“致命错误:在展开可选值时意外发现 nil"是什么意思?意思? [英] What does "Fatal error: Unexpectedly found nil while unwrapping an Optional value" mean?

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问题描述

我的 Swift 程序因 EXC_BAD_INSTRUCTION 和以下类似错误之一而崩溃.这个错误是什么意思，我该如何解决?

<块引用>

致命错误:在展开可选值时意外发现 nil

或

<块引用>

致命错误:在隐式解包可选值时意外发现 nil

<小时>

本帖旨在收集意外发现nil"问题的答案，以免分散和难以找到.随意添加您自己的答案或

这种崩溃可以通过两种不同的强制解包发生:

1.显式强制展开

这是通过可选的 ! 运算符完成的.例如:

let anOptionalString: String?打印(anOptionalString！)//<-崩溃

<块引用>

致命错误:在展开可选值时意外发现 nil

因为 anOptionalString 在这里是 nil，你会在强制解包的那一行崩溃.

2.隐式展开的可选项

这些是用 ! 定义的，而不是在类型后面的 ?.

var optionalDouble:双倍！//这个值在任何使用它的地方都被隐式解包

假定这些选项包含一个值.因此，无论何时您访问一个隐式解包的可选项，它都会自动为您强制解包.如果它不包含一个值，它就会崩溃.

print(optionalDouble)//<- 崩溃

<块引用>

致命错误:在隐式解包可选值时意外发现 nil

为了找出导致崩溃的变量，您可以按住⌥同时单击以显示定义，您可能会在其中找到可选类型.

IBOutlets，尤其是，通常是隐式解包的选项.这是因为您的 xib 或故事板将在运行时链接出口，在初始化之后.因此，您应该确保在加载之前没有访问插座.您还应该检查故事板/xib 文件中的连接是否正确，否则这些值在运行时将为 nil，因此当它们被隐式解包时会崩溃.修复连接时，尝试删除定义插座的代码行，然后重新连接它们.

我什么时候应该强制打开一个 Optional?

显式强制展开

作为一般规则，您永远不应该使用 ! 运算符显式强制解包可选项.在某些情况下，使用 ! 是可以接受的——但只有当你 100% 确定可选包含一个值时，你才应该使用它.

虽然可能在某些情况下您可以使用强制解包，但正如您所知道的事实一个可选项包含一个值 - 没有 单个地方，你不能安全地解开那个可选的.

隐式展开的可选项

这些变量的设计目的是让您可以将它们的分配推迟到您的代码中.您有责任在您访问它们之前确保它们具有价值.然而，因为它们涉及强制展开，所以它们本质上仍然是不安全的——因为它们假设你的值是非 nil，即使分配 nil 是有效的.

您应该只将隐式解包的选项用作最后的手段.如果您可以使用 懒惰变量，或提供默认值 变量 - 您应该这样做，而不是使用隐式解包的可选值.

但是，有一个隐式解包的可选项在少数情况下是有益的，您仍然可以使用各种方式安全地拆开它们，如下所列——但您应该始终谨慎地使用它们.

我如何安全地处理 Optionals?

检查可选项是否包含值的最简单方法是将其与 nil 进行比较.

if anOptionalInt != nil {打印(包含一个值！")} 别的 {打印(不包含值.")}

然而，在使用可选项时，99.9% 的情况下，您实际上想要访问它包含的值，如果它包含一个值的话.为此，您可以使用可选绑定.

可选绑定

Optional Binding 允许您检查一个 optional 是否包含一个值——并允许您将解包的值分配给一个新的变量或常量.它使用语法 if let x = anOptional {...} 或 if var x = anOptional {...}，具体取决于您是否需要修改绑定后的新变量.

例如:

if let number = anOptionalInt {打印(包含一个值！它是(数字)！")} 别的 {打印(不包含数字")}

这首先检查可选项是否包含一个值.如果它确实，那么解包"值被分配给一个新变量(number)——然后你可以自由地使用它，就好像它是非可选的一样.如果可选的不包含一个值，那么else子句将被调用，正如你所期望的那样.

