Correct Answer:

Explanation:
* RedCircleView()
* GreenTriangleView()
* BlueSquareView()
* BlueSquareView()
* GreenTriangleView()
This question belongs to View Building with SwiftUI , specifically arranging views with HStack , VStack , and ZStack . In SwiftUI, an HStack lays views out horizontally, a VStack lays them out vertically, and a ZStack overlays views front-to-back. Apple's stack layout guidance describes these three containers exactly this way.
To match the canvas, the main HStack must show three items from left to right: a red circle , a green triangle
, and then a right-side vertical group. That means the first two blanks inside HStack are RedCircleView() and GreenTriangleView(). On the right side, the VStack shows a blue square on top, so the next blank is BlueSquareView(). Under that, the lower-right shape is made by layering a green triangle on top of a blue square , which means the ZStack must contain BlueSquareView() first as the background and GreenTriangleView() second as the foreground. SwiftUI's documentation notes that ZStack aligns and overlays its children in depth order, which is why the square goes before the triangle.
So the correct placement order is:
HStack {
RedCircleView()
GreenTriangleView()
VStack {
BlueSquareView()
ZStack {
BlueSquareView()
GreenTriangleView()
}
}
}
That arrangement reproduces the exact layout shown in the canvas.

Correct Answer:

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.

Correct Answer: C,D  🗳️

Correct Answer:

Explanation:

This question belongs to Swift Programming Language , specifically the domain covering structs, properties, methods, and initializers .
A computed property does not store a value directly. Instead, it returns a value calculated from other data.
That is why description is a computed property: it returns a string based on content.
A memberwise initializer is automatically provided by Swift for structs when their stored properties are initialized through parameters. So Document(content: " Greetings! " ) is using the struct's memberwise initializer.
A type property belongs to the type itself rather than to an instance. In Swift, static var docCount = 0 is a type property because it is declared with static.
An instance method is a function that belongs to an instance of the struct or class. The display() method uses the instance's content, so it is an instance method.
A type method is a method declared with static and belongs to the type itself. So static func increment() is a type method because it changes the shared type property docCount.

Correct Answer:

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 .