You have an optional dictionary [String: Int?] and want to safely print the value for key "score" using force unwrapping. Which code safely avoids a crash?

hard📝 Application Q8 of 15

Swift - Optionals

You have an optional dictionary [String: Int?] and want to safely print the value for key "score" using force unwrapping. Which code safely avoids a crash?

Aprint(dict["score"]!)

Bif let dict = dict, let val = dict["score"] { print(val!) }

Cprint(dict!["score"]!)

Dprint(dict["score"] ?? 0)

Step-by-Step Solution

Solution:

Step 1: Understand dictionary optional and nested optional
The dictionary itself and its values are optional, so multiple unwraps needed.
Step 2: Analyze options for safe unwrapping
if let dict = dict, let val = dict["score"] { print(val!) } uses optional binding to unwrap dictionary and value safely before force unwrapping inner optional.
Final Answer:
if let dict = dict, let val = dict["score"] { print(val!) } -> Option B
Quick Check:
Safe nested unwrap = optional binding + force unwrap inner [OK]

Quick Trick: Use optional binding before force unwrapping nested optionals [OK]

Common Mistakes:

Force unwrapping dictionary or value directly
Ignoring nested optional structure
Using ?? without unwrapping inner optional

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More Swift Quizzes

You have an optional dictionary [String: Int?] and want to safely print the value for key "score" using force unwrapping. Which code safely avoids a crash?

Step 1: Understand dictionary optional and nested optional

Step 2: Analyze options for safe unwrapping

Final Answer:

Quick Check:

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