可选绑定的巧妙之处在于，您可以同时解开多个可选绑定.您可以只用逗号分隔语句.如果所有的选项都被解包，这个语句就会成功.

var anOptionalInt : Int?var anOptionalString : 字符串?如果让 number = anOptionalInt，让 text = anOptionalString {print("anOptionalInt 包含一个值:(number).anOptionalString 也是如此，它是:(text)")} 别的 {print(一个或多个选项不包含值")}

另一个巧妙的技巧是，您还可以在解包后使用逗号检查值的特定条件.

if let number = anOptionalInt, number >0 {print("anOptionalInt 包含一个值:(number)，并且它大于零！")}

在 if 语句中使用可选绑定的唯一问题是您只能从语句的范围内访问解包的值.如果需要从语句范围之外访问值，可以使用guard 语句.

A guard 语句 允许您定义成功的条件——并且当前作用域仅在满足该条件时才会继续执行.它们使用语法 guard condition else {...} 定义.

因此，要将它们与可选绑定一起使用，您可以这样做:

guard let number = anOptionalInt else {返回}

(请注意，在guard body 内，您必须使用控制转移语句 以退出当前执行代码的范围).上>

如果 anOptionalInt 包含一个值，它将被解包并分配给新的 number 常量.守卫之后的代码将继续执行.如果它不包含值——guard 会执行括号内的代码，这会导致控制权的转移，从而导致后面的代码不会被执行.

关于guard 语句的真正巧妙之处在于，现在可以在语句后面的代码中使用解包的值(因为我们知道，如果可选值具有值，将来的代码仅才能执行).这非常适合消除通过嵌套多个 if 语句创建的 '末日金字塔'.

例如:

guard let number = anOptionalInt else {返回}print("anOptionalInt 包含一个值，它是:(number)!")

Guards 还支持与 if 语句支持的相同的巧妙技巧，例如同时解开多个可选项和使用 where 子句.

是否使用 if 或 guard 语句完全取决于将来的代码是否要求可选值包含值.

无合并运算符

Nil Coalescing Operator 是 三元条件运算符，主要用于将可选项转换为非可选项.它的语法为 a ??b，其中 a 是可选类型，b 与 a 的类型相同(尽管通常是非可选的).

它本质上让你说如果 a 包含一个值，打开它.如果没有，则返回 b 代替".例如，您可以这样使用它:

let number = anOptionalInt ??0

这将定义一个 Int 类型的 number 常量，它将包含 anOptionalInt 的值，如果它包含一个值，或者0 否则.

这只是以下内容的简写:

let number = anOptionalInt != nil ?一个OptionalInt！: 0

可选链

您可以使用 Optional Chaining 以调用一个方法或访问一个可选的属性.这只需在使用时在变量名后加上 ? 即可完成.

例如，假设我们有一个变量 foo，类型为可选的 Foo 实例.

var foo : Foo?

如果我们想在 foo 上调用一个不返回任何东西的方法，我们可以简单地做:

foo?.doSomethingInteresting()

如果 foo 包含一个值，这个方法将被调用.如果没有，就不会发生什么坏事——代码将继续执行.

(这类似于在 Objective-C 中向 nil 发送消息)

因此这也可用于设置属性和调用方法.例如:

foo?.bar = Bar()

同样，如果 foo 是 nil，这里不会发生任何坏事.您的代码将继续执行.

可选链可以让你做的另一个巧妙的技巧是检查设置属性或调用方法是否成功.您可以通过将返回值与 nil 进行比较来实现.

(这是因为在不返回任何内容的方法上，可选值将返回 Void? 而不是 Void)

例如:

if (foo?.bar = Bar()) != nil {打印(栏设置成功")} 别的 {打印(栏没有设置成功")}

然而，当尝试访问属性或调用返回值的方法时，事情变得有点棘手.因为 foo 是可选的，所以从它返回的任何东西也是可选的.为了解决这个问题，您可以使用上述方法之一解包返回的选项 - 或者在访问方法或调用返回值的方法之前解包 foo 本身.

