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In swift, structs have an automatically generated memberwise initializer.
This means the following struct can be initialised without me having to write an init.
struct Activity {
let name: String
let desc: String
let category: Category
let subcategory: Subcategory
let emoji: Character
let coordinate: CLLocationCoordinate2D
let creationTime: Date = Date()
let activityTime: Date
let id: UUID = UUID()
var comments: [Comment] = []
}
I have one single property called emojiwhich is computed by the subcategory. In other words, the value for emoji depends on the value of subcategory.
However this means that the value of emoji can only be assigned after the initialisation of subcategory.
How should I do this in code?
Approach 1:
Provide my own initialiser
init(name: String, desc: String, category: Category, subcategory: Subcategory,
coordinate: CLLocationCoordinate2D, activityTime: Date) {
self.name = name
self.desc = desc
self.category = category
self.subcategory = subcategory
self.coordinate = coordinate
self.activityTime = activityTime
self.emoji = AllCategories.categories[category]?[subcategory] ?? "❌"
}
I don't like this approach as it adds a lot of unecessary code that will only grow if I add more properties... I would like to use the generated initialiser of the struct. On the other hand, the code is still very simple.
Approach 2:
Use a lazy varthat is only computed when called.
lazy var emoji: Character = {
AllCategories.categories[category]?[subcategory] ?? "❌"
}()
I also don't really like this approach, I find it overly complex for what I am trying to do. Also this makes emojia varinstead of letwhich it is not, I want it to remain a constant. On the other hand, I can continue using the automatically generated initialiser.
Questions:
What other possibilities do I have?
If there are none, which of the 2 approches is the best?
This sounds like a great chance to use computed properties:
var emoji: Character {
AllCategories.categories[category]?[subcategory] ?? "❌"
}
Although it is declared a var, you can't actually set it. It's just how computed properties must be declared.
The expression AllCategories.categories[category]?[subcategory] ?? "❌" will be evaluated every time you use the property. It's not a too time-consuming expression, so IMO it's fine.
I know optional chain is like:
if let name = student.info?.name {
// if student has info & info contains name, it comes here.
}
Questions:
1. But what is the difference between doing above and doing the following:
let name = student.info?.name
Is it so that the let name would be nil if student doesn't have info or if info doesn't contains name?
2. What would happen to app if info doesn't contain name with following code:
var student: Student? {
didSet {
// 'nameLabel' is an outlet for UILabel
nameLabel.text = student.info?.name
}
}
will app crash in above case if student doesn't have info or if info doesn't have name when setting the student variable?
Optional chaining means that you can safely check the values of nested optionals. If any of the optionals in the chain is nil, the values are not checked further, the return value of the whole chain becomes nil. You need to mark all optional values with the ? character in an optional chain, not just the last one, in your question you only marked the last optional.
The code in your question wouldn't actually compile, since when accessing properties of an optional, you need to either force unwrap it or use optional chaining.
student!.info?.name compiles, but results in a runtime crash if student == nil. student?.info?.name compiles and doesn't lead to a crash, but returns nil if either student == nil or student?.info ==
nil. student.info?.name doesn't compile, since you don't unwrap the optional student variable at all.
This is optional binding and you should use it in most cases when you need to work with optional values.
if let name = student?.info?.name {
//name is non-optional here, so it can safely be used inside the if statement
}
If you use optional chaining without optional binding, your variable type will be optional as well:
let name = student?.info?.name //name is optional here
For your last question, the fact that didSet is called doesn't mean that student is not nil. In fact, the property observer will be called every time a value is assigned to the property and since student is optional, nil is a valid value, hence student can be nil inside the property observer. Moreover, even the fact that student has a non-nil value doesn't guarantee that student.info has a non-nil value as well. However, since UILabel.text has type String?, assigning a nil value to it doesn't lead to a crash unless later on you force unwrap nameLabel.text.
var student: Student? {
didSet {
// 'nameLabel' is an outlet for UILabel
nameLabel.text = student?.info?.name
}
}
For more information, have a look at The Swift Programming Language Guide - Optionals part.
If info property is nil, name will become nil. Otherwise name will be assigned name property of info.
Conclusion, "name" depends on the info property, not the info.name property.
The app won't crash, beacause UILabel.text is of optional type. text will be nil if info is nil, otherwise it will be info.name.
If you are trying to access a variable in an object that is not defined as you do, your app will crash.
If info wasn't optional but name was, then it would be put nil value in the nameLabel.text, but your example will certainly crash.
You seem to know how to deal with this, but I'll add a solution for that :
var student: Student? {
didSet {
if let studentInfo = student.info {
nameLabel.text = studentInfo.name
} else {
nameLabel.text = "unknown"
}
}
}
if let name = student.info?.name {
But what is the difference between doing above and doing the following:
let name = student.info?.name
A huge difference.
The second one perpetuates optionality: name is still an Optional, and you still have the problem of unwrapping it safely, i.e. without crashing in case it happens to be nil.
The first one removes optionality: if we dive into the curly braces, name is not an Optional, because the original Optional has been wrapped safely.
