you are forcing the downcast of that variable as though you are certain it is of that type, therefore safe to operate on and not crash your application. Great explanation. Though, your final solution and the original one outputs different results (see my edit). at runtime, the debug console will print Optional("foobar"). would result in b havi… Check if a String Contains Another String in Swift. This does not mean that you are working with a type of String. print (someValue!) Nice explanation of the optional types. Rink Rats is a blog of observations, predictions and commentary. Optional Binding is a mess topic to understand, but you can! Does Swift have dynamic dispatch and virtual methods? '?' To address this, we could do a nil check before we force unwrap an optional. An implicitly unwrapped optional is a normal optional behind the scenes, but can also be used like a non optional value, so yes you are correct. Like before when we forcibly unwrapped an optional we used the ! In order to understand this article, ... We know we can easily unwrap an optional using the if condition but the code will get messy if there are more than one optional that we need to unwrap … ], then you'd know you had a (potentially empty) Dictionary with integer subscripts that yield wrapped contents of any old type. Adjust the arrows between the nodes of two matrices. This cannot go wrong at runtime, and there is no scope for surprises. Optional variables are an important concept in Swift. Unwrapping w/ Type Casting. And it must be able to be downcast as Int, otherwise, like using the bang operator when the value is nil your application will crash. if let unwrapped = name { print("\ (unwrapped.count) letters") } else { print("Missing name.") More annoying are the times when there is no sensible alternative. This means that you are working with a type of String? This will generate a new type and assign the value that resided within that optional to the new non-optional type. The “if let” allows us to unwrap optional values safely only when there is a value, and if not, the code block will not run Swift places a high premium on type safety. Cool. Optional(10) 2. However, in a handful of circumstances you might want to consider force unwrapping the optionals rather than using if let or similar. Instead, force unwraps should be reserved for times when your code is absolutely guaranteed to be safe, as demonstrated above. Update Policy             You might know the type of the something. Now, some people disagree and instead argue that force unwrapping should never happen. imagine all the null pointer exceptions in Java crashing a program because somebody forgot to check for null. acts on the unwrapped value. Let’s look at a situation where optionals are useful. Such optional variables will unwrap automatically and you do not need to use any further exclamation mark at the end of the variable to get the assigned value. NEW: Start my new Ultimate Portfolio App course with a free Hacking with Swift+ trial! Unwrapping Hey Swift folks!, We are going to talk about “Optional Binding” in Swift. means optional chaining expression Optional Binding. bang operator to do so. Automatic Unwrapping. unwraps the optional and outputs 5 on the screen. You must unwrap the value of an Optional instance before you can use it in many contexts. When you’re writing tests there is a nice way to work with optionals … Same with Swift. By default, variables in Swift cannot be set to nil, unless they are defined as an Optional type. Helped clear things up for me. These examples are drawn from real-world Apple APIs that you might not have met yet, but that’s OK – you’ll get the idea. Cannot subscript a value of type '[NSObject : AnyObject]' with an index of type 'String'. For example, we could rewrite our Direction enum to this: With that in place, everywhere we need to get a random direction no longer needs a force unwrap: Before moving on, I want to reiterate one key point here: if you’re only 99% certain that a force unwrap is safe, you shouldn’t use it. How do I call Objective-C code from Swift? Remember, what you don’t want to do is risk a crash – you don’t want to force unwrap an optional only to find that it’s nil. When you cast a type with the as! Note: A variable or type Int is not the same as Int?. Both usernameField and its text property are optionals, so we’ll need to unwrap the optional to get to the value. Swift Force Unwrapping In swift if we define variable as optional means we need to unwrap the variable value using exclamation mark (!) = 10 print(x) Above code will get below result. Because at the end of the day the optional may in fact be nil. You cannot get to that value (in this case “moo”) without unwrapping it. Optional types always return either a wrapped value (the bit with the braces) or nil. We also add both of the parameters to disposeBag to make sure we won't have any memory issues. A great approach is to use if let, like this: if let username = usernameField?.text { // Do something with `username`} In the above code, we can safely use the username constant to authenticate the user. Is this kitchen wall parallel with the joists load-bearing? In this case you will want to use the as keyword to determine this: Note: The as operator is how you type cast in Swift. In other words, you have to assign a non-nil value to the variable. In older languages, like C++ or Java, all pointers are optional, but all basic types are not — unless they’re wrapped in some structure… This is the most dangerous of all the