此外，顾名思义，您可以将这些语句链接"在一起.这意味着如果 foo 有一个可选属性 baz，它有一个属性 qux – 你可以写如下:

let optionalQux = foo?.baz?.qux

同样，因为 foo 和 baz 是可选的，从 qux 返回的值将始终是可选的，无论 qux 本身是可选的.

map 和 flatMap

一个经常未被充分利用的可选功能是能够使用 map 和 flatMap 函数.这些允许您将非可选转换应用于可选变量.如果可选项具有值，则可以对其应用给定的转换.如果它没有值，它将保持nil.

例如，假设您有一个可选字符串:

让一个OptionalString:String?

通过对它应用 map 函数——我们可以使用 stringByAppendingString 函数将它连接到另一个字符串.

因为 stringByAppendingString 接受一个非可选字符串参数，我们不能直接输入我们的可选字符串.然而，通过使用map，我们可以使用allow stringByAppendingString，如果anOptionalString 有一个值.

例如:

var anOptionalString:String?=酒吧"anOptionalString = anOptionalString.map {unwrappedString in返回foo".stringByAppendingString(unwrappedString)}打印(anOptionalString)//可选(foobar")

然而，如果 anOptionalString 没有值，map 将返回 nil.例如:

var anOptionalString:String?anOptionalString = anOptionalString.map {unwrappedString in返回foo".stringByAppendingString(unwrappedString)}打印(anOptionalString)//零

flatMap 与 map 的工作方式类似，不同之处在于它允许您从闭包主体中返回另一个可选.这意味着您可以将可选输入到需要非可选输入的流程中，但可以输出可选本身.

试试！

Swift 的错误处理系统可以安全地与 Do-Try-Catch:

做{让结果 = 尝试 someThrowingFunc()} 抓住 {打印(错误)}

如果 someThrowingFunc() 抛出错误，该错误将被安全地捕获在 catch 块中.

您在 catch 块中看到的 error 常量尚未被我们声明 - 它是由 catch 自动生成的.

您也可以自己声明error，它的优点是可以将其转换为有用的格式，例如:

做{让结果 = 尝试 someThrowingFunc()} catch let error as NSError {打印(错误.调试说明)}

使用 try 这种方式是尝试、捕获和处理来自抛出函数的错误的正确方法.

还有 try? 吸收错误:

if let result = try?someThrowingFunc() {//凉爽的} 别的 {//处理失败，但没有可用的错误信息}

但是 Swift 的错误处理系统也提供了一种强制尝试"的方法.使用 try!:

让结果=尝试！someThrowingFunc()

本文中解释的概念也适用于此处:如果抛出错误，应用程序将崩溃.

你应该只使用 try! 如果你能证明它的结果在你的上下文中永远不会失败 - 这非常罕见.

大多数情况下，您将使用完整的 Do-Try-Catch 系统 - 以及可选的 try?，在极少数情况下，处理错误并不重要.

资源

My Swift program is crashing with EXC_BAD_INSTRUCTION and one of the following similar errors. What does this error mean, and how do I fix it?

Fatal error: Unexpectedly found nil while unwrapping an Optional value

or

Fatal error: Unexpectedly found nil while implicitly unwrapping an Optional value

This post is intended to collect answers to "unexpectedly found nil" issues, so that they are not scattered and hard to find. Feel free to add your own answer or edit the existing wiki answer.

解决方案

This answer is community wiki. If you feel it could be made better, feel free to edit it!

Background: What’s an Optional?

In Swift, Optional<Wrapped> is an option type: it can contain any value from the original ("Wrapped") type, or no value at all (the special value nil). An optional value must be unwrapped before it can be used.

Optional is a generic type, which means that Optional<Int> and Optional<String> are distinct types — the type inside <> is called the Wrapped type. Under the hood, an Optional is an enum with two cases: .some(Wrapped) and .none, where .none is equivalent to nil.

