स्विफ्ट 5.2 - नया क्या है?

KeyPath एक्सप्रेशंस को फ़ंक्शंस के रूप में उपयोग करना

struct User {
    let name: String
    let age: Int
    let bestFriend: String?

    var canVote: Bool {
        age >= 18
    }
}

let eric = User(name: "Eric Effiong", age: 18, bestFriend: "Otis Milburn")
let maeve = User(name: "Maeve Wiley", age: 19, bestFriend: nil)
let otis = User(name: "Otis Milburn", age: 17, bestFriend: "Eric Effiong")
let users = [eric, maeve, otis]

let userNames = users.map(\.name)
print(userNames)

let oldUserNames = users.map { $0.name }

let voters = users.filter(\.canVote)

let bestFriends = users.compactMap(\.bestFriend)

उपयोगकर्ता-परिभाषित प्रकारों के लिए मान

struct Dice {
    var lowerBound: Int
    var upperBound: Int

    func callAsFunction() -> Int {
        (lowerBound...upperBound).randomElement()!
    }
}

let d6 = Dice(lowerBound: 1, upperBound: 6)
let roll1 = d6()
print(roll1)

let d12 = Dice(lowerBound: 1, upperBound: 12)
let roll2 = d12.callAsFunction()
print(roll2)

struct StepCounter {
    var steps = 0

    mutating func callAsFunction(count: Int) -> Bool {
        steps += count
        print(steps)
        return steps > 10_000
    }
}

var steps = StepCounter()
let targetReached = steps(count: 10)

सदस्यता डिफ़ॉल्ट तर्क घोषित कर सकती है

struct PoliceForce {
    var officers: [String]

    subscript(index: Int, default default: String = "Unknown") -> String {
        if index >= 0 && index < officers.count {
            return officers[index]
        } else {
            return `default`
        }
    }
}

let force = PoliceForce(officers: ["Amy", "Jake", "Rosa", "Terry"])
print(force[0])
print(force[5])

print(force[-1, default: "The Vulture"])

नए और बेहतर निदान

struct ContentView: View {
    @State private var name = 0

    var body: some View {
        VStack {
            Text("What is your name?")
            TextField("Name", text: $name)
                .frame(maxWidth: 300)
        }
    }
}