techniques discussed in this tutorial. Of course, like any if statement, you may include an else block. The exclamation mark tells the compiler to unwrap the value in the optional and to use it. Otherwise, the optional value is unwrapped, and This means your force unwrapping can be stored near to the place where its behavior is clarified, and the vast majority of your code doesn’t need to force unwrap directly. With Optional type you can't forget to deal with nil. Rink Rats is now viewed in Europe, Canada, South America, and the United States. rev 2021.1.18.38333, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Cezary's answer is spot on. SPONSORED Would you describe yourself as knowledgeable, but struggling when you have to come up with your own code? Optionals provide us with safety in Swift by stopping us from accessing data that doesn’t exist. is a different story. The randomElement() method returns an optional because that’s how it works, but we know it will always succeed because our enum has cases. Well, in Swift we luckily have a feature called Optionals. You need to unwrap optionals to safely use them On a basic level, Swift code consists of variables and functions. Hello, Swift 4! Writing unit tests for optionals. The "Optional()" part indicates that this variable may or may not have a value at runtime, but just so happens to currently be containing the string "foobar". Implicitly unwrapped optionals are optionals that are automatically unwrapped if needed for an expression to compile. In Swift world, it is always Christmas, and there are always presents — or at least variables — to unwrap. The compiler forces you to deal with it. Unexpectedly found nil while unwrapping an Optional value (Cannot figure out which part). Optionals provide us with safety in Swift by stopping us from accessing data that doesn’t exist. It is particularly relevant when it comes to class properties. expression is nil. The code included in my question is quoted from Apple's tutorial (link in the question), and I guess it's supposed to work as-is (and it does). For a discussing on avoiding force unwrapping, check out this blog post from Alexandre Colucci: Glossary             Think of optionals as being like birthday presents that come in stiff, hard, colored wrapping. In Swift 3, that was exactly how they worked: declarations like var a: Int? symbol. (String Optional, or Optional String). That might save you a line or two of code, but “that should never crash” isn’t a good enough reason for a force unwrap, and we can all do better. If we were to decide we never wanted to use force unwraps, what value could we use instead – what default value could we give randomDirection that allowed our code to continue safely? Related Q&A : What is an optional value in Swift? However, in a handful of circumstances you might want to consider force unwrapping the optionals rather than using if let or similar. I'd love to see someone write up an example of these showing how they help make code shorter and safer than the equivalent code that doesn't use them. You can unwrap an optional by appending ! Sponsor Hacking with Swift and reach the world's largest Swift community! Xcode 6.1 GM seed 2; iOS Simulator (iPhone6 / iOS 8.1) Optional 型とは? Optional 型 - nil の代入を許す 非 optional 型 - nil の代入を許さない Optional 型の宣言方法 How do I unwrap an Optional when pattern matching tuples in Swift? The entire Swift language was designed with safety in mind. Join Stack Overflow to learn, share knowledge, and build your career. The "Optional()" part indicates that this variable may or may not have a value at runtime, but just so happens to currently be containing the string "foobar". like methods, properties, and subscripting. This way you can perform operations on that variable as it has been guaranteed by the compiler to have a solid value. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. And from there you may safely use the non-optional in the if block. instead of an Int. If you try to set a nil value to a non-optional, the compiler will say, “hey you can’t set a nil value!”. I haven’t included it here because it doesn’t really matter, but it’s still some code that isn’t visible right here so we can’t guarantee that items will have any items by the time we call randomElement() on it. Following is the simple example of using Optional with variables to assign nil value in a swift … Thanks for your support, Kristaps Grinbergs! in your code example, Thanks for confusing me.... What kind of programmer gives a Birthday Present that's empty anyways? after the type name rather than a ?. I’m going to go into more detail in a moment, but I want to give you some examples of code that I’m happy to force unwrap and code that should not be force unwrapped. to the end of a type. As we said earlier that while you are unwrapping the optional you are actually casting to a non-optional type, you can also cast the non-optional to a different type. For example, if you have a string … But Swift provides a better alternative called optional binding. This is dangerous for several reasons. after the name of the type in which you are declaring as optional you are essentially casting this not as the type in which is before the ?, but instead as the optional type. That’s it – that’s the entire rule. Maximum useful resolution for scanning 35mm film. The question mark indicates the fact that canBeNil can be nil. everything after the ? ", love the analogy! You can use a conditional check on an optional to execute some code only if the optional type holds