A ? mark, is like opening a pizza box to see if there is a pizza inside of it.
Pizza box, is optional variable. (A variable with ?, ex. var student: Student?)
and the pizza is the student object it self.
Let's say you want to access the name of the var student. It's like you want a slice of pizza from the pizza. Because the pizza is in a box, you have to open the box first, same with optional variable, to access its member, you have to open it first.
But wait, as you open the pizza box, there is no pizza inside of it, and of course you can't have the slice pizza you wanted to eat. The same can be applied to the var student, because it's optional, after you open it up, it can be empty (nil).
That's why if you have var student: Student?, and you want to access name from student, you have to open it up first with student? followed by the member you want to access student?.name. Of course, if the student is nil the return value is nil.
Question 1
What's the different between if let and just let, well, if let is a conditional to ask if the object in the box exist? If there is an object, then do this.
if let name = student?.name {
print(name)
}
from the above code, if student is not nil and just say the the var name contain joe then joe will be printed. But, if the student is nil then the print(name) is never get executed.
let name = student?.name
print(name)
The above code, remove the if statement, so this is not asking if the object is exist, but regardless the object exist or not, it want something. So if student is nil, of course the name will return nil, and we get nil printed in the debug view.
Question 2
You should first see if the nameLabel.text is optional. We can see from your code that student is optional, but we don't know if the .text is optional. Assign an optional into an optional is never a problem. If the variable is nil then the assigned var also become nil.
But if you assign an optional into a non optional variable, swift compiler will tell you that you can't do that, because if the optional variable is nil then the non optional variable will crash because it can't be nil
Looking at this example of conditionals I am confused.
Here is the code and my interpretation
var animal = Animal(name: "Lenny", species: "lemur", tailLength: 12)
animal = Animal(name: "Gilbert", species: "Gorilla", tailLength: nil )
if let tailLength = animal.tail?.length {
print("\(animal.name)'s tail is \(tailLength) long")
} else {
print("\(animal.name) doesn't have a tail.")
}
We have a variable "Animal". Not all animals have tails and so some tailLength values will return nil.
To me, an optional is something with options - in this case it can be an Int or nil (In the case of Gilbert above).
Now, when we want to unwrap it, we are using chaining to check if it will return nil. If it returns an Int, we return it. If it returns nil, we move to the else statement.
Why is the syntax -
if let tailLength = animal.tail?.length
rather than
if let tailLength = animal.taillength?
or
if let tailLength = animal.tail.length?
Note: New to programming. I'm aware that perhaps the answer requires some prerequisite knowledge of other languages and common syntax.
Let's clear this up.
First some prerequisites…
Optionals
Optionals do not necessarily imply options. Instead, "optional" simply implies that the value may or may not exist, hence it is "optional". Think of Optionals as containers sized just right so that they may only hold one specific type. For example, an Optional created specifically to hold an Int could never hold a String, and vice versa. By this definition, Optionals sound exactly the same as variables and constants, but the big difference to note is that unlike regular variables and constants, the containers for Optionals may contain "no value at all", AKA nil or they may contain a value of their specified type. There are no other options.
Optionals are used anytime it makes sense for there to be no value at all, since nil is a better representation of "nothing" than say, -1 for an Integer.
For example,
class Contact {
var name: String
var phoneNumber: String?
var emailAddress: String?
init(name: String, number: String?, emailAddress: String?) {
self.name = name
self.phoneNumber = number
self.emailAddress = emailAddress
}
}
This is a vastly simplified version of what you might use to represent an entry in a contacts app. For simplicity, we'll use Strings to represent each of these properties. The question mark after the type name means that we do not want these properties to be of type String, but instead String?, or "Optional String". The app requires the user to enter their name, but the user may or may not choose to provide a phone number or email address, so the corresponding properties are declared as optional Strings. If the user provides an email address, for example, then the emailAddress property will hold a String representing the address. If the user had not provided an address, the emailAddress property would not hold a String but instead nil, or the absence of a value.
Behind the scenes, an Optional is just an enumeration with two cases: None, and Some with a wrapped value of the type the optional was declared. If you're not familiar with Swift enumerations or this is confusing to you, please ignore this paragraph. It is not important in order to understand Optionals.
Optionals always have only two options. Either nil or a value of the type they were declared.
Unwrapping Optionals
Now suppose we want to access an Optional at some point. We can't just reference it because there is a possibility that it could be nil and our calculation wouldn't function properly. For example, if we wanted to access the user's email address, we might try something like
let user = Contact(name: "Matt", number: nil, emailAddress: nil)
sendEmail(user.emailAddress)
Assuming that the sendEmail function requires a String representing the address as an argument, Swift will report an error and not let the code compile as is. The problem is that we cannot guarantee that there will actually be a String to pass sendEmail. sendEmail requires a String but we are trying to pass it a String?. These are not the same type. If we know that the value will indeed exist, we can just append an exclamation mark immediately after the value that is optional. So, continuing with our example,
let user = Contact(name: "Matt", number: nil, emailAddress: nil)
sendEmail(user.emailAddress!)
