CODEWITHSUNDEEP
swiftswift4codabledecodable
Melodius
Melodius

Reputation: 2775

Codable class does not conform to protocol Decodable

Why am I getting a "Type 'Bookmark' does not conform to protocol 'Decodable'" error message?

class Bookmark: Codable {
   weak var publication: Publication?
   var indexPath: [Int]
   var locationInText = 0

   enum CodingKeys: String, CodingKey {
      case indexPath
      case locationInText
   }

   init(publication: Publication?, indexPath: [Int]) {
      self.publication = publication
      self.indexPath = indexPath
   }
}

I do not wish to save the publication var since the Publication owns the Bookmark but the bookmark needs to know which Publication it belongs to. The decode init of Publication will set the bookmark reference to itself.

Upvotes: 119

Views: 190594

Answers (12)

Amisha Italiya
Amisha Italiya

Reputation: 246

If you have tried all the above solutions and are still unable to fix the error then I think it could be because of the data type you are using for your data class fields.

In the question they have used 1 field weak var publication: Publication?, and if you have the same kind of class structure then maybe you should check if that Publication data class conforms to the Codable class or not.

Because it is mandatory to conform same protocol for the child class as well, as it's fields are also should be encodable and decodable.

Upvotes: 1

bikram sapkota
bikram sapkota

Reputation: 1124

Any class to be a codeable must have it's all property codeable. Standard library types like String, Int, Double and Foundation types like Date, Data, and UR confirm the codeable protocol but some doesn't.

For e.g below Note class have all properties of string which confirm codable protocol so no error: String has codable protocol so class has no error

But UIImage don't confirm codable protocol so it throw error: Uiimage don't have codable it shows error

Upvotes: 15

Amit Baderia
Amit Baderia

Reputation: 4882

In a similar scenario, I was getting the same issue because the variable name in my CodingKeys was not the same as a class variable. See below

enter image description here

Upvotes: 38

Declan McKenna
Declan McKenna

Reputation: 4870

Bit of a daft one but in case it helps someone else. I got this error because I put enum CodingKeys: CodingKeys instead of enum CodingKeys: CodingKey.

Upvotes: 2

Saif
Saif

Reputation: 2958

You can omit a property from coding keys enum, only if it has a default value.

From apple docs

Omit properties from the CodingKeys enumeration if they won't be present when decoding instances, or if certain properties shouldn't be included in an encoded representation. A property omitted from CodingKeys needs a default value in order for its containing type to receive automatic conformance to Decodable or Codable.

Upvotes: 6

Wissa
Wissa

Reputation: 1592

Simply because your CodingKeys enum is not exhaustive, add publication property to the enum to achieve that.

try this: 

class Bookmark: Codable {
   weak var publication: Publication?
   var indexPath: [Int]
   var locationInText = 0

   // All your properties should be included
   enum CodingKeys: String, CodingKey {
      case indexPath
      case locationInText
      case publication   // this one was missing
   }
}

You wont need the init method anymore as the implementation now can be synthesized.

Upvotes: 25

kalpesh jetani
kalpesh jetani

Reputation: 1803

In-short, while implementing Codable, all properties which are non-primitive data type (mean class type or may it can be objective-c class) must be Codable.

   weak var publication: Publication?

in this case publication is of type class so Publication must have implemented Codable

Upvotes: 4

iOSer
iOSer

Reputation: 2326

On the hindsight, I received a similar error when trying to set Codable to my class which consisted of NSNumber type variables. See image below:

enter image description here

Changing NSNumber to primitive data type Int resolved the issue. See below:

enter image description here

I'm guessing this might be true for other datatypes that require bridging to Swift Standard Library Value Types such as NSString, NSArray and so on

Upvotes: 33

mikepj
mikepj

Reputation: 1338

Another reason you could get this message is if your CodingKeys enum isn't exhaustive. If you have three properties in the data type, then your CodingKeys enum needs to have three property/name cases as well.

Upvotes: 41

ThunderGrad
ThunderGrad

Reputation: 89

I had a similar issue which I stumbled upon this fix to. As I am new to Swift, I am unsure as to why it works! If anyone knows I would appreciate the knowledge.

I changed this:

let id, type: Int

to this:

let id: Int
let type: Int

Upvotes: 0

matt
matt

Reputation: 534885

Why am I getting a "Type 'Bookmark' does not conform to protocol 'Decodable'" error message

It's either because Publication isn't Decodable (you have not shown what it is, so it's hard to tell) or because of the weak designation on publication.

Either way, it's easy to fix: you just need to implement init(from:) to complete the implementation of Decodable; the compiler is simply telling you that this implementation cannot be synthesized.

Upvotes: 51

David Pasztor
David Pasztor

Reputation: 54706

The compiler cannot synthesise the required init(from:) method due to the weak reference, so you need to write it yourself.

class Bookmark: Codable {
    weak var publication: Publication?
    var indexPath: [Int]
    var locationInText = 0

    private enum CodingKeys: String, CodingKey {
        case indexPath
        case locationInText
    }

    init(publication: Publication?, indexPath: [Int]) {
        self.publication = publication
        self.indexPath = indexPath
    }

    required init(from decoder:Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        indexPath = try values.decode([Int].self, forKey: .indexPath)
        locationInText = try values.decode(Int.self, forKey: .locationInText)
    }
}

Upvotes: 73

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