Name Convention for Unwrapped Value in Swift

Name convention for unwrapped value in Swift

You can assign the same name to an unwrapped variable as the optional.

Preferred:

var subview: UIView?
var volume: Double?

// later on...
if let subview = subview, let volume = volume {
  // do something with unwrapped subview and volume
}

Not preferred:

var optionalSubview: UIView?
var volume: Double?

if let unwrappedSubview = optionalSubview {
  if let realVolume = volume {
    // do something with unwrappedSubview and realVolume
  }
}

Taken from The Official raywenderlich.com Swift Style Guide.
However these are just guidelines and other conventions may be just fine.

Naming convention for optional binding

Just use the same name:

if let bananasInBarrel = bananasInBarrel {
  print("We have \(bananasInBarrel) bananas in the barrel.")
}

Don't use hungarian notation - the compiler will complain if you are using an unwrapped optional.

Why create Implicitly Unwrapped Optionals , since that implies you know there's a value?

Consider the case of an object that may have nil properties while it's being constructed and configured, but is immutable and non-nil afterwards (NSImage is often treated this way, though in its case it's still useful to mutate sometimes). Implicitly unwrapped optionals would clean up its code a good deal, with relatively low loss of safety (as long as the one guarantee held, it would be safe).

(Edit) To be clear though: regular optionals are nearly always preferable.

Swift core data: why do I need to unwrap when setting optional attribute values?

It seems you're equivocating on the word "optional".
The word "optional" in the attribute description optional="YES" is not the same as, and has nothing to do with, the Swift Optional enum type.

The former is merely the ordinary English word "optional", meaning "not required" — for a particular entity, there might or might not be a value for this attribute, which is a Float, but even if there isn't, it's a valid entity. Nothing in the story says that this attribute's type is Optional<Float>. Its type is Float. And you can't assign a Swift Optional where its wrapped type is expected; you have to unwrap it.

Indeed, this point is made very explicitly by the documentation:

Optional attributes aren’t required to have a value when saved to the persistent store. Attributes are optional by default.
Core Data optionals aren’t the same as Swift optionals. [My italics.]

How to unwrap double optionals?

Given a double optional such as this doubly wrapped String:

let a: String?? = "hello"
print(a as Any) // "Optional(Optional("hello"))\n"

@Leo, showed that you could use optional binding twice:

if let temp = a, let value = temp {
    print(value) // "hello\n"
}

or force unwrap twice:

print(value!!)  // don't do this - you're just asking for a crash

Here are 5 more methods you can use to safely unwrap a double optional:

Method 1:

You can also use pattern matching:

if case let value?? = a {
    print(value) // "hello\n"
}

As @netigger noted in their answer, this can also be written as:

if case .some(.some(let value)) = a {
    print(value) // "hello\n"
}

which while less concise might be a bit easier to read.

Method 2:

Alternatively, you can use the nil coalescing operator ?? twice:

print((a ?? "") ?? "")  // "hello\n"

Note: Unlike the other methods presented here, this will always produce a value. "" (empty String) is used if either of the optionals is nil.

Method 3:

Or you can use the nil coalescing operator ?? with optional binding:

if let value = a ?? nil {
    print(value)  // "hello\n"
}

How does this work?

With a doubly wrapped optional, the value held by the variable could be one of 3 things: Optional(Optional("some string")), Optional(nil) if the inner optional is nil, or nil if the outer optional is nil. So a ?? nil unwraps the outer optional. If the outer optional is nil, then ?? replaces it with the default value of nil. If a is Optional(nil), then ?? will unwrap the outer optional leaving nil. At this point you will have a String? that is nil if either the inner or outer optional is nil. If there is a String inside, you get Optional("some string").

Finally, the optional binding (if let) unwraps Optional("some string") to get "some string" or the optional binding fails if either of the optionals is nil and skips the block.

Method 4:

Also, you can use flatMap with optional binding:

if let value = a.flatMap({ $0 }) {
    print(value)  // "hello\n"
}

Method 5:

Conditionally cast the value to the type. Surprisingly, this will remove all levels of optionals:

let a: String?? = "hello"
let b: String??????? = "bye"

if let value = a as? String {
    print(value)  // "hello\n"
}

print(b as Any)  // "Optional(Optional(Optional(Optional(Optional(Optional(Optional("bye")))))))\n"

if let value = b as? String {
    print(value)  // "bye\n"
}

Why don't you need to unwrap an optional type value when being set?

An optional is a box. The box can either contain nothing (which is called nil) or it can contain something of a specific type (a String in your example). You unwrap an optional to access the value inside in the box.

When you are assigning a value to an optional, you just need to assign the value to the box itself. There is no need to unwrap anything, because you are just putting a value into the box. Swift either empties the box, if you assign nil, or it wraps the value by putting it into the box.

Unwrapping is for accessing a value that is already in the box.

Answering your question from a comment on another answer...

But why Optional binding don't need unwrapping? i think i if let constantName = some Optional is kind of assignment too

Optional binding is an unwrapping operation and an assignment operation. It says "if there is a value inside the box, assign it to this new variable and enter the then clause, otherwise proceed to the else clause if it exists".

var optionalValue: String? = "hello"

if let value = optionalValue {
    // value is the contents of the box, it has type String
    print("value is \(value)")
} else {
    // the optional binding failed because the box is empty
    print("optionalValue is nil")
}

What is the swift underscore's official name?

It's called a wildcard pattern:

Wildcard Pattern
A wildcard pattern matches and ignores any value and consists of an underscore (_). Use a wildcard pattern when you don’t care about the values being matched against. For example, the following code iterates through the closed range 1...3, ignoring the current value of the range on each iteration of the loop:
for _ in 1...3 {
   // Do something three times.
}

https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Patterns.html#//apple_ref/doc/uid/TP40014097-CH36-ID420

What is an optional value in Swift?

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

Name Convention for Unwrapped Value in Swift