This is because you are trying to use a value which can possible be nil, but you decided you don't want to have to check it, but instead assume its set when you uses it and define it as !, there are different philosophies on use of variable set as force unwrap, some people are against there use at all, I personal think they are ok for things that will crash all the time and are simple to reason about, usually references to resource, like outlets to xib files, or uses of images with you app that are part of your assets, if these are not set up properly, you app is going to crash straight away, for a very obvious reason, you can get into difficult when the order of objects being created can be uncertain, and trying to reason solutions to this can be difficult, it usually means a bad design as even it you make them optional, calls to you optional variable may not ever be executed, some projects can demand use of force unwraps for security reasons, things like banking apps, because they want the app to crash rather then continue to work in an unplanned way.

TL;博士回答

除了极少数例外，这条规则是黄金法则:

避免使用!

声明变量是可选的(?)，而不是隐式地打开可选的(IUO) (!)

换句话说，应该使用: var nameOfDaughter:字符串?

而不是: var nameOfDaughter:字符串!

使用if let或guard let展开可选变量

要么像这样展开变量:

if let nameOfDaughter = nameOfDaughter {
    print("My daughters name is: \(nameOfDaughter)")
}

或者像这样:

guard let nameOfDaughter = nameOfDaughter else { return }
print("My daughters name is: \(nameOfDaughter)")

这个答案是简洁的，为了充分理解阅读接受的答案

资源

避免强制展开

当我试图从prepareforsegue方法中设置outlet值时，我有过这样的错误:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // This line pops up the error
            destination.nameLabel.text = item.name
        }
    }
}

然后我发现我不能设置目标控制器出口的值，因为控制器还没有加载或初始化。

所以我是这样解决的:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // Created this method in the destination Controller to update its outlets after it's being initialized and loaded
            destination.updateView(itemData:  item)
        }
    }
}

目的地控制器:

// This variable to hold the data received to update the Label text after the VIEW DID LOAD
var name = ""

// Outlets
@IBOutlet weak var nameLabel: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    // Do any additional setup after loading the view.
    nameLabel.text = name
}

func updateView(itemDate: ObjectModel) {
    name = itemDate.name
}

我希望这个答案能帮助那些有同样问题的人，因为我发现标记的答案对理解可选选项及其工作方式是很好的资源，但并没有直接解决问题本身。

我在从表视图控制器到视图控制器进行segue时遇到了这个错误，因为我忘记在主故事板中为视图控制器指定自定义类名。

一些简单的东西值得检查，如果其他看起来都没问题

基本上，你试图在Swift只允许非nil值的地方使用nil值，通过告诉编译器信任你，那里永远不会有nil值，从而允许你的应用程序编译。

有几种情况会导致这种致命错误:

forced unwraps: let user = someVariable! If someVariable is nil, then you'll get a crash. By doing a force unwrap you moved the nil check responsibility from the compiler to you, basically by doing a forced unwrap you're guaranteeing to the compiler that you'll never have nil values there. And guess what it happens if somehow a nil value ends in in someVariable? Solution? Use optional binding (aka if-let), do the variable processing there: if user = someVariable { // do your stuff } forced (down)casts: let myRectangle = someShape as! Rectangle Here by force casting you tell the compiler to no longer worry, as you'll always have a Rectangle instance there. And as long as that holds, you don't have to worry. The problems start when you or your colleagues from the project start circulating non-rectangle values. Solution? Use optional binding (aka if-let), do the variable processing there: if let myRectangle = someShape as? Rectangle { // yay, I have a rectangle } Implicitly unwrapped optionals. Let's assume you have the following class definition: class User { var name: String! init() { name = "(unnamed)" } func nicerName() { return "Mr/Ms " + name } } Now, if no-one messes up with the name property by setting it to nil, then it works as expected, however if User is initialized from a JSON that lacks the name key, then you get the fatal error when trying to use the property. Solution? Don't use them :) Unless you're 102% sure that the property will always have a non-nil value by the time it needs to be used. In most cases converting to an optional or non-optional will work. Making it non-optional will also result in the compiler helping you by telling the code paths you missed giving a value to that property Unconnected, or not yet connected, outlets. This is a particular case of scenario #3. Basically you have some XIB-loaded class that you want to use. class SignInViewController: UIViewController { @IBOutlet var emailTextField: UITextField! } Now if you missed connecting the outlet from the XIB editor, then the app will crash as soon as you'll want to use the outlet. Solution? Make sure all outlets are connected. Or use the ? operator on them: emailTextField?.text = "my@email.com". Or declare the outlet as optional, though in this case the compiler will force you to unwrap it all over the code. Values coming from Objective-C, and that don't have nullability annotations. Let's assume we have the following Objective-C class: @interface MyUser: NSObject @property NSString *name; @end Now if no nullability annotations are specified (either explicitly or via NS_ASSUME_NONNULL_BEGIN/NS_ASSUME_NONNULL_END), then the name property will be imported in Swift as String! (an IUO - implicitly unwrapped optional). As soon as some swift code will want to use the value, it will crash if name is nil. Solution? Add nullability annotations to your Objective-C code. Beware though, the Objective-C compiler is a little bit permissive when it comes to nullability, you might end up with nil values, even if you explicitly marked them as nonnull.

首先，您应该知道什么是Optional值。 详细信息请参见《Swift编程语言》。

其次，您应该知道可选值有两个状态。一个是全值，另一个是空值。因此，在实现一个可选值之前，应该检查它是哪个状态。

你可以用if let…或者守卫让…Else等等。

还有一种方法，如果你不想在实现之前检查变量的状态，你也可以使用var buildingName = buildingName ??“buildingName”。