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| Section | Objectives |
|---|---|
| Topic 1: Xcode Developer Tools | - Identify and use the features of the Xcode interface
|
| Topic 2: View Building with SwiftUI | - Create multiple Views to implement app logic - Position and/or layout a single SwiftUI View with standard Views and modifiers - 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 - Use @State, @Binding, @Environment, and/or Observable to share data between Views |
| Topic 3: Swift Programming Language | - Demonstrate proper use of structs, classes
|
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NEW QUESTION # 10
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
Answer:
Explanation:
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
NEW QUESTION # 11
Review the code snippet.
The code snippet does not compile.
Which two actions will fix the errors? (Choose 2.)
Answer: B,C
Explanation:
This question belongs to Swift Programming Language , especially the domains covering basic Swift types
, operators , and constants versus variables .
There are two compile problems in the snippet. First, unitPrice and shipping are inferred as Double, while quantity is inferred as Int. In Swift, arithmetic operands must have compatible types; Swift does not automatically mix Int and Double in one arithmetic expression. So unitPrice * quantity fails unless quantity is changed to Double or explicitly converted. That makes A a correct fix.
Second, the line totalCost += ... uses the compound assignment operator +=, which stores a new value back into the left-hand side. Swift requires the left-hand side of += to be mutable, so totalCost must be declared with var, not let. That makes D the second correct fix.
The other choices do not solve the actual compile issues. B is unnecessary because totalCost is already explicitly declared as Double, so 0 is valid there. C would still leave shipping as Double, so the mixed-type arithmetic problem remains. E is irrelevant because shipping is never reassigned. Therefore, the two correct answers are A and D
NEW QUESTION # 12
Review the code snippet.
What value does the code output?
Answer:
Explanation:
Answer the question by typing in the box.
2
Explanation:
This question belongs to Swift Programming Language , specifically the objectives covering functions , control flow , and default parameter values . The function is declared as func getAgeCategory(_ age: Int =
20) - > Int, which means if no argument is supplied, Swift uses the default value 20. Apple's Swift documentation explains that you can define a default value for any parameter, and that value is used when the caller omits that argument. Since the code calls getAgeCategory() with no parameter, the function executes using age = 20.
The conditional logic is then evaluated in order:
* if age > 64 # false, because 20 is not greater than 64
* else if age > 19 # true, because 20 is greater than 19
* so the function returns 2
Because Swift's if / else if control flow stops at the first true condition, the later checks are never reached once age > 19 succeeds. Apple describes Swift as supporting standard control flow including conditional branching, and this example is a direct use of that branching behavior.
Therefore, print(getAgeCategory()) outputs 2 , which corresponds to option B .
NEW QUESTION # 13
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 # 14
When you press ' Show Button ' on your app. a modal View appears.
Complete the code by selecting the correct option from each drop-down list.
Note: You will receive partial credit for each correct answer.
Answer:
Explanation:
Explanation:
This question belongs to View Building with SwiftUI , specifically the domain on creating a multi-view app with navigation stacks, links, and sheets .
To present a modal view in SwiftUI when a Boolean state changes, the correct modifier is .sheet . The matching sheet API for a Boolean binding is:
sheet(isPresented: $showInfo) {
// modal content
}
So the first blank must be .sheet , and the second blank must be (isPresented: .
The logic works like this:
* @State stores the local Boolean that controls presentation.
* Pressing the button calls showInfo.toggle(), changing the value from false to true.
* When that Boolean becomes true, the .sheet(isPresented:) modifier presents the modal view.
* When the modal is dismissed, SwiftUI updates the Boolean back as needed.
There is also a typing issue in the screenshot: the state variable appears as ShowInfo, while the button and binding use showInfo. Swift is case-sensitive, so those names must match. The corrected code should use the same identifier consistently, such as:
@State var showInfo = false
Therefore, the correct dropdown selections are:
sheet
(isPresented:
NEW QUESTION # 15
......
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