What if you could write one piece of code that works perfectly with many different things, no matter how different they are?
Why Protocol as types in Swift? - Purpose & Use Cases
Imagine you have different kinds of devices like phones, tablets, and laptops. You want to write code that works with any device, but each device has its own unique features and ways to do things.
If you try to write separate code for each device type everywhere, your code becomes messy and hard to manage.
Writing separate code for each device means repeating yourself a lot. It's easy to make mistakes because you have to remember all the differences. When you add a new device, you must change many parts of your code, which is slow and error-prone.
Using protocols as types lets you define a common set of features that all devices share. You can write code that works with any device type through this shared protocol, without worrying about the specific device. This keeps your code clean, flexible, and easy to update.
func useDevice(device: Phone) { device.call() }
func useDevice(device: Tablet) { device.call() }protocol Device { func call() }
func useDevice(device: Device) { device.call() }This lets you write one simple function that works with many different types, making your code easier to read, maintain, and extend.
Think of a music app that plays songs on different devices. Using protocols as types, the app can play music on phones, tablets, or smart speakers without changing the main code.
Protocols let you define shared behavior for different types.
Using protocols as types simplifies code by treating different objects the same way.
This approach makes your code flexible and easier to maintain.
Practice
protocol as a type in Swift?Solution
Step 1: Understand what a protocol defines
A protocol defines a set of rules or features that a type must implement.Step 2: Using protocol as a type
When a protocol is used as a type, any object that implements those rules can be assigned to that type.Final Answer:
Any object that follows the protocol's rules can be used as that type. -> Option AQuick Check:
Protocol as type means flexible object use = A [OK]
- Thinking protocols only apply to classes
- Believing protocols define stored properties
- Confusing protocol usage with inheritance
Drawable?Solution
Step 1: Understand protocol type declaration
To declare a variable of a protocol type, use the protocol name without parentheses or brackets.Step 2: Analyze options
var shape: Drawable correctly declares a variable of type Drawable. Options A and B incorrectly try to instantiate a protocol, which is not allowed. var shape: Drawable[] uses invalid syntax.Final Answer:
var shape: Drawable -> Option AQuick Check:
Protocol type variable = var name: ProtocolName [OK]
- Trying to instantiate a protocol directly
- Using parentheses after protocol name
- Using array brackets incorrectly
protocol Greetable {
func greet() -> String
}
struct Person: Greetable {
func greet() -> String {
"Hello"
}
}
let greeter: Greetable = Person()
print(greeter.greet())Solution
Step 1: Understand protocol conformance
Person struct conforms to Greetable and implements greet() returning "Hello".Step 2: Using protocol as type and calling method
Variable greeter is of type Greetable but holds a Person instance, so calling greet() returns "Hello".Final Answer:
Hello -> Option DQuick Check:
Protocol method call returns implemented string = Hello [OK]
- Expecting protocol name or method name as output
- Thinking protocol cannot be used as variable type
- Confusing compile error with runtime behavior
protocol Vehicle {
func start()
}
class Car: Vehicle {
func start() {
print("Car started")
}
}
var myVehicle: Vehicle = Vehicle()
myVehicle.start()Solution
Step 1: Check protocol instantiation
Protocols cannot be instantiated directly, so Vehicle() causes an error.Step 2: Analyze variable assignment
Variable myVehicle must hold an instance of a type conforming to Vehicle, like Car(), not Vehicle() itself.Final Answer:
Cannot instantiate protocol directly with Vehicle() -> Option CQuick Check:
Protocols can't be created with () = Error [OK]
- Trying to create an instance of a protocol
- Confusing protocol with class instantiation
- Ignoring method return types
Animal in an array and calls sound() on each?protocol Animal {
func sound() -> String
}
class Dog: Animal {
func sound() -> String { "Bark" }
}
class Cat: Animal {
func sound() -> String { "Meow" }
}Solution
Step 1: Understand protocol array usage
To store different types conforming to Animal, declare array as [Animal].Step 2: Call method correctly
Use animal.sound() with parentheses to call the method. let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } does this correctly.Final Answer:
let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } -> Option BQuick Check:
Protocol array with method calls = let animals: [Animal] = [Dog(), Cat()] for animal in animals { print(animal.sound()) } [OK]
- Using concrete type array instead of protocol type
- Calling method without parentheses
- Mixing incompatible types in array
Ready to prove it?
Pass the test and this pattern moves unknown