Here, the exclamation mark tells Swift that we are sure there will be a value in the emailAddress property, so Swift goes ahead and converts the String? to a String with the value. If at runtime it turns out that emailAddress evaluates to nil, we will get a fatal error. Of course, in this case, we will get a fatal error at runtime because emailAddress did indeed contain nil.
Much of the time, however, and especially in our Contact example, we will not be able to guarantee that a emailAddress exists. Now introducing Optional Binding…
Optional Binding
For times when we simply do not know whether a value exists, but would like to use that value for something if it does, we can use Optional Binding.
Suppose we want to send the user an email if they provided their address. Well, we would need to ensure the value is not nil, assign it to a constant if it isn't, and do some workaround if it is. Here's a perfectly valid way of doing so:
if user.emailAddress != nil {
let address = user.emailAddress!
sendEmail(address)
} else {
promptForEmailAddress()
}
Here, we are checking to make sure the address is not nil. If it is, we will ask the user to give it to us. If the address does exist (if it's not nil), we will assign it to a constant by force unwrapping it with an exclamation mark. We can do this because if the if statement resolves to true, we can be absolutely certain that user.emailAddress will not be nil. sendEmail is then satisfied because address is of type String and not String?, since we unwrapped it first.
But because this is such a common pattern in Swift, the language defines a simplified way of doing this (Optional Binding). The following code snippet is exactly equivalent to the snippet above.
if let address = user.emailAddress {
sendEmail(address)
} else {
promptForEmailAddress()
}
With optional binding, you combine the assignment operation with the if statement. If whatever is on the right of the equals operator (assignment operator) evaluates to nil, the else clause is called and the address constant is never created. If, however, the expression on the right of the equals operator does indeed contain a value, in this case a String, then the address constant is created and initialized with whatever String value was stored in emailAddress. The type of address is therefore String and not String? because if the constant is created, we know it is not nil.
Optional Chaining
Now, sometimes you need to access properties that are at the end of a long chain of objects. For example, suppose emailAddress was not a String but instead a special Email object that held several properties of its own, including a domainName declared as a 'String'.
visitWebAddress(user.emailAddress.domainName)
If one of these values (user, emailAddress, or domainName) was declared as an Optional, this would not be a safe call and Swift would yell at us because our imaginary visitWebAddress function expects a String, but got a String?. Actually, this is not even valid code because we should have a ? after emailAddress. What's going on here? Well, remember that emailAddress was declared as a String?, or "optional string". Anytime we want to access a property using dot syntax on an already optional value, we must use either an exclamation mark (which forces the Optional unwrapped and returns the value which must exist) or a question mark. Whether we choose ! or ?, the symbol must immediately follow the property that was declared to be Optional.
visitWebAddress(user.emailAddress?.domainName)
This statement is syntactically correct. The question mark tells Swift that we realize emailAddress could be nil. If it is, the value of the entire statement user.emailAddress?.domainName will be nil. Regardless of whether emailAddress actually is nil or not, the resulting type of user.emailAddress?.domainName will be String?, not String. Therefore, Swift will still yell at us because visitWebAddress expected a String, but got a String?.
Therefore, we could use a combination of Optional Binding and Optional Chaining to create a simple and elegant solution to this problem:
if let domain = user.emailAddress?.domainName {
visitWebAddress(domain)
}
Remember, the expression on the right side of the equal operator is evaluated first. As already discussed, user.emailAddress?.domainName will evaluate to a String? that will be nil if the email address was not specified. If this is the case and the expression evaluates to nil, no constant is ever created and we are done with the if statement. Otherwise, the value is bound (hence "Optional Binding") to a created constant domain (which will have type String) and we can use it inside the if clause.
As a side note, Optional Chaining can sometimes result in no action at all taking place when a value is nil. For example,
textField?.text = "Hello"
Here, if textField evaluates to nil (for example if the UI has not loaded yet), the entire line is skipped and no action is taken.
Now for your exact question,
Code Analysis
Based on your posted code sample, this is what is likely going on. The Animal class probably looks something like this at a minimum:
class Animal {
let name: String
let species: String
let tail: Tail?
init(name: String, species: String, tailLength: Int?) {
self.name = name
self.species = species
if let lengthOfTail = tailLength {
self.tail = Tail(length: lengthOfTail)
}
}
}
Now, you're probably wondering what is up with all this mention of a Tail object, so this is probably also existent somewhere behind the scenes, looking like this at a minimum:
class Tail {
let length: Int
init(length: Int) {
self.length = length
}
}
(Now going back and looking at Animal should make more sense.)
Here, we have a Tail object with one property called length, of type Int (not optional). The Animal class, however, contains a property called 'tail' which is of type Tail?, or "optional Tail". This means that an Animal may or may not have a tail, but if it does it is guaranteed to have a length of type Int.
So here…
if let tailLength = animal.tail?.length {
print("\(animal.name)'s tail is \(tailLength) long")
} else {
print("\(animal.name) doesn't have a tail.")
}
…a ? immediately follows tail because it is the property that is optional. The statement animal.tail?.length has a type of Int? and if it is nil because tail is nil, the else clause is called. Otherwise, a constant tailLength is created and the tail length as an Int is bound to it.
