可选链(Optional Chaining)

class Car { var price = 0 } 

class Dog { var weight = 0 } 

class Person {

    var name: String = ""

    var dog: Dog = Dog()

    var car: Car? = Car()

    func age() -> Int { 18 }

    func eat() { print("Person eat") } 

    subscript(index: Int) -> Int { index }

}

var person: Person? = Person() 

var age1 = person!.age() // Int 

var age2 = person?.age() // Int? 

var name = person?.name // String? 

var index = person?[6] // Int?

func getName() -> String { "jack" } // 如果person是nil，不会调用getName() person?.name = getName()

    如果可选项为nil，调用方法、下标、属性失败，结果为nil

    如果可选项不为nil，调用方法、下标、属性成功，结果会被包装成可选项 

    如果结果本来就是可选项，不会进行再次包装

if let _ = person?.eat() { // ()? 

    print("eat调用成功")

} else { 

    print("eat调用失败")

}

var dog = person?.dog // Dog?

var weight = person?.dog.weight // Int? 

var price = person?.car?.price // Int?

    多个?可以链接在一起

    如果链中任何一个节点是nil，那么整个链就会调用失败

可选链

var scores = ["Jack": [86, 82, 84], "Rose": [79, 94, 81]]

scores["Jack"]?[0] = 100

scores["Rose"]?[2] += 10

scores["Kate"]?[0] = 88

var num1: Int? = 5

num1? = 10 // Optional(10)

var num2: Int? = nil

num2? = 10 // nil

var dict: [String : (Int, Int) -> Int] = [

    "sum" : (+),

    "difference" : (-)

]

var result = dict["sum"]?(10, 20) // Optional(30), Int?