ios – 任意操作可选择快速链接？

class Person {
  var residence: Residence?
}

class Residence {
  var numberOfRooms = 1
}

let john = Person()

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

想象一下尝试用一些算术运算来调整条件.这会导致编译器错误,因为模运算符不支持选项.

if john.residence?.numberOfRooms % 2 == 0 { 
  // compiler error: Value of optional type int? not unwrapped
  println("John has an even number of rooms")
} else {
  println("John has an odd number of rooms")
}

当然,您总是可以执行以下操作,但它缺乏可选链接的简单性和简洁性.

if let residence = john.residence {
  if residence.numberOfRooms % 2 == 0  {
    println("John has an even number of rooms")
  }else{
    println("John has an odd number of rooms")
  }
} else {
  println("John has an odd number of rooms")
}

是否有任何Swift语言功能可以提供更好的解决方案？