Fatal error: swift 2.3 when unwrapping value

Question

I'm trying to send a saved string to the apple watch with WCsession methode. But when I do this I get an error in swift 2.3: fatal error: unexpectedly found nil while unwrapping an optional value !!

func Reloadip() {


    let ip =  nsdefauts.object(forKey: saved)


    let requestValues = ["send" : "A" , "IP" : ip as! String ]
    print(requestValues)

    if(WCSession.isSupported()){
        session!.sendMessage(requestValues, replyHandler: nil, errorHandler: nil)
        print("sended ip")
    }

}

the error is made at the "ip as! String line" how can I fix this. ?

regards Quinn

just remove ! from "as! String" – Mitul Marsoniya Jun 17 '16 at 07:29 — Mitul Marsoniya, Jun 17 '16 at 07:29

Shadow Of · Accepted Answer · 2016-06-17T08:00:49.580

2

You should use more safe code instead

func Reloadip() {

    guard let ip = nsdefauts.objectForKey(saved) as? String else {
        print("there is no saved ip")
        return
    }

    let requestValues = ["send" : "A" , "IP" : ip]
    print(requestValues)

    if(WCSession.isSupported()){
        session?.sendMessage(requestValues, replyHandler: nil, errorHandler: nil)
        print("sended ip")
    }  
}

edited Jun 17 '16 at 08:00

answered Jun 17 '16 at 07:28

Shadow Of

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This is the right approach, but I'd prefer a guard over an if for something like this. – overactor Jun 17 '16 at 07:35
@overactor probably guard more readable; I replaced to it in my answer. – Shadow Of Jun 17 '16 at 08:02

Fatal error: swift 2.3 when unwrapping value

1 Answers1