Optionals can be declared using the named type Optional<T>, or (most commonly) as a shorthand with a ? suffix.

var anInt: Int = 42
var anOptionalInt: Int? = 42
var anotherOptionalInt: Int?  // `nil` is the default when no value is provided
var aVerboseOptionalInt: Optional<Int>  // equivalent to `Int?`

anOptionalInt = nil // now this variable contains nil instead of an integer

Optionals are a simple yet powerful tool to express your assumptions while writing code. The compiler can use this information to prevent you from making mistakes. From The Swift Programming Language:

Swift is a type-safe language, which means the language helps you to be clear about the types of values your code can work with. If part of your code requires a String, type safety prevents you from passing it an Int by mistake. Likewise, type safety prevents you from accidentally passing an optional String to a piece of code that requires a non-optional String. Type safety helps you catch and fix errors as early as possible in the development process.

Some other programming languages also have generic option types: for example, Maybe in Haskell, option in Rust, and optional in C++17.

In programming languages without option types, a particular "sentinel" value is often used to indicate the absence of a valid value. In Objective-C, for example, nil (the null pointer) represents the lack of an object. For primitive types such as int, a null pointer can't be used, so you would need either a separate variable (such as value: Int and isValid: Bool) or a designated sentinel value (such as -1 or INT_MIN). These approaches are error-prone because it's easy to forget to check isValid or to check for the sentinel value. Also, if a particular value is chosen as the sentinel, that means it can no longer be treated as a valid value.

Option types such as Swift's Optional solve these problems by introducing a special, separate nil value (so you don't have to designate a sentinel value), and by leveraging the strong type system so the compiler can help you remember to check for nil when necessary.

Why did I get "Fatal error: Unexpectedly found nil while unwrapping an Optional value"?

In order to access an optional’s value (if it has one at all), you need to unwrap it. An optional value can be unwrapped safely or forcibly. If you force-unwrap an optional, and it didn't have a value, your program will crash with the above message.

Xcode will show you the crash by highlighting a line of code. The problem occurs on this line.

This crash can occur with two different kinds of force-unwrap:

1. Explicit Force Unwrapping

This is done with the ! operator on an optional. For example:

let anOptionalString: String?
print(anOptionalString!) // <- CRASH

Fatal error: Unexpectedly found nil while unwrapping an Optional value

As anOptionalString is nil here, you will get a crash on the line where you force unwrap it.

2. Implicitly Unwrapped Optionals

These are defined with a !, rather than a ? after the type.

var optionalDouble: Double!   // this value is implicitly unwrapped wherever it's used

These optionals are assumed to contain a value. Therefore whenever you access an implicitly unwrapped optional, it will automatically be force unwrapped for you. If it doesn’t contain a value, it will crash.

print(optionalDouble) // <- CRASH

Fatal error: Unexpectedly found nil while implicitly unwrapping an Optional value

In order to work out which variable caused the crash, you can hold ⌥ while clicking to show the definition, where you might find the optional type.

IBOutlets, in particular, are usually implicitly unwrapped optionals. This is because your xib or storyboard will link up the outlets at runtime, after initialization. You should therefore ensure that you’re not accessing outlets before they're loaded in. You also should check that the connections are correct in your storyboard/xib file, otherwise the values will be nil at runtime, and therefore crash when they are implicitly unwrapped. When fixing connections, try deleting the lines of code that define your outlets, then reconnect them.

When should I ever force unwrap an Optional?

Explicit Force Unwrapping

As a general rule, you should never explicitly force unwrap an optional with the ! operator. There may be cases where using ! is acceptable – but you should only ever be using it if you are 100% sure that the optional contains a value.

While there may be an occasion where you can use force unwrapping, as you know for a fact that an optional contains a value – there is not a single place where you cannot safely unwrap that optional instead.