Note that in the Animal initializer, it simply took an Int? for tailLength, but behind the scenes it checked to see if it was nil and only if it wasn't did it create a Tail object with the specified length. Otherwise, the tail property on the Animal was left nil, since nil is the default value of a non-initialized optional variable or constant.
Hope that clears up all of your questions on Optionals in Swift. The Swift Language Guide has some great content on this, so check it out if needed.
In Swift:
1) If you provide a default value for all of the stored properties in a class, then you inherit the default initializer, ie - init().
-- AND --
2) A property of any optional type defaults to the value of nil, ie - var shouldBeNill: String? //should initially be nill
-- THEREFORE --
I would expect this code to work:
class Product {
let name: String?
}
let product = Product()
But when I type it in as a playground, I get the error: "class Product has no initializers".
Why isn't Product inheriting the default initializer init()? I know I can make this work by explicitly setting let name: String? = nil, but I'm unsure why I have to do this. Is this an error on Swift's side, or is there something I am not quite grasping?
You are on the right track. The issue here is actually the let vs var.
let declares the property constant. In this case Product would have an optional constant name of type String with no initial value, and this of course makes no sense.
The compiler complains about a lacking init() function because let properties are allowed to be set once during init(), as part of object construction, if not defined already in declaration eg.
let name: String = "Im set!" // OK
let name: String? = nil // OK, but very weird :)
let name = "Im set!" // OK, type not needed, implicit.
let name: String // OK, but needs to be set to a string during init()
let name: String? // OK, but needs to be set to string or nil during init()
let name // Not OK
The Swift Programming Language - Constants and Variables
From Apple's documentation:
You can use if and let together to work with values that might be missing. These values are represented as optionals. An optional value either contains a value or contains nil to indicate that the value is missing. Write a question mark (?) after the type of a value to mark the value as optional.
Why would you want to use an optional value?
An optional in Swift is a type that can hold either a value or no value. Optionals are written by appending a ? to any type:
var name: String? = "Bertie"
Optionals (along with Generics) are one of the most difficult Swift concepts to understand. Because of how they are written and used, it's easy to get a wrong idea of what they are. Compare the optional above to creating a normal String:
var name: String = "Bertie" // No "?" after String
From the syntax it looks like an optional String is very similar to an ordinary String. It's not. An optional String is not a String with some "optional" setting turned on. It's not a special variety of String. A String and an optional String are completely different types.
Here's the most important thing to know: An optional is a kind of container. An optional String is a container which might contain a String. An optional Int is a container which might contain an Int. Think of an optional as a kind of parcel. Before you open it (or "unwrap" in the language of optionals) you won't know if it contains something or nothing.
You can see how optionals are implemented in the Swift Standard Library by typing "Optional" into any Swift file and ⌘-clicking on it. Here's the important part of the definition:
enum Optional<Wrapped> {
case none
case some(Wrapped)
}
Optional is just an enum which can be one of two cases: .none or .some. If it's .some, there's an associated value which, in the example above, would be the String "Hello". An optional uses Generics to give a type to the associated value. The type of an optional String isn't String, it's Optional, or more precisely Optional<String>.
Everything Swift does with optionals is magic to make reading and writing code more fluent. Unfortunately this obscures the way it actually works. I'll go through some of the tricks later.
Note: I'll be talking about optional variables a lot, but it's fine to create optional constants too. I mark all variables with their type to make it easier to understand type types being created, but you don't have to in your own code.
How to create optionals
To create an optional, append a ? after the type you wish to wrap. Any type can be optional, even your own custom types. You can't have a space between the type and the ?.
var name: String? = "Bob" // Create an optional String that contains "Bob"
var peter: Person? = Person() // An optional "Person" (custom type)
// A class with a String and an optional String property
class Car {
var modelName: String // must exist
var internalName: String? // may or may not exist
}
Using optionals
You can compare an optional to nil to see if it has a value:
var name: String? = "Bob"
name = nil // Set name to nil, the absence of a value
if name != nil {
print("There is a name")
}
if name == nil { // Could also use an "else"
print("Name has no value")
}
This is a little confusing. It implies that an optional is either one thing or another. It's either nil or it's "Bob". This is not true, the optional doesn't transform into something else. Comparing it to nil is a trick to make easier-to-read code. If an optional equals nil, this just means that the enum is currently set to .none.
Only optionals can be nil
If you try to set a non-optional variable to nil, you'll get an error.
var red: String = "Red"
red = nil // error: nil cannot be assigned to type 'String'
Another way of looking at optionals is as a complement to normal Swift variables. They are a counterpart to a variable which is guaranteed to have a value. Swift is a careful language that hates ambiguity. Most variables are define as non-optionals, but sometimes this isn't possible. For example, imagine a view controller which loads an image either from a cache or from the network. It may or may not have that image at the time the view controller is created. There's no way to guarantee the value for the image variable. In this case you would have to make it optional. It starts as nil and when the image is retrieved, the optional gets a value.
