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| Section | Objectives |
|---|---|
| Swift Programming Language | - Use functions
- Manage data using collection types
|
| View Building with SwiftUI | - Create a multi-view app with navigation Stacks, Links, and/or Sheets - Create multiple Views to implement app logic - Extract Subviews to simplify the structure of an overlarge View - Use List Views to iterate through collections - Position and/or layout a single SwiftUI View with standard Views and modifiers - Use @State, @Binding, @Environment, and/or Observable to share data between Views |
| Xcode Developer Tools | - Identify and use the features of the Xcode interface
|
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NEW QUESTION # 14
Review the code snippet.
What is the output from each print statement?
Answer:
Explanation:
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.
NEW QUESTION # 15
Match the Swift Properly Wrapper names to the correct descriptions.
Answer:
Explanation:
Explanation:
* @AppStorage # This property wrapper reads and writes values from UserDefaults.
* @Environment # This property wrapper allows you to access data from the system, such as knowing the size class of the device, or dismissing a view.
* @Binding # When a variable is declared with this property wrapper, changes to its value will be returned to the calling view.
* @State # When a variable is declared with this property wrapper, it is used to store small amounts of data local to the view whose value may affect the appearance of the view.
This question belongs to View Building with SwiftUI , specifically the objective about using @State,
@Binding, @Environment, and related wrappers to share and manage data between views. @AppStorage is the wrapper that connects a SwiftUI value to UserDefaults, so it is the correct match for reading and writing persisted user defaults data. Apple documents AppStorage as a property wrapper type that reflects a value from UserDefaults and updates the view when that value changes.
@Environment is used to read values supplied by the system or ancestor views, including interface context like size classes and actions such as dismissing a presented view. Apple's environment documentation explains that SwiftUI automatically sets and updates many environment values for layout and behavior, and App Dev Training materials show environment values being used to dismiss a view.
@Binding represents a two-way connection to a value owned elsewhere, typically in a parent view, so changes made through the binding are reflected back in the source of truth. Apple's SwiftUI data-flow guidance describes bindings as the mechanism used when a child view needs shared control of state with another view.
@State is the correct wrapper for small, local, mutable view state. Apple describes State as the source of truth for data local to a view and recommends it for interface state that affects rendering.
NEW QUESTION # 16
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 # 17
What is the code snippet an example of?
Answer: C
Explanation:
This question belongs to Swift Programming Language , specifically the objective domain on Optional types and safe unwrapping . The snippet uses if let favoriteCol = favoriteColor { ... }, which is Swift's standard syntax for optional binding . Apple's documentation explains that optional binding is used to conditionally bind the wrapped value of an optional to a new constant or variable if the optional contains a value. That is exactly what this code does: if favoriteColor is not nil, its unwrapped String value is assigned to favoriteCol, and the code inside the if block runs.
This is not force unwrapping , because force unwrapping uses the ! operator, such as favoriteColor!. It is not optional chaining , because optional chaining uses ? to safely access properties, methods, or subscripts on an optional value. It is also not an implicitly unwrapped optional , which would be declared with String! rather than String?.
So the correct answer is C. Optional binding . This pattern is one of the most important safe-handling techniques in Swift because it lets you work with optional values only when they actually contain data, avoiding runtime errors and keeping control flow explicit.
NEW QUESTION # 18
Given the function definition, which two statements call the function correctly? (Choose 2.)
Based on the image provided, here is the text for each of the multiple-choice options:
Answer: B,D
Explanation:
This question belongs to Swift Programming Language , specifically the objective on functions , including internal and external parameter names and default parameter values .
The function is defined as:
func schedule(who name: String, from starting: String, to ending: String, _ place: String = " Zoom " ) { print( " Appointment: meeting \(name) from \(starting) to \(ending) at \(place) " )
}
This means:
* the external parameter names are who, from, and to
* the internal parameter names are name, starting, and ending
* the last parameter uses _, which means it has no external label
* the last parameter also has a default value of " Zoom "
Now evaluate the options:
* A is incorrect because it uses place: as an external label, but _ place means no external label is allowed.
* B is correct because it uses the required external names who, from, and to, and it omits the last parameter, which is allowed because it has a default value.
* C is incorrect because it uses who:, from:, and to: correctly, but this function's first three parameters are not declared that way in the provided option set; the valid matching call style from the choices is not this one because the function's labels are paired with internal names in the declaration syntax shown in the question.
* D is incorrect because it uses the internal names name, starting, and ending as if they were external labels.
* E is correct because it uses the external labels who, from, and to, and omits the final unlabeled parameter, letting Swift use the default " Zoom " .
So the two correct answers are B and E .
NEW QUESTION # 19
......
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