Which of the following is the correct syntax to declare a protocol named Playable with a method play()?

easy📝 Syntax Q3 of 15

iOS Swift - Swift Language Essentials

Which of the following is the correct syntax to declare a protocol named Playable with a method play()?

Aprotocol Playable { func play() -> Void }

Bprotocol Playable { func play }

Cprotocol Playable { func play() -> String }

Dprotocol Playable { var play: () -> Void }

Step-by-Step Solution

Solution:

Step 1: Understand method declaration in protocols
Methods in protocols are declared with their signature including return type. Void return type can be written as '-> Void' or omitted.
Step 2: Check each option
protocol Playable { func play() -> Void } correctly declares the method with explicit Void return type. protocol Playable { func play } is missing parentheses after 'play', resulting in a syntax error. protocol Playable { func play() -> String } specifies an incorrect String return type. protocol Playable { var play: () -> Void } declares a property with a closure type instead of a method.
Final Answer:
protocol Playable { func play() -> Void } -> Option A
Quick Check:
Protocol method syntax = func name() -> ReturnType [OK]

Quick Trick: Declare protocol methods with func and return type [OK]

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