Using an optional reveals the programmers intent. Compared to Objective-C, where any object could be nil, Swift needs you to be clear about when a value can be missing and when it's guaranteed to exist.
To use an optional, you "unwrap" it
An optional String cannot be used in place of an actual String. To use the wrapped value inside an optional, you have to unwrap it. The simplest way to unwrap an optional is to add a ! after the optional name. This is called "force unwrapping". It returns the value inside the optional (as the original type) but if the optional is nil, it causes a runtime crash. Before unwrapping you should be sure there's a value.
var name: String? = "Bob"
let unwrappedName: String = name!
print("Unwrapped name: \(unwrappedName)")
name = nil
let nilName: String = name! // Runtime crash. Unexpected nil.
Checking and using an optional
Because you should always check for nil before unwrapping and using an optional, this is a common pattern:
var mealPreference: String? = "Vegetarian"
if mealPreference != nil {
let unwrappedMealPreference: String = mealPreference!
print("Meal: \(unwrappedMealPreference)") // or do something useful
}
In this pattern you check that a value is present, then when you are sure it is, you force unwrap it into a temporary constant to use. Because this is such a common thing to do, Swift offers a shortcut using "if let". This is called "optional binding".
var mealPreference: String? = "Vegetarian"
if let unwrappedMealPreference: String = mealPreference {
print("Meal: \(unwrappedMealPreference)")
}
This creates a temporary constant (or variable if you replace let with var) whose scope is only within the if's braces. Because having to use a name like "unwrappedMealPreference" or "realMealPreference" is a burden, Swift allows you to reuse the original variable name, creating a temporary one within the bracket scope
var mealPreference: String? = "Vegetarian"
if let mealPreference: String = mealPreference {
print("Meal: \(mealPreference)") // separate from the other mealPreference
}
Here's some code to demonstrate that a different variable is used:
var mealPreference: String? = "Vegetarian"
if var mealPreference: String = mealPreference {
print("Meal: \(mealPreference)") // mealPreference is a String, not a String?
mealPreference = "Beef" // No effect on original
}
// This is the original mealPreference
print("Meal: \(mealPreference)") // Prints "Meal: Optional("Vegetarian")"
Optional binding works by checking to see if the optional equals nil. If it doesn't, it unwraps the optional into the provided constant and executes the block. In Xcode 8.3 and later (Swift 3.1), trying to print an optional like this will cause a useless warning. Use the optional's debugDescription to silence it:
print("\(mealPreference.debugDescription)")
What are optionals for?
Optionals have two use cases:
Things that can fail (I was expecting something but I got nothing)
Things that are nothing now but might be something later (and vice-versa)
Some concrete examples:
A property which can be there or not there, like middleName or spouse in a Person class
A method which can return a value or nothing, like searching for a match in an array
A method which can return either a result or get an error and return nothing, like trying to read a file's contents (which normally returns the file's data) but the file doesn't exist
Delegate properties, which don't always have to be set and are generally set after initialization
For weak properties in classes. The thing they point to can be set to nil at any time
A large resource that might have to be released to reclaim memory
When you need a way to know when a value has been set (data not yet loaded > the data) instead of using a separate dataLoaded Boolean
Optionals don't exist in Objective-C but there is an equivalent concept, returning nil. Methods that can return an object can return nil instead. This is taken to mean "the absence of a valid object" and is often used to say that something went wrong. It only works with Objective-C objects, not with primitives or basic C-types (enums, structs). Objective-C often had specialized types to represent the absence of these values (NSNotFound which is really NSIntegerMax, kCLLocationCoordinate2DInvalid to represent an invalid coordinate, -1 or some negative value are also used). The coder has to know about these special values so they must be documented and learned for each case. If a method can't take nil as a parameter, this has to be documented. In Objective-C, nil was a pointer just as all objects were defined as pointers, but nil pointed to a specific (zero) address. In Swift, nil is a literal which means the absence of a certain type.
Comparing to nil
You used to be able to use any optional as a Boolean:
let leatherTrim: CarExtras? = nil
if leatherTrim {
price = price + 1000
}
In more recent versions of Swift you have to use leatherTrim != nil. Why is this? The problem is that a Boolean can be wrapped in an optional. If you have Boolean like this:
var ambiguous: Boolean? = false
it has two kinds of "false", one where there is no value and one where it has a value but the value is false. Swift hates ambiguity so now you must always check an optional against nil.
You might wonder what the point of an optional Boolean is? As with other optionals the .none state could indicate that the value is as-yet unknown. There might be something on the other end of a network call which takes some time to poll. Optional Booleans are also called "Three-Value Booleans"
Swift tricks
Swift uses some tricks to allow optionals to work. Consider these three lines of ordinary looking optional code;
var religiousAffiliation: String? = "Rastafarian"
religiousAffiliation = nil
if religiousAffiliation != nil { ... }
None of these lines should compile.
The first line sets an optional String using a String literal, two different types. Even if this was a String the types are different
The second line sets an optional String to nil, two different types
The third line compares an optional string to nil, two different types
I'll go through some of the implementation details of optionals that allow these lines to work.
