Question

如果我创建一个结构数组，一切都按预期工作，但是如果我创建一个结构字典，它们的属性不再可访问，是否有一个额外的步骤来访问它们？请参阅下面的简单示例：

struct SceneStruct {
   var number: Int

   init(number: Int) {
       self.number = number
   }
}
let aStruct = SceneStruct(number: 0)  //shows in assistant editor as:  {number 0}
let bStruct = SceneStruct(number: 26) //shows in assistant editor as: {number 26}

//EXPECTED BEHAVIOR:
let arr = [aStruct,bStruct] //shows in assistant editor as: [{number 0}, {number 26}]
var i = arr[0] //shows in assistant editor as: {number 0}
var j = i.number //shows in assistant editor as: 0

//BROKEN BEHAVIOR:
let dict = ["one": aStruct, "two": bStruct] //shows in assistant editor as: ["one": {number 0}, "two": {number 26}]
var x = dict["one"] //shows in assistant editor as: {{number 0}}
var y = x.number // Fails: "error: 'SceneStruct?' does not have a member named 'number'"

似乎对象以某种方式包装在字典中，但对于我的生活，我找不到有关此行为的任何信息。

Answer 1

对于数组，下标运算符返回值。

var i = arr[0] // <- i is of type SceneStruct

对于字典，下标运算符返回一个可选字段。

var x = dict["one"] // <- x is of type SceneStruct? (Optional< SceneStruct>)

原因是arr[100]抛出异常，因为数组超出范围。鉴于dict["invalid"]返回nil，查找失败。

您可以通过几种方式解决这种差异。

使用默认值

进行评估

var defaultValue = SceneStruct(number: 0)
var x = dict["one"] ?? defaultValue // provide a default value if dict["one"] is nil.
var y = x.number

条件评估

if let x = dict["one"] { // conditionally set x
    var y = x.number
} else {
    // dict["one"] is nil
}

强制解包

var x = dict["one"]! // force unwrap dict["one"], this will throw an exception if dict["one"] is nil. 
var y = x.number

Answer 2

通过键对字典值进行下标的结果是一个Optional值（在您的情况下为SceneStruct?）。你需要处理那个可选项。

let dict = ["one": aStruct, "two": bStruct]
let x = dict["one"] // x is SceneStruct?
let y = x?.number // y is Int?

Swift：访问结构字典中的属性

2 个答案: