在任意操作中快速选择链接？

Question

Apple提供了一个简洁的可选链接示例

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？

Answer 1

我能想到的最好的办法是一个名为 omap（可选映射）的可选扩展。

extension Optional{
  func omap<K:Any>(optionalBlock:(T)->(K?))->K?{
    if self{
      return optionalBlock(self!)
    }
    return nil
  }
}

与 map 类似，omap 只是将可选值的实例返回给所提供的闭包，但会适当地处理 nil 情况。

这允许您将可选值与任意操作链接起来，如下所示：

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

请参阅：https ://gist.github.com/Grantismo/3e1ba0412e911dfd7cfe