Solution:Step 1: Assign optional to IUOThe optional optionalString has value "Hello" and is assigned to IUO implicitlyUnwrapped.Step 2: Access IUO propertyAccessing implicitlyUnwrapped.uppercased() unwraps IUO safely and returns "HELLO".Final Answer:HELLO -> Option BQuick Check:IUO unwraps optional

[Solved] Given this code:var optionalString: String? = "Hello" var implicitlyUnwrapped: String! = optionalString print(implicitlyUnwrapped.uppercased())What is the output? | Swift

Swift - Optionals

Given this code:

var optionalString: String? = "Hello"
var implicitlyUnwrapped: String! = optionalString
print(implicitlyUnwrapped.uppercased())

What is the output?

ARuntime crash

BHELLO

COptional("HELLO")

DCompile error

Step-by-Step Solution

Solution:

Step 1: Assign optional to IUO
The optional optionalString has value "Hello" and is assigned to IUO implicitlyUnwrapped.
Step 2: Access IUO property
Accessing implicitlyUnwrapped.uppercased() unwraps IUO safely and returns "HELLO".
Final Answer:
HELLO -> Option B
Quick Check:
IUO unwraps optional value automatically = HELLO [OK]

Quick Trick: IUO unwraps optional value automatically when accessed [OK]

Common Mistakes:

Expecting Optional("HELLO") output
Assuming runtime crash
Thinking compile error due to assignment

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Given this code:

Step 1: Assign optional to IUO

Step 2: Access IUO property

Final Answer:

Quick Check:

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