Implicitly Unwrapped Optionals

These variables are designed so that you can defer their assignment until later in your code. It is your responsibility to ensure they have a value before you access them. However, because they involve force unwrapping, they are still inherently unsafe – as they assume your value is non-nil, even though assigning nil is valid.

You should only be using implicitly unwrapped optionals as a last resort. If you can use a lazy variable, or provide a default value for a variable – you should do so instead of using an implicitly unwrapped optional.

However, there are a few scenarios where implicitly unwrapped optionals are beneficial, and you are still able to use various ways of safely unwrapping them as listed below – but you should always use them with due caution.

How can I safely deal with Optionals?

The simplest way to check whether an optional contains a value, is to compare it to nil.

if anOptionalInt != nil {
    print("Contains a value!")
} else {
    print("Doesn’t contain a value.")
}

However, 99.9% of the time when working with optionals, you’ll actually want to access the value it contains, if it contains one at all. To do this, you can use Optional Binding.

Optional Binding

Optional Binding allows you to check if an optional contains a value – and allows you to assign the unwrapped value to a new variable or constant. It uses the syntax if let x = anOptional {...} or if var x = anOptional {...}, depending if you need to modify the value of the new variable after binding it.

For example:

if let number = anOptionalInt {
    print("Contains a value! It is (number)!")
} else {
    print("Doesn’t contain a number")
}

What this does is first check that the optional contains a value. If it does, then the ‘unwrapped’ value is assigned to a new variable (number) – which you can then freely use as if it were non-optional. If the optional doesn’t contain a value, then the else clause will be invoked, as you would expect.

What’s neat about optional binding, is you can unwrap multiple optionals at the same time. You can just separate the statements with a comma. The statement will succeed if all the optionals were unwrapped.

var anOptionalInt : Int?
var anOptionalString : String?

if let number = anOptionalInt, let text = anOptionalString {
    print("anOptionalInt contains a value: (number). And so does anOptionalString, it’s: (text)")
} else {
    print("One or more of the optionals don’t contain a value")
}

Another neat trick is that you can also use commas to check for a certain condition on the value, after unwrapping it.

if let number = anOptionalInt, number > 0 {
    print("anOptionalInt contains a value: (number), and it’s greater than zero!")
}

The only catch with using optional binding within an if statement, is that you can only access the unwrapped value from within the scope of the statement. If you need access to the value from outside of the scope of the statement, you can use a guard statement.

A guard statement allows you to define a condition for success – and the current scope will only continue executing if that condition is met. They are defined with the syntax guard condition else {...}.

So, to use them with an optional binding, you can do this:

guard let number = anOptionalInt else {
    return
}

(Note that within the guard body, you must use one of the control transfer statements in order to exit the scope of the currently executing code).

If anOptionalInt contains a value, it will be unwrapped and assigned to the new number constant. The code after the guard will then continue executing. If it doesn’t contain a value – the guard will execute the code within the brackets, which will lead to transfer of control, so that the code immediately after will not be executed.

The real neat thing about guard statements is the unwrapped value is now available to use in code that follows the statement (as we know that future code can only execute if the optional has a value). This is a great for eliminating ‘pyramids of doom’ created by nesting multiple if statements.

For example:

guard let number = anOptionalInt else {
    return
}

print("anOptionalInt contains a value, and it’s: (number)!")

Guards also support the same neat tricks that the if statement supported, such as unwrapping multiple optionals at the same time and using the where clause.

Whether you use an if or guard statement completely depends on whether any future code requires the optional to contain a value.

Nil Coalescing Operator

The Nil Coalescing Operator is a nifty shorthand version of the ternary conditional operator, primarily designed to convert optionals to non-optionals. It has the syntax a ?? b, where a is an optional type and b is the same type as a (although usually non-optional).

It essentially lets you say "If a contains a value, unwrap it. If it doesn’t then return b instead". For example, you could use it like this:

let number = anOptionalInt ?? 0

This will define a number constant of Int type, that will either contain the value of anOptionalInt, if it contains a value, or 0 otherwise.

