Given a dictionary [String: String?] where values are optionals, how can you use if let to safely print all non-nil values?

hard📝 Application Q15 of 15

Swift - Optionals

Given a dictionary [String: String?] where values are optionals, how can you use if let to safely print all non-nil values?

Afor (key, value) in dict { print(key + ": " + value!) }

Bfor (key, value) in dict { if let val = value { print(key + ": " + val) } }

Cfor (key, value) in dict { if value != nil { print(key + ": " + value) } }

Dfor (key, value) in dict { if let value { print(key + ": " + value) } }

Step-by-Step Solution

Solution:

Step 1: Understand dictionary with optional values
Values are optionals, so we must unwrap before printing to avoid errors.
Step 2: Use if let to unwrap safely
for (key, value) in dict { if let val = value { print(key + ": " + val) } } unwraps value into val safely and prints only non-nil values.
Step 3: Check other options
for (key, value) in dict { print(key + ": " + value!) } force unwraps which can crash; C tries to print optional directly; D syntax is invalid.
Final Answer:
for (key, value) in dict { if let val = value { print(key + ": " + val) } } -> Option B
Quick Check:
Use if let val = value to unwrap optionals [OK]

Quick Trick: Use if let to unwrap optionals before printing [OK]

Common Mistakes:

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