Creating an optional
Using ? to create an optional is syntactic sugar, enabled by the Swift compiler. If you want to do it the long way, you can create an optional like this:
var name: Optional<String> = Optional("Bob")
This calls Optional's first initializer, public init(_ some: Wrapped), which infers the optional's associated type from the type used within the parentheses.
The even longer way of creating and setting an optional:
var serialNumber:String? = Optional.none
serialNumber = Optional.some("1234")
print("\(serialNumber.debugDescription)")
Setting an optional to nil
You can create an optional with no initial value, or create one with the initial value of nil (both have the same outcome).
var name: String?
var name: String? = nil
Allowing optionals to equal nil is enabled by the protocol ExpressibleByNilLiteral (previously named NilLiteralConvertible). The optional is created with Optional's second initializer, public init(nilLiteral: ()). The docs say that you shouldn't use ExpressibleByNilLiteral for anything except optionals, since that would change the meaning of nil in your code, but it's possible to do it:
class Clint: ExpressibleByNilLiteral {
var name: String?
required init(nilLiteral: ()) {
name = "The Man with No Name"
}
}
let clint: Clint = nil // Would normally give an error
print("\(clint.name)")
The same protocol allows you to set an already-created optional to nil. Although it's not recommended, you can use the nil literal initializer directly:
var name: Optional<String> = Optional(nilLiteral: ())
Comparing an optional to nil
Optionals define two special "==" and "!=" operators, which you can see in the Optional definition. The first == allows you to check if any optional is equal to nil. Two different optionals which are set to .none will always be equal if the associated types are the same. When you compare to nil, behind the scenes Swift creates an optional of the same associated type, set to .none then uses that for the comparison.
// How Swift actually compares to nil
var tuxedoRequired: String? = nil
let temp: Optional<String> = Optional.none
if tuxedoRequired == temp { // equivalent to if tuxedoRequired == nil
print("tuxedoRequired is nil")
}
The second == operator allows you to compare two optionals. Both have to be the same type and that type needs to conform to Equatable (the protocol which allows comparing things with the regular "==" operator). Swift (presumably) unwraps the two values and compares them directly. It also handles the case where one or both of the optionals are .none. Note the distinction between comparing to the nil literal.
Furthermore, it allows you to compare any Equatable type to an optional wrapping that type:
let numberToFind: Int = 23
let numberFromString: Int? = Int("23") // Optional(23)
if numberToFind == numberFromString {
print("It's a match!") // Prints "It's a match!"
}
Behind the scenes, Swift wraps the non-optional as an optional before the comparison. It works with literals too (if 23 == numberFromString {)
I said there are two == operators, but there's actually a third which allow you to put nil on the left-hand side of the comparison
if nil == name { ... }
Naming Optionals
There is no Swift convention for naming optional types differently from non-optional types. People avoid adding something to the name to show that it's an optional (like "optionalMiddleName", or "possibleNumberAsString") and let the declaration show that it's an optional type. This gets difficult when you want to name something to hold the value from an optional. The name "middleName" implies that it's a String type, so when you extract the String value from it, you can often end up with names like "actualMiddleName" or "unwrappedMiddleName" or "realMiddleName". Use optional binding and reuse the variable name to get around this.
The official definition
From "The Basics" in the Swift Programming Language:
Swift also introduces optional types, which handle the absence of a value. Optionals say either “there is a value, and it equals x” or “there isn’t a value at all”. Optionals are similar to using nil with pointers in Objective-C, but they work for any type, not just classes. Optionals are safer and more expressive than nil pointers in Objective-C and are at the heart of many of Swift’s most powerful features.
Optionals are an example of the fact that Swift is a type safe language. Swift helps you to be clear about the types of values your code can work with. If part of your code expects a String, type safety prevents you from passing it an Int by mistake. This enables you to catch and fix errors as early as possible in the development process.
To finish, here's a poem from 1899 about optionals:
Yesterday upon the stair
I met a man who wasn’t there
He wasn’t there again today
I wish, I wish he’d go away
Antigonish
More resources:
The Swift Programming Guide
Optionals in Swift (Medium)
WWDC Session 402 "Introduction to Swift" (starts around 14:15)
More optional tips and tricks
Let's take the example of an NSError, if there isn't an error being returned you'd want to make it optional to return Nil. There's no point in assigning a value to it if there isn't an error..
var error: NSError? = nil
This also allows you to have a default value. So you can set a method a default value if the function isn't passed anything
func doesntEnterNumber(x: Int? = 5) -> Bool {
if (x == 5){
return true
} else {
return false
}
}
You can't have a variable that points to nil in Swift — there are no pointers, and no null pointers. But in an API, you often want to be able to indicate either a specific kind of value, or a lack of value — e.g. does my window have a delegate, and if so, who is it? Optionals are Swift's type-safe, memory-safe way to do this.