It’s just shorthand for:

let number = anOptionalInt != nil ? anOptionalInt! : 0

Optional Chaining

You can use Optional Chaining in order to call a method or access a property on an optional. This is simply done by suffixing the variable name with a ? when using it.

For example, say we have a variable foo, of type an optional Foo instance.

var foo : Foo?

If we wanted to call a method on foo that doesn’t return anything, we can simply do:

foo?.doSomethingInteresting()

If foo contains a value, this method will be called on it. If it doesn’t, nothing bad will happen – the code will simply continue executing.

(This is similar behaviour to sending messages to nil in Objective-C)

This can therefore also be used to set properties as well as call methods. For example:

foo?.bar = Bar()

Again, nothing bad will happen here if foo is nil. Your code will simply continue executing.

Another neat trick that optional chaining lets you do is check whether setting a property or calling a method was successful. You can do this by comparing the return value to nil.

(This is because an optional value will return Void? rather than Void on a method that doesn’t return anything)

For example:

if (foo?.bar = Bar()) != nil {
    print("bar was set successfully")
} else {
    print("bar wasn’t set successfully")
}

However, things become a little bit more tricky when trying to access properties or call methods that return a value. Because foo is optional, anything returned from it will also be optional. To deal with this, you can either unwrap the optionals that get returned using one of the above methods – or unwrap foo itself before accessing methods or calling methods that return values.

Also, as the name suggests, you can ‘chain’ these statements together. This means that if foo has an optional property baz, which has a property qux – you could write the following:

let optionalQux = foo?.baz?.qux

Again, because foo and baz are optional, the value returned from qux will always be an optional regardless of whether qux itself is optional.

map and flatMap

An often underused feature with optionals is the ability to use the map and flatMap functions. These allow you to apply non-optional transforms to optional variables. If an optional has a value, you can apply a given transformation to it. If it doesn’t have a value, it will remain nil.

For example, let’s say you have an optional string:

let anOptionalString:String?

By applying the map function to it – we can use the stringByAppendingString function in order to concatenate it to another string.

Because stringByAppendingString takes a non-optional string argument, we cannot input our optional string directly. However, by using map, we can use allow stringByAppendingString to be used if anOptionalString has a value.

For example:

var anOptionalString:String? = "bar"

anOptionalString = anOptionalString.map {unwrappedString in
    return "foo".stringByAppendingString(unwrappedString)
}

print(anOptionalString) // Optional("foobar")

However, if anOptionalString doesn’t have a value, map will return nil. For example:

var anOptionalString:String?

anOptionalString = anOptionalString.map {unwrappedString in
    return "foo".stringByAppendingString(unwrappedString)
}

print(anOptionalString) // nil

flatMap works similarly to map, except it allows you to return another optional from within the closure body. This means you can input an optional into a process that requires a non-optional input, but can output an optional itself.

try!

Swift's error handling system can be safely used with Do-Try-Catch:

do {
    let result = try someThrowingFunc() 
} catch {
    print(error)
}

If someThrowingFunc() throws an error, the error will be safely caught in the catch block.

The error constant you see in the catch block has not been declared by us - it's automatically generated by catch.

You can also declare error yourself, it has the advantage of being able to cast it to a useful format, for example:

do {
    let result = try someThrowingFunc()    
} catch let error as NSError {
    print(error.debugDescription)
}

Using try this way is the proper way to try, catch and handle errors coming from throwing functions.

There's also try? which absorbs the error:

if let result = try? someThrowingFunc() {
    // cool
} else {
    // handle the failure, but there's no error information available
}

But Swift's error handling system also provides a way to "force try" with try!:

let result = try! someThrowingFunc()

The concepts explained in this post also apply here: if an error is thrown, the application will crash.

You should only ever use try! if you can prove that its result will never fail in your context - and this is very rare.

Most of the time you will use the complete Do-Try-Catch system - and the optional one, try?, in the rare cases where handling the error is not important.

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