BLACK FRIDAY SALE: Save big on all my Swift books and bundles! >>

Typecasting

Swift must always know the type of each of your variables, but sometimes you know more information than Swift does. For example, here are three classes:

class Animal { }
class Fish: Animal { }

class Dog: Animal {
    func makeNoise() {
        print("Woof!")
    }
}

We can create a couple of fish and a couple of dogs, and put them into an array, like this:

let pets = [Fish(), Dog(), Fish(), Dog()]

Swift can see both Fish and Dog inherit from the Animal class, so it uses type inference to make pets an array of Animal.

If we want to loop over the pets array and ask all the dogs to bark, we need to perform a typecast: Swift will check to see whether each pet is a Dog object, and if it is we can then call makeNoise().

This uses a new keyword called as?, which returns an optional: it will be nil if the typecast failed, or a converted type otherwise.

Here’s how we write the loop in Swift:

for pet in pets {
    if let dog = pet as? Dog {
        dog.makeNoise()
    }
}
Hacking with Swift is sponsored by RevenueCat

SPONSORED In-app subscriptions are a pain to implement, hard to test, and full of edge cases. RevenueCat makes it straightforward and reliable so you can get back to building your app. Oh, and it's free if your app makes less than $10k/mo.

Learn more

Sponsor Hacking with Swift and reach the world's largest Swift community!

BUY OUR BOOKS
Buy Pro Swift Buy Pro SwiftUI Buy Swift Design Patterns Buy Testing Swift Buy Hacking with iOS Buy Swift Coding Challenges Buy Swift on Sundays Volume One Buy Server-Side Swift Buy Advanced iOS Volume One Buy Advanced iOS Volume Two Buy Advanced iOS Volume Three Buy Hacking with watchOS Buy Hacking with tvOS Buy Hacking with macOS Buy Dive Into SpriteKit Buy Swift in Sixty Seconds Buy Objective-C for Swift Developers Buy Beyond Code

Was this page useful? Let us know!

Average rating: 4.7/5

 
Unknown user

You are not logged in

Log in or create account
 

Link copied to your pasteboard.