I made a short answer, that sums up most of the above, to clean the uncertainty that was in my head as a beginner:
Opposed to Objective-C, no variable can contain nil in Swift, so the Optional variable type was added (variables suffixed by "?"):
var aString = nil //error
The big difference is that the Optional variables don't directly store values (as a normal Obj-C variables would) they contain two states: "has a value" or "has nil":
var aString: String? = "Hello, World!"
aString = nil //correct, now it contains the state "has nil"
That being, you can check those variables in different situations:
if let myString = aString? {
println(myString)
}
else {
println("It's nil") // this will print in our case
}
By using the "!" suffix, you can also access the values wrapped in them, only if those exist. (i.e it is not nil):
let aString: String? = "Hello, World!"
// var anotherString: String = aString //error
var anotherString: String = aString!
println(anotherString) //it will print "Hello, World!"
That's why you need to use "?" and "!" and not use all of them by default. (this was my biggest bewilderment)
I also agree with the answer above: Optional type cannot be used as a boolean.
In objective C variables with no value were equal to 'nil'(it was also possible to use 'nil' values same as 0 and false), hence it was possible to use variables in conditional statements (Variables having values are same as 'TRUE' and those with no values were equal to 'FALSE').
Swift provides type safety by providing 'optional value'. i.e. It prevents errors formed from assigning variables of different types.
So in Swift, only booleans can be provided on conditional statements.
var hw = "Hello World"
Here, even-though 'hw' is a string, it can't be used in an if statement like in objective C.
//This is an error
if hw
{..}
For that it needs to be created as,
var nhw : String? = "Hello World"
//This is correct
if nhw
{..}
Optional value allows you to show absence of value. Little bit like NULL in SQL or NSNull in Objective-C. I guess this will be an improvement as you can use this even for "primitive" types.
// Reimplement the Swift standard library's optional type
enum OptionalValue<T> {
case None
case Some(T)
}
var possibleInteger: OptionalValue<Int> = .None
possibleInteger = .Some(100)”
Excerpt From: Apple Inc. “The Swift Programming Language.” iBooks. https://itun.es/gb/jEUH0.l
An optional means that Swift is not entirely sure if the value corresponds to the type: for example, Int? means that Swift is not entirely sure whether the number is an Int.
To remove it, there are three methods you could employ.
1) If you are absolutely sure of the type, you can use an exclamation mark to force unwrap it, like this:
// Here is an optional variable:
var age: Int?
// Here is how you would force unwrap it:
var unwrappedAge = age!
If you do force unwrap an optional and it is equal to nil, you may encounter this crash error:
This is not necessarily safe, so here's a method that might prevent crashing in case you are not certain of the type and value:
Methods 2 and three safeguard against this problem.
2) The Implicitly Unwrapped Optional
if let unwrappedAge = age {
// continue in here
}
Note that the unwrapped type is now Int, rather than Int?.
3) The guard statement
guard let unwrappedAge = age else {
// continue in here
}
From here, you can go ahead and use the unwrapped variable. Make sure only to force unwrap (with an !), if you are sure of the type of the variable.
Good luck with your project!
When i started to learn Swift it was very difficult to realize why optional.
Lets think in this way.
Let consider a class Person which has two property name and company.
class Person: NSObject {
var name : String //Person must have a value so its no marked as optional
var companyName : String? ///Company is optional as a person can be unemployed that is nil value is possible
init(name:String,company:String?) {
self.name = name
self.companyName = company
}
}
Now lets create few objects of Person
var tom:Person = Person.init(name: "Tom", company: "Apple")//posible
var bob:Person = Person.init(name: "Bob", company:nil) // also Possible because company is marked as optional so we can give Nil
But we can not pass Nil to name
var personWithNoName:Person = Person.init(name: nil, company: nil)
Now Lets talk about why we use optional?.
Lets consider a situation where we want to add Inc after company name like apple will be apple Inc. We need to append Inc after company name and print.
print(tom.companyName+" Inc") ///Error saying optional is not unwrapped.
print(tom.companyName!+" Inc") ///Error Gone..we have forcefully unwrap it which is wrong approach..Will look in Next line
print(bob.companyName!+" Inc") ///Crash!!!because bob has no company and nil can be unwrapped.
Now lets study why optional takes into place.
if let companyString:String = bob.companyName{///Compiler safely unwrap company if not nil.If nil,no unwrap.
print(companyString+" Inc") //Will never executed and no crash!!!
}
Lets replace bob with tom
if let companyString:String = tom.companyName{///Compiler safely unwrap company if not nil.If nil,no unwrap.
print(companyString+" Inc") //Will executed and no crash!!!
}
And Congratulation! we have properly deal with optional?
So the realization points are
We will mark a variable as optional if its possible to be nil
If we want to use this variable somewhere in code compiler will
remind you that we need to check if we have proper deal with that variable
if it contain nil.
