printer2 point to the same Printer instance.
Next, this line changes the shared object:
printer2.copies = 10
Because printer2 refers to the same instance as printer1, changing printer2.copies also changes printer1.
copies. Therefore, when the code executes:
print(printer1.copies)
the output is 10 .
This question tests one of the most important Swift concepts: classes are reference types , while structs are value types . If Printer had been a struct instead of a class, the result would have been different because assignment would copy the value rather than share the same instance.
질문 # 38
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