How do I encode enum using NSCoder in swift?

Question

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
        }

    // ...

    }

vacawama · Accepted Answer · 2015-07-11 03:17:51Z

70

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" )

edited Jul 11, 2015 at 3:17

answered Oct 12, 2014 at 15:26

vacawama

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2 Comments

Oren Over a year ago

If you have as! then the String can never be nil and ?? .DisplayAll is useless. Shouldn't it be as? instead?

vacawama Over a year ago

The as! is casting the String which does exist so the cast will succeed. Stage(rawValue: "someString") returns an optional because the string may not define a valid enum value. You have to unwrap that optional. The nil coalescing operator replaces that optional with an unwrapped version if it exists or .DisplayAll if it doesn't.

Jared Cleghorn · Accepted Answer · 2018-07-13 17:33:57Z

10

Here is a solution for Swift 4.2. As stated in the other answers, the problem is that you try to directly assign the stage variable with a decoded string, and you try to cast the stage variable to a string in the encodeWithCoder method. You need to use raw values instead.

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.decodeInteger(forKey: "idx")
        self.stage = Stage(rawValue: aDecoder.decodeObject(forKey: "stage") as String)
    }

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

    // ...

}

answered Jul 13, 2018 at 17:33

Jared Cleghorn

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1 Comment

Hemang Over a year ago

What if the enum don't have the row values? In my app, it is just an enum: enum SomeType { case1, case2, case3 } like this? How to handle for that case?

Adrian Sluyters · Accepted Answer · 2015-03-10 13:10:53Z

9

Update for Xcode 6.3, Swift 1.2:

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

note the as!

answered Mar 10, 2015 at 13:10

Adrian Sluyters

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