Thank you...Happy Coding
Lets Experiment with below code Playground.I Hope will clear idea what is optional and reason of using it.
var sampleString: String? ///Optional, Possible to be nil
sampleString = nil ////perfactly valid as its optional
sampleString = "some value" //Will hold the value
if let value = sampleString{ /// the sampleString is placed into value with auto force upwraped.
print(value+value) ////Sample String merged into Two
}
sampleString = nil // value is nil and the
if let value = sampleString{
print(value + value) ///Will Not execute and safe for nil checking
}
// print(sampleString! + sampleString!) //this line Will crash as + operator can not add nil
From https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/OptionalChaining.html:
Optional chaining is a process for querying and calling properties, methods, and subscripts on an optional that might currently be nil. If the optional contains a value, the property, method, or subscript call succeeds; if the optional is nil, the property, method, or subscript call returns nil. Multiple queries can be chained together, and the entire chain fails gracefully if any link in the chain is nil.
To understand deeper, read the link above.
Well...
? (Optional) indicates your variable may contain a nil value while ! (unwrapper) indicates your variable must have a memory (or value) when it is used (tried to get a value from it) at runtime.
The main difference is that optional chaining fails gracefully when the optional is nil, whereas forced unwrapping triggers a runtime error when the optional is nil.
To reflect the fact that optional chaining can be called on a nil value, the result of an optional chaining call is always an optional value, even if the property, method, or subscript you are querying returns a nonoptional value. You can use this optional return value to check whether the optional chaining call was successful (the returned optional contains a value), or did not succeed due to a nil value in the chain (the returned optional value is nil).
Specifically, the result of an optional chaining call is of the same type as the expected return value, but wrapped in an optional. A property that normally returns an Int will return an Int? when accessed through optional chaining.
var defaultNil : Int? // declared variable with default nil value
println(defaultNil) >> nil
var canBeNil : Int? = 4
println(canBeNil) >> optional(4)
canBeNil = nil
println(canBeNil) >> nil
println(canBeNil!) >> // Here nil optional variable is being unwrapped using ! mark (symbol), that will show runtime error. Because a nil optional is being tried to get value using unwrapper
var canNotBeNil : Int! = 4
print(canNotBeNil) >> 4
var cantBeNil : Int = 4
cantBeNil = nil // can't do this as it's not optional and show a compile time error
Here is basic tutorial in detail, by Apple Developer Committee: Optional Chaining
An optional in Swift is a type that can hold either a value or no value. Optionals are written by appending a ? to any type:
var name: String?
You can refer to this link to get knowledge in deep: https://medium.com/#agoiabeladeyemi/optionals-in-swift-2b141f12f870
There are lots of errors which are caused by people trying to use a value which is not set, sometime this can cause a crash, in objective c trying to call the methods of a nil object reference would just be ignored, so some piece of your code not executing and the compiler or written code has no way of telling your why. An optional argument let you have variables that can never be nil, and if you try to do build it the compiler can tell you before your code has even had a chance to run, or you can decide that its appropriate for the object to be undefined, and then the compiler can tell you when you try to write something that doesn't take this into account.
In the case of calling a possible nil object you can just go
object?.doSomthing()
You have made it explicit to the compiler and any body who reads your code, that its possible object is nil and nothing will happen. Some times you have a few lines of code you only want to occur if the value exists, so you can do
if let obj = object {
obj.doSomthing()
doSomethingto(obj)
}
The two statements will only execute if object is something, simarly you may want to stop the rest of the entire block of code if its not something
guard let obj = object {
return
}
obj.doSomthing()
doSomethingto(obj)
This can be simpler to read if everything after is only applicable if object is something, another possiblity is you want to use a default value
let obj = object ?? <default-object>
obj.doSomthing()
doSomethingto(obj)
Now obj will be assigned to something even if its a default value for the type
options are useful in situation where a value may not gain a value until some event has occurred or you can use setting an option to nil as a way to say its no longer relevant or needs to be set again and everything that uses it has no point it doing anything with it until it is set, one way I like to use optionals is to tell me something has to be done or if has already been done for example
func eventFired() {
guard timer == nil else { return }
timer = scheduleTimerToCall(method, in: 60)
}
func method() {
doSomthing()
timer = nil
}
This sudo code can call eventFired many times, but it's only on the first call that a timer is scheduled, once the schedule executes, it runs some method and sets timer back to nil so another timer can be scheduled.
Once you get around your head around variables being in an undefined state you can use that for all sort of thing.
It's very simple. Optional (in Swift) means a variable/constant can be nullable. You can see that Kotlin language implements the same thing but never calls it an 'optional'. For example:
var lol: Laugh? = nil
is equivalent to this in Kotlin:
var lol: Laugh? = null
or this in Java:
#Nullable Laugh lol = null;
In the very first example, if you don't use the ?symbol in front of the object type, then you will have an error. Because the question mark means that the variable/constant can be null, therefore being called optional.
Here is an equivalent optional declaration in Swift:
var middleName: String?
This declaration creates a variable named middleName of type String. The question mark (?) after the String variable type indicates that the middleName variable can contain a value that can either be a String or nil. Anyone looking at this code immediately knows that middleName can be nil. It's self-documenting!
If you don't specify an initial value for an optional constant or variable (as shown above) the value is automatically set to nil for you. If you prefer, you can explicitly set the initial value to nil:
var middleName: String? = nil
for more detail for optional read below link
http://www.iphonelife.com/blog/31369/swift-101-working-swifts-new-optional-values