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| Section | Objectives |
|---|---|
| Xcode Developer Tools | - Demonstrate how to build and run an app
|
| View Building with SwiftUI | - Use @State, @Binding, @Environment, and/or Observable to share data between Views - Position and/or layout a single SwiftUI View with standard Views and modifiers - Create multiple Views to implement app logic - Extract Subviews to simplify the structure of an overlarge View - Use List Views to iterate through collections - Create a multi-view app with navigation Stacks, Links, and/or Sheets |
| Swift Programming Language | - Know how and when to apply control flow and loops
- Declare and use basic Swift types
|
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16. Frage
Review the code snippet.
What will print after the final line of code is executed?
Antwort:
Begründung:
Answer the question by typing in the box.
2
Explanation:
This question belongs to Swift Programming Language , specifically the objective on variable scope and shadowing .
The code first declares:
let even = 2
This creates a constant named even in the outer scope with the value 2.
Inside the for loop, the code declares another constant with the same name:
let even = num * 2
This inner even exists only inside the loop body. It shadows the outer even, which means that within the loop, the name even refers to the loop's local constant, not the original one. However, that inner constant goes out of scope at the end of each loop iteration.
After the loop finishes, the inner even no longer exists. So when the final line runs:
print(even)
Swift uses the original outer constant, which is still 2.
So the output is:
2
This question tests two important Swift concepts:
* Scope : where a variable or constant can be accessed
* Shadowing : when a local declaration temporarily hides another declaration with the same name Therefore, the correct answer is 2 .
17. Frage
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.)
Antwort: B,E
18. Frage
You need to create a Watchpoint in Xcode. In which order should you complete the actions? Move all the actions to the answer area and place them in the correct order.
Antwort:
Begründung:
Explanation:
This question belongs to Xcode Developer Tools , specifically the objective on using debugging techniques including breakpoints, watchpoints, and logging to resolve errors . A watchpoint monitors a variable or memory location during a debugging session, so you first need the program to stop while being debugged.
That is why the correct order begins with setting a breakpoint and then running the code so execution pauses at a useful point. Apple's debugging guidance describes debugging as something done at runtime using the debugger, and LLDB's watchpoint documentation explains that watchpoints are part of the debugger workflow rather than something you set before the program is stopped.
Once execution is paused, you use the debug area to inspect the current variables. After locating the variable you want to monitor, you right-click the variable and select Watch to create the watchpoint. This sequence is consistent with how Xcode and LLDB expose watchpoint functionality during an active debug session.
LLDB also describes watchpoints as objects you create to stop execution when a watched value changes, which only makes sense after the debugger has access to the running program state.
19. Frage
You are creating or updating human resource records for your employees. For each identifier, select whether it is a Constant or a Variable Note: You will receive partial credit for each correct answer.
Antwort:
Begründung:
Explanation:
* age - Variable
* birthDate - Constant
* socialSecurityNumber - Constant
* salary - Variable
* currentDepartment - Variable
This question belongs to Swift Programming Language , specifically the objective on demonstrating when to use constants and variables . In Swift, a constant is declared with let and is used for values that should not change after they are set. A variable is declared with var and is used for values that may change over time.
Birth date is a constant because a person's date of birth does not change. Social security number is also a constant because it is intended to be a fixed identifier for that employee record. By contrast, age is a variable because it changes over time. Salary is a variable because compensation can be adjusted. Current department is also a variable because an employee may transfer to another department.
This matches Swift best practice: use let for fixed data and var for mutable data. So in a human resources record, identifiers that are permanent should be constants, while values that can change during employment should be variables.
20. Frage
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?
Antwort: B
Begründung:
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.
21. Frage
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