## Optional Chaining You specify optional chaining by placing a question mark (?) after the optional value on which you wish to call a property, method or subscript if the optional is non-nil. ~~~ class Person { var residence: Residence? } class Residence { var numberOfRooms = 1 } let john = Person() let roomCount = john.residence!.numberOfRooms // this triggers a runtime error if let roomCount = john.residence?.numberOfRooms { // 返回的是Int? println("John's residence has \(roomCount) room(s).") } else { println("Unable to retrieve the number of rooms.") } // prints "Unable to retrieve the number of rooms.” john.residence = Residence() if let roomCount = john.residence?.numberOfRooms { println("John's residence has \(roomCount) room(s).") } else { println("Unable to retrieve the number of rooms.") } // prints "John's residence has 1 room(s)." ~~~ The fact that it is queried through an optional chain means that the call to numberOfRooms will always return an Int? instead of an Int. You cannot set a property’s value through optional chaining. ~~~ if john.residence?.printNumberOfRooms() { println("It was possible to print the number of rooms.") } else { println("It was not possible to print the number of rooms.") } // prints "It was not possible to print the number of rooms.” if let firstRoomName = john.residence?[0].name { println("The first room name is \(firstRoomName).") } else { println("Unable to retrieve the first room name.") } // prints "Unable to retrieve the first room name." ~~~ If you try to retrieve an Int value through optional chaining, an Int? is always returned, no matter how many levels of chaining are used. Similarly, if you try to retrieve an Int? value through optional chaining, an Int? is always returned, no matter how many levels of chaining are used.