How do I encode enum using NSCoder in swift?

kfmfe04 picture kfmfe04 · Oct 12, 2014 · Viewed 17.6k times · Source

Background

I am trying to encode a String-style enum using the NSCoding protocol, but I am running into errors converting to and back from String.

I get the following errors while decoding and encoding:

String is not convertible to Stage

Extra argument ForKey: in call

Code

    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
        }

    // ...

    }

Answer

vacawama picture vacawama · Oct 12, 2014

You need to convert the enum to and from the raw value. In Swift 1.2 (Xcode 6.3), this would look like this:

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), uses as instead of as!:

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

Swift 1.0 (Xcode 6.0) uses toRaw() and fromRaw() like this:

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

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