a value. Isn't the result supposed to be identical in both case? It prints out the value of myString that was set as "foobar". All optionals in Swift are demarcated with the ? Before you can use the contents of an optional value in any context you need to unwrap it. declares your Int as a completely different data type all together. Forced unwrapping uses the force unwrapping operator and we write it as an exclamation mark (!) @ToddLehman Apple created this to help you write much safer code. Swift, the Swift logo, Swift Playgrounds, Xcode, Instruments, Cocoa Touch, Touch ID, AirDrop, iBeacon, iPhone, iPad, Safari, App Store, watchOS, tvOS, Mac and macOS are trademarks of Apple Inc., registered in the U.S. and other countries. A common way of unwrapping optionals is with if let syntax, which unwraps with a condition. By setting the ? The optional Int is accessed through optional binding to unwrap the integer and assign the non-optional value to the roomCount constant. We welcome your comments and questions. So, what is optional? I can see their point of view, and I’ve certainly thought that myself previous. is nil, everything after the ? July 18, 2018 By Jay Mayu Leave a Comment. For instance, in Swift only valid number data types of the same kind may be operated on each other. This "Optional()" indication will remain unless you do what is called "unwrapping" the optional value. From section one, we have know what is swift optional variable. I hate unwrapping optionals with if let ”, but there’s a problem: if you try to use a value that contains nil your code will crash. The following examples use this dictionary of image names and file paths: Maybe we are using generics or maybe there is some other logic that is going on that will cause this to change. Stack Overflow for Teams is a private, secure spot for you and To declare an optional that’s implicitly unwrapped, place a ! Notice the difference between the two as keywords. so, is there a better way to unwrap the boxes then? Help identifying pieces in ambiguous wall anchor kit, CEO is pressing me regarding decisions made by my former manager whom he fired, One class classifier vs binary classifier. Optional Binding. Hopefully you can see the difference between the two: one situation is guaranteed to work 100% of the time, and the other “should” work or “will nearly always work”. The unwrapped version just shows the straight object because it is, well, unwrapped. In fact, as? Swift : Unwrapping Like a Boss. Unwrapping an optional means that you are now casting that type as non-optional. If there is something inside, it could be yet another present, which is also wrapped, and which also might contain nothing. Additing processing script to processing toolbox by PyQGIS3. Compare that to using an unknown array: What does that that isLuckyNumber() function do? You can declare optional variables using exclamation mark instead of a question mark. is ignored and the value of the whole An Optional can either be nil or some value.In the Swift language this is represented by an enum with two cases: None or Some.Optional looks like this: would result in a having type Optional, and declarations like var b: String! Note. This will let you set s to nil: s = nil. Forcibly Unwrapping is done by employing what is known as the ! In the first and second cases, the object is not being automatically unwrapped, so you see two "layers" ({{...}}), whereas in the third case, you see only one layer ({...}) because the object is being automatically unwrapped. Of course the error message is not so user-friendly but it’s like “Could not find an overload for ‘__co… operator. It is one of the hallmarks of Swift and one that you should welcome with open arms. if you use Enhance Ability: Cat's Grace on a creature that rolls initiative, does that creature lose the better roll when the spell ends? This uncovered some places where we needed to perform proper optional handling, so we refactored accordingly. E.g. Also participate in our monthly poll. Hacking with Swift is ©2021 Hudson Heavy Industries. This "Optional()" indication will remain unless you do what is called "unwrapping" the optional value. So, let me help those fellow "right-brained" (visual thinking) developers out by giving a different perspective in addition to the correct answer. Nonetheless casting with as! 'means is force-value. (optional). You can also declare optionals to automatically unwrap by using an exclamation mark instead of a question mark. Second, this code creates a closed range from 1 through 10, then picks a random number from that range using the randomElement() method: randomElement() returns an optional value because it’s possible the range or array we’re using it with was empty. From Cezary's answer, there's a slight difference between the output of the original code and the final solution (tested on playground) : The superclass' properties are shown in the output in the second case, while there's an empty object in the first case. Force Unwrapping. } This is represented syntactically by appending a ? character at the end of the variable/constant as in the next line print (someValue!). How To Unwrap Optional Variable Value In Swift. When is it useful? But if you declare a value as implicitly unwrapped, it is equivalent to force unwrapping it at every use. What was the first microprocessor to overlap loads with ALU ops? For example, if you declared the variable as being type, [Int:Any? Heck, even if you’re 99.999% certain that it’s safe, you shouldn’t be using it – that equates to one crash in 100,000, which is really bad. Remove Unwrapped and make Optional in Swift? What does children mean in “Familiarity breeds contempt - and children.