What will be the output of this Swift code?

medium📝 Predict Output Q13 of 15

Swift - Optionals

var name: String? = "Alice"
print(name)

Anil

BAlice

COptional("Alice")

DError: Cannot print optional directly

Step-by-Step Solution

Solution:

Step 1: Understand printing an optional variable
When printing an optional variable directly, Swift shows it wrapped as Optional(value) if it has a value.
Step 2: Check the variable value
The variable name holds the string "Alice", so printing it shows Optional("Alice").
Final Answer:
Optional("Alice") -> Option C
Quick Check:
Printing optional shows Optional(value) [OK]

Quick Trick: Printing optional shows Optional(...) wrapper [OK]

Common Mistakes:

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