Swift Variable Nil causes fatal error: unexpectedly found nil while unwrapping an Optional value

Question

Firstly, I'm very new to Swift, so keep that in mind before you berate my question for being an easy problem to solve.

I'm trying to use a button to send the contents of text field to a table. This bit of code defines the variable and you can see the sender:

import UIKit

var bookTitle:String!

class secondViewController: UIViewController {

@IBOutlet weak var titleField: UITextField!

@IBAction func addBook(sender: AnyObject) {
    bookTitle = (titleField.text)!
}

This is the code where I'm receiving the error message:

import UIKit

var cellContent = ["Book 1", "Book 2", "Book 3", "Book 4"]

class firstViewController: UIViewController, UITableViewDelegate {

@IBOutlet weak var table: UITableView!

override func viewDidLoad() {
    super.viewDidLoad()
    // Do any additional setup after loading the view, typically from a nib.

    cellContent.insert(bookTitle, atIndex: 0)
//THIS IS THE ERROR:  EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)
}

func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {

    return cellContent.count
}

From searching online, I think the problem is that the variable "bookTitle" is nil, but I'm not sure how to fix that.

With `var bookTitle:String!` you've made your variable an *implicitly unwrapped optional* (this is what you want to search and read about). It means it must not be nil in any case. If you never actioned `addBook`, then `bookTitle` is nil, so the app crashes when you try to use it. — Eric Aya, Apr 20 '16 at 21:22
[My answer here](http://stackoverflow.com/a/36360605/2976878) may be helpful in explaining how to properly deal with optionals. Essentially, you should only ever be using implicitly unwrapped optionals as a last resort. If you can make that variable non-optional with a default value, a lazy variable or a normal optional – you should do so. You should only ever be using IUOs if you know beyond doubt that it will *always* be non-nil after you initialise it and you never access it before initialising it. — Hamish, Apr 20 '16 at 21:43

score 1 · Answer 1 · 2016-04-20T21:44:32.003

It's because of your

var bookTitle:String!

Because you are accessing it before it was set. You can check that using the didSet.

var bookTitle: String! {
   didSet {
      print("did set called")
   }
}

You should use this declaration instead:

var bookTitle: String?

It's much safer because you have to unwrap your variable when accessing it. You should read the Swift documentation chapter about optional and explicit types.

For reference:

Because the value of an implicitly unwrapped optional is automatically unwrapped when you use it, there’s no need to use the ! operator to unwrap it. That said, if you try to use an implicitly unwrapped optional that has a value of nil, you’ll get a runtime error.

See: https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Types.html#//apple_ref/doc/uid/TP40014097-CH31-ID445

Chapter: Implicitly Unwrapped Optional Type

Swift Variable Nil causes fatal error: unexpectedly found nil while unwrapping an Optional value

1 Answers1