“? Why are good absorbers also good emitters? For example, if you make an enum conform to the CaseIterable protocol, Swift will generate a special allCases array property for your enum that contains all the cases in the order they were defined. I'm learning Swift for iOS 8 / OSX 10.10 by following this tutorial, and the term "unwrapped value" is used several times, as in this paragraph (under Objects and Class): When working with optional values, you can write ? For example, if you have an assignment, var s: String = “”, you would change String to String?. @AdamSmith — Coming from a Java background, I can definitely see the use of the optionals... but coming also (more recently) from an Objective-C background, I'm still having trouble seeing the use of it, because Objective-C explicitly allows you to send messages to nil objects. The fact that it is queried through an optional chain means that the call to numberOfRooms will always return an Int? Here is a good analogy that helped me a lot. Put simply, an optional value is one that may or may not exist, which means Swift won't let you use it by accident – you need to either check whether it has a value and unwrap it, or force unwrap. < When to use guard let rather than if let, Why does Swift need both implicitly unwrapped optionals and regular optionals? >,, Click here to visit the Hacking with Swift store >>. This will generate a new type and assign the value that resided within that optional … If there was a value inside the optional then you can use it, but if there wasn’t the condition fails. So, we could create a Direction enum and pick a random value like this: We’ve defined an enum with four cases, and we’re asking to pick a random case. A mental model many people have for implicitly unwrapped optionals is that they are a type, distinct from regular optionals. otherwise we will get variable value with “Optional” keyword. If it is not nil, there will be a newly created constant variable that will be assigned the value and unwrapped into the non-optional constant. I’ve mentioned optionals in the previous post but didn’t go into the details. And if you ever tried to do your homework writing something like. value of the whole expression is an optional value. They are two different types that can not be operated on each other. Unlike Objective-C, whenever an attempt is made to forcibly unwrap an optional, and the optional has not been set and is nil, when you try to unwrap the optional to see what's inside your application will crash. This is less safe but still allows your code to compile. Or weird behaviour on Objective-C because you forgot to check for nil. In these cases, Swift lets you force unwrap the optional: convert it from an optional type to a non-optional type. Here you will do the same thing but instead of casting as just a non-optional you are casting it also as Int. the'! Privacy Policy             Note: You can give your conditionally unwrapped constant the same name as the optional variable you are unwrapping. var x:Int? But instead of printing 23, it shows… Swift Optionals - Different ways of unwrapping, “unexpectedly found nil while unwrapping an Optional value” in Swift 1.2, swift if-let doesn't unwrap optional NSDictionary, swift reflection causes impossible nil value for any. Instead, a better idea is to create a handful of functions and extensions that isolate your force unwraps in one place. The only way a force unwrap would be unsafe here would be if we deleted the four cases from Direction, but even then Swift shows a warning that we’re going to get unexpected behavior. For Implicitly unwrapped Optionals there are 4 main reasons to do that. your coworkers to find and share information. Force Unwrapping is the programmer taking responsibility and telling the compiler that they know for sure that some value will exist and in no way when they force unwrap that variable will they get nil. Why would a land animal need to move continuously to stay alive? We will get into which way is better for what context. I'm not saying they're not useful; I'm just saying I don't yet "get it. If the value before the ? If-let. What is the simplest proof that the density of primes goes to zero? So, I think force unwrapping is sometimes a good idea, and sometimes even required. Refund Policy             An optional represents two possibilities: Either there is a value, and you can unwrap the optional to access that value, or there isn’t a value at all. Even after you unwrap the variable, you are still like Brad Pitt in the last scene in SE7EN (warning: spoilers and very R-rated foul language and violence), because even after you have unwrapped the present, you are in the following situation: you now have nil, or a box containing something (but you don't know what). It will assist in the development of clean, readable code and help keep your application from crashing. You don't know if there's anything inside the wrapping until you unwrap the present — maybe there is nothing at all inside! Swift provides multiple ways to unwrap optionals safely. First, you have to understand what an Optional type is. The first way to unwrap an optional in Swift is something called force unwrapping. Compare that to using string interpolation: That isn’t safe, and can’t be verified, and if given a malformed web address could crash. Why would I want to use an optional value?