Non-optional values in Swift-structure returns optional values

Question

I have a structure in one ViewController which has a couple of non-optional members like:

struct MyStruct {
    let id: Int64!
    let name: String!
    func checkValues() -> Bool {
        if id == nil || name == nil {
            return false
        } else {
            print("id: \(id) name: \(name)"); return true
        }
    }
}

I create an instance of the struct in another VC (let temp = ViewController.MyStruct(id: 10, name: "Example")) and pass it to the one where I created the struct via a Segue. When executing temp.checkValues() I get

id: Optional(10) name: Optional("Example")

From my understanding of (non-)optionals I shouldn't get the Optional() since I declared the variables with !. Did I get it wrong? Right now I just put an exclamation mark whenever I use it. I also tried using init in the struct but that didn't change anything. Since this is not as intended (if I'm right) I fear I didn't get the full concept... Please tell me what is wrong with my code / me.

score 7 · Accepted Answer · answered Sep 06 '17 at 13:18

7

Did I get it wrong?

Yes. Your variables are optionals - the ! operator is a force unwrap operator, so your variables are implicitly unwrapped. If you don't want your variables to be optionals, just drop !.

The ! operator exists in Swift to facilitate bridging from Objective-C, for instance for IBOutlet variables. Using ! operator aside from that is a bad practice.

answered Sep 06 '17 at 13:18

mag_zbc

6,801
14
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Thanks a lot! That makes sense... I had the wrong impression that when adding the `!` I would make sure that they cannot be optionals... – Michael Sep 06 '17 at 13:25

score 5 · Answer 2 · answered Sep 06 '17 at 13:17

5

! is an implictly unwrapped optional. Everything variable declaration's type that ends with ! is an optional. You dont have to declare them to be Optionals. Just use them normally and the initializer will take care of them:

struct MyStruct {
    let id: Int64
    let name: String
    func checkValues() -> Bool {
        print("id: \(id) name: \(name)"); return true
    }
}

answered Sep 06 '17 at 13:17

Papershine

4,995
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Yes, thanks! That makes sense... – Michael Sep 06 '17 at 13:23

Non-optional values in Swift-structure returns optional values

2 Answers2