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swift - 如何在 swift 中使用 NSCoder 对枚举进行编码?

转载 作者:IT王子 更新时间:2023-10-29 04:58:13 24 4
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背景

我正在尝试使用 NSCoding 协议(protocol)对 String 样式的枚举进行编码,但在与 String 相互转换时遇到错误。

我在解码和编码时遇到以下错误:

String 不能转换为 Stage

额外参数 ForKey: 在调用中

代码

    enum Stage : String
{
case DisplayAll = "Display All"
case HideQuarter = "Hide Quarter"
case HideHalf = "Hide Half"
case HideTwoThirds = "Hide Two Thirds"
case HideAll = "Hide All"
}

class AppState : NSCoding, NSObject
{
var idx = 0
var stage = Stage.DisplayAll

override init() {}

required init(coder aDecoder: NSCoder) {
self.idx = aDecoder.decodeIntegerForKey( "idx" )
self.stage = aDecoder.decodeObjectForKey( "stage" ) as String // ERROR
}

func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger( self.idx, forKey:"idx" )
aCoder.encodeObject( self.stage as String, forKey:"stage" ) // ERROR
}

// ...

}

最佳答案

您需要将枚举与原始值相互转换。在 Swift 1.2 (Xcode 6.3) 中,这看起来像这样:

class AppState : NSObject, NSCoding
{
var idx = 0
var stage = Stage.DisplayAll

override init() {}

required init(coder aDecoder: NSCoder) {
self.idx = aDecoder.decodeIntegerForKey( "idx" )
self.stage = Stage(rawValue: (aDecoder.decodeObjectForKey( "stage" ) as! String)) ?? .DisplayAll
}

func encodeWithCoder(aCoder: NSCoder) {
aCoder.encodeInteger( self.idx, forKey:"idx" )
aCoder.encodeObject( self.stage.rawValue, forKey:"stage" )
}

// ...

}

Swift 1.1 (Xcode 6.1),使用 as 而不是 as!:

    self.stage = Stage(rawValue: (aDecoder.decodeObjectForKey( "stage" ) as String)) ?? .DisplayAll

Swift 1.0 (Xcode 6.0) 像这样使用 toRaw()fromRaw():

    self.stage = Stage.fromRaw(aDecoder.decodeObjectForKey( "stage" ) as String) ?? .DisplayAll

aCoder.encodeObject( self.stage.toRaw(), forKey:"stage" )

关于swift - 如何在 swift 中使用 NSCoder 对枚举进行编码?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26326645/

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