properties – “无法赋值给该表达式的结果” – 如何设置可选变量的属性

import Foundation

class MyClass: NSObject {

    var parent: MyClass?
    var String: String?

    init() {
        super.init()

    }

}

let variable : MyClass = MyClass()
variable.String = "variable"
variable.parent?.String = "parent"

variable.parent?.String = "parent"

您无法在不存在的父级中分配字符串.你将如何避免这种情况取决于你.有一点可以肯定,要么你要确保父绑定(比如在一些init()方法中绑定它),要么每次都必须使用if variable.parent(即使你不想这样)或者你必须定义一个像assignParentString()这样的方法,用于检查父项.

func assignParentString (String: String) {
  if let parent = self.parent {
    parent.String = String
  }
}

例如：

3> class MyClass {
  4.     var parent : MyClass?
  5.     var name   : String?
  6.     func aps (name:string) {
  7.         if let parent = self.parent {
  8.             parent.name = name
  9.         }   
 10.     }   
 11. }   
 12> var mc = MyClass()
mc: MyClass = {
  parent = nil
  name = nil
}
 13> mc.name = "mc"
 14> mc.aps ("mcp")           // no error,name did not take
 15> mc
$R3: MyClass = {
  parent = nil
  name = "mc"
}
 16> mc.parent = MyClass()
 17> mc.aps ("mcp")
 18> mc
$R6: MyClass = {
  parent = Some {
    parent = nil
    name = "mcp"
  }
  name = "mc"
}