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| Section | Objectives |
|---|---|
| App Lifecycle and Deployment Concepts | - Understanding app structure
|
| App Logic and Data Handling | - Data management in apps
|
| User Interface Development | - Building iOS interfaces
|
| Introduction to App Development with Swift | - Swift programming fundamentals
|
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Topexamが提供したAppleのApp-Development-with-Swift-Certified-Userトレーニング資料を持っていたら、美しい未来を手に入れるということになります。Topexamが提供したAppleのApp-Development-with-Swift-Certified-Userトレーニング資料はあなたの成功への礎になれることだけでなく、あなたがIT業種でもっと有効な能力を発揮することも助けられます。このトレーニングはカバー率が高いですから、あなたの知識を豊富させる以外、操作レベルを高められます。もし今あなたがAppleのApp-Development-with-Swift-Certified-User「App Development with Swift Certified User Exam」試験にどうやって合格することに困っているのなら、心配しないでください。Topexamが提供したAppleのApp-Development-with-Swift-Certified-Userトレーニング資料はあなたの問題を解決することができますから。
質問 # 27
Review the code snippet.
What value does the code output?
正解：
解説：
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 .
質問 # 28
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
正解：
解説：
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
質問 # 29
Review the code snippet.
What will print after the final line of code is executed?
正解：
解説：
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 .
質問 # 30
Review the code snippet.
The code snippet does not compile.
Which two actions will fix the errors? (Choose 2.)
正解：C、E
解説：
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
質問 # 31
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.
正解：
解説：
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.
質問 # 32
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