[Solved] Which of the following Swift code snippets correctly unwraps an optional String? using if let? — Ans: if let unwrapped = optionalString { print(unwrapped) } | Swift

Swift - Optionals

Which of the following Swift code snippets correctly unwraps an optional String? using if let?

Aif optionalString != nil { print(optionalString!) }

Bif let unwrapped = optionalString { print(unwrapped) }

Cif let optionalString { print(optionalString) }

Dif optionalString = nil { print(optionalString) }

Step-by-Step Solution

Solution:

Step 1: Understand optional binding syntax
Optional binding with if let requires declaring a new constant to unwrap the optional safely.
Step 2: Analyze each option
- A: Correct syntax: declares unwrapped to hold the unwrapped value.
- B: Uses force unwrap ! inside the block, which is unsafe.
- C: Missing assignment in if let, invalid syntax.
- D: Uses assignment = instead of binding, invalid syntax.
Final Answer:
if let unwrapped = optionalString { print(unwrapped) } -> Option B
Quick Check:
Proper if let syntax requires a new constant assignment [OK]

Quick Trick: Use 'if let constant = optional' to unwrap safely [OK]

Common Mistakes:

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Which of the following Swift code snippets correctly unwraps an optional String? using if let?