Option可选值（一）

//: Playground - noun: a place where people can play

import Cocoa

class Person {

var residence:
Residence?//供选连接

}

class Residence {

var rooms = [Room]()

var numberOfRooms:Int {

return
rooms.count

}

subscript(i:Int) ->Room
{

return
rooms[i]

}

func printNumberOfRooms() {

println("The number of rooms is\(numberOfRooms)")

}

var address:
Address?

}

class Room {

let name:
String

init(name:
String) {

self.name = name

}

}

class Address {

var buildingName:String?

var buildingNubmer:String?

var street:
String?

func buildingIdentifier() ->String?

{

if (buildingName !=nil)
{

return
buildingName

}else
if (buildingNubmer !=
nil) {

returnbuildingNubmer

}else {

return
nil

}

}

}

let john =Person()

//let johnsHouse = Residence()

//johnsHouse.rooms[0] = Room(name: "Living Room")

//john.residence = johnsHouse

/*

你能够将多层供选链接连接在一起,能够掘取模型内更下层的属性方法和角标。然而多层供选链接不能再加入比已经返回的供选值很多其它的层。
也就是说:

假设你试图获得类型不是供选类型,因为供选链接它将变成供选类型。假设你试图获得的类型已经是供选类型,因为供选链接它也不会提高供选性。因此:

假设你试图通过供选链接获得 Int 值,不论使用了多少层链接返回的总是 Int?。相似的,假设你试图通过供选链接获得
Int?

值,不论使用了多少层链接返回的总是 Int?。

*/

let johnsAddress =Address()

johnsAddress.buildingName ="The"

johnsAddress.street ="Laurel"

john.residence!.address =johnsAddress

//链接供选返回值的方法

//if let buildingIdentifier = john.residence?

.address?.buildingIdentifier()?

.uppercaseString {

//    println("John's building identifier is \(buildingIdentifier).")

//}

//连接多层链接

//if let johnsStreet = john.residence?.address?.street {

//    println("John's street name is \(johnsStreet).")

//} else {

//    println("Unable to retrieve the address.")

//}

//使用供选链接调用角标

//if let firstRoomName = john.residence?[0].name {

//    println("The first room name is \(firstRoomName).")

//} else {

//    println("Unable to retrieve the first room name.")

//}