How to handle this: Swift.ImplicitlyUnwrappedOptional.some

Question

Since Xcode 9.3 I get my string variables in my models wrapped with "Swift.ImplicitlyUnwrappedOptional.some"

I don't know how this is happening but its ruined my apps!

I create my models like this:

struct MyModel {
  var myString:String!
  init(){} // for creating empty instances.
  init(son:JSON){
     myString = son["theStringKey"].string
  }
}

Till Xcode 9.3 when I printed my model I got pure string variable from it.

But after Xcode 9.3 it gives me wrapped string inside of Swift.ImplicitlyUnwrappedOptional.some("MyStringValue")

Any Idea what is this?

Outside of this being the weirdest `init()` I've seen, it would help to see what the expected JSON payload would look like. — SirCJ, Apr 10 '18 at 07:12
@Sweeper the print output is exactly like this : Swift.ImplicitlyUnwrappedOptional.some("MyStringValue") — Ahmadreza, Apr 10 '18 at 07:15
@Alfi No, I mean the line of code starting with `print(...`. Can you show that? — Sweeper, Apr 10 '18 at 07:15
@Sweeper: like this: print(MyModel(son: JSONVARIABLE).myString) — Ahmadreza, Apr 10 '18 at 07:39
You're using SwiftyJSON. This library has many different accessors for its objects. The one you're using, `.string`, returns an optional string. — Eric Aya, Apr 10 '18 at 07:52

score 2 · Accepted Answer · edited Oct 12 '19 at 08:20

Today i have facing same error in Xcode 9.3 V

Just add option value ?? "" your problem is solved

  let paramData:Dictionary<String,Any> = [ PARAM.REQUEST_ID : requestId ?? ""]

Here you can pass json replace in dictionary

struct MyModel {
    var myString:String!
    init(){} // for creating empty instances.
    init(son:Dictionary<String,Any>){
        myString = son["theStringKey"] as! String
    }
}
print(MyModel.init(son: ["theStringKey":"Harshil"]).myString ?? "")

How to handle this: Swift.ImplicitlyUnwrappedOptional.some

1 Answers1