swift - 可选链

可选链代替强制展开

// 你可以通过在你希望如果可选项为非 nil 就调用属性、方法或者脚本的可选值后边使用问号（ ? ）来明确可选链。这和在可选值后放叹号（ ! ）来强制展开它的值非常类似。主要的区别在于可选链会在可选项为 nil 时得体地失败，而强制展开则在可选项为 nil 时触发运行时错误。

class Person {
    var residence: Residence?
}
 
class Residence {
    var numberOfRooms = 1
}

// 强制展开导致崩溃
let john = Person()
let roomCount = john.residence!.numberOfRooms
// this triggers a runtime error

// 可选链
if let roomCount = john.residence?.numberOfRooms {
    print("John's residence has \(roomCount) room(s).")
} else {
    print("Unable to retrieve the number of rooms.")
}
// Prints "Unable to retrieve the number of rooms."

john.residence = Residence()
if let roomCount = john.residence?.numberOfRooms {
    print("John's residence has \(roomCount) room(s).")
} else {
    print("Unable to retrieve the number of rooms.")
}
// Prints "John's residence has 1 room(s)."

可选链定义模型类的优势

class Person {
    var residence: Residence?
}
// Residence 类比以前要复杂一些。这次， Residence 类定义了一个叫做 rooms 的变量属性，它使用一个空的 [Room] 类型空数组初始化
class Room {
    let name: String
    init(name: String) { self.name = name }
}

class Address {
    var buildingName: String?
    var buildingNumber: String?
    var street: String?
    func buildingIdentifier() -> String? {
        if buildingName != nil {
            return buildingName
        } else if buildingNumber != nil && street != nil {
            return "\(buildingNumber) \(street)"
        } else {
            return nil
        }
    }
}

class Residence {
    var rooms = [Room]()
    var numberOfRooms: Int {
        return rooms.count
    }
    subscript(i: Int) -> Room {
        get {
            return rooms[i]
        }
        set {
            rooms[i] = newValue
        }
    }
    func printNumberOfRooms() {
        print("The number of rooms is \(numberOfRooms)")
    }
    var address: Address?
}

通过可选链访问属性

// 由于 john.residence 是 nil ，这个可选链调用与之前一样失败了。
let john = Person()
if let roomCount = john.residence?.numberOfRooms {
    print("John's residence has \(roomCount) room(s).")
} else {
    print("Unable to retrieve the number of rooms.")
}
// Prints "Unable to retrieve the number of rooms."

// 给 john.residence 的 address 属性赋值会失败，因为 john.residence 目前是 nil 。
let someAddress = Address()
someAddress.buildingNumber = "29"
someAddress.street = "Acacia Road"
john.residence?.address = someAddress

func createAddress() -> Address {
    print("Function was called.")
    
    let someAddress = Address()
    someAddress.buildingNumber = "29"
    someAddress.street = "Acacia Road"
    
    return someAddress
}
john.residence?.address = createAddress()
// 你可以看到 createAddress() 函数没有被调用，因为没有任何东西打印出来。

通过可选链调用方法

// Residence 类中的 printNumberOfRooms() 方法打印了当前 numberOfRooms 的值。方法看起来长这样：
func printNumberOfRooms() {
    print("The number of rooms is \(numberOfRooms)")
}

if john.residence?.printNumberOfRooms() != nil {
    print("It was possible to print the number of rooms.")
} else {
    print("It was not possible to print the number of rooms.")
}
// Prints "It was not possible to print the number of rooms."

if (john.residence?.address = someAddress) != nil {
    print("It was possible to set the address.")
} else {
    print("It was not possible to set the address.")
}
// Prints "It was not possible to set the address."

通过可选链访问下标

if let firstRoomName = john.residence?[0].name {
    print("The first room name is \(firstRoomName).")
} else {
    print("Unable to retrieve the first room name.")
}
// Prints "Unable to retrieve the first room name."

john.residence?[0] = Room(name: "Bathroom")

let johnsHouse = Residence()
johnsHouse.rooms.append(Room(name: "Living Room"))
johnsHouse.rooms.append(Room(name: "Kitchen"))
john.residence = johnsHouse
 
if let firstRoomName = john.residence?[0].name {
    print("The first room name is \(firstRoomName).")
} else {
    print("Unable to retrieve the first room name.")
}
// Prints "The first room name is Living Room."

访问可选类型的下标

var testScores = ["Dave": [86, 82, 84], "Bev": [79, 94, 81]]
testScores["Dave"]?[0] = 91
testScores["Bev"]?[0] += 1
testScores["Brian"]?[0] = 72
// the "Dave" array is now [91, 82, 84] and the "Bev" array is now [80, 94, 81]

链的多层连接

1. 如果你访问的值不是可选项，它会因为可选链而变成可选项；
2. 如果你访问的值已经是可选的，它不会因为可选链而变得更加可选。
   //因此：
3. 如果你尝试通过可选链取回一个 Int 值，就一定会返回 Int? ，不论通过了多少层的可选链；
4. 类似地，如果你尝试通过可选链访问 Int? 值， Int? 一定就是返回的类型，无论通过了多少层的可选链。

if let johnsStreet = john.residence?.address?.street {
    print("John's street name is \(johnsStreet).")
} else {
    print("Unable to retrieve the address.")
}
// Prints "Unable to retrieve the address."

let johnsAddress = Address()
johnsAddress.buildingName = "The Larches"
johnsAddress.street = "Laurel Street"
john.residence?.address = johnsAddress
 
if let johnsStreet = john.residence?.address?.street {
    print("John's street name is \(johnsStreet).")
} else {
    print("Unable to retrieve the address.")
}
// Prints "John's street name is Laurel Street.

用可选返回值链接方法

if let buildingIdentifier = john.residence?.address?.buildingIdentifier() {
    print("John's building identifier is \(buildingIdentifier).")
}
// Prints "John's building identifier is The Larches."

if let beginsWithThe =
    john.residence?.address?.buildingIdentifier()?.hasPrefix("The") {
        if beginsWithThe {
            print("John's building identifier begins with \"The\".")
        } else {
            print("John's building identifier does not begin with \"The\".")
        }
}
// Prints "John's building identifier begins with "The"."