I'm trying to convert height in feet with a decimal to height in feet and inches.
I'm trying to use the .round(.down) method to get the feet, and multiply the decimal by 12 for the inches. I'm getting all kinds of errors, like so:
var heightInFeet: Float = 5.45
let feetRounded = heightInFeet.round(.down) // feetRounded is "type ()." What is that?
percentToNextFoot = heightInFeet - feetRounded // Error: Binary operator '-' cannot be applied to operands of type 'Float' and '()'
I tried the following and got another error:
percentToNextFoot = heightInFeet - Float(feetRounded) // Cannot invoke initializer for type 'Float' with an argument list of type '(())'
I finally got it working by avoiding the .round() method, but I'm still really confused by the value type (). Can anyone explain what's going on here?:
var heightInFeet: Float = 5.45
var feet = Int(heightInFeet) // 5
var percentToNextFoot = heightInFeet - Float(feet) // 0.45
let heightInFeetAndInches = "\(feet)ft \(Int(percentToNextFoot * 12))in" // 5ft 5in
() is shorthand for void. It means "no value is possible here".
In this example, it means that the .round() method does not return anything - it is a mutating function called on its receiver. So assigning its void return to a var causes that var's type to be inferred to be void. Void vars can be useful, sometimes, rarely*, but not in this case.
Methods on value types often come in pairs: a verb like round, and a passive verb e.g. rounded. The first operates directly on, and modifies, its target; the second returns a modified version of its target. For another example, see sort() and sorted() on collections, or append(_) and appending(_) on strings, etc.
(* note: filter is an annoying exception; it means "filtered", and there is no handy "filter in place".)
To get the effect you were going for in the first example, rounded() is what you want.
--
(* To answer the tangential question in your title: how would one actually use a void variable? Well, here's a way I use them sometimes:
In an object with some setup that I would like to happen sometime after init, but guaranteed at most once per instance, I used to use Objective-C's dispatch_once. That's not available in Swift, so now I'll make a lazy void member like so:
class Foo {
lazy var setup: () = {
// do some complicated stuff I couldn't do in `init` for some reason
// this will only run once
}()
func doSomethingUseful() {
_ = setup // runs the setup initializer just the first time we get here
// do useful things that depend on setup having happened
}
}
I'll leave it to the comments to judge whether we're "meant to" use such a technique. :)
Welcome to stack overflow!
Double has two rounding methods:
Double.round(), which rounds a Double value by mutating it in-place. This is one you called. It doesn't return anything, which strictly speaking means it returns Void, a.k.a. (), the empty tuple.
Double.rounded(), which rounds a Double value by returning a new Double. This is the one you probably intended to call.
By calling the first, and trying to assign the value to a variable, you end up with a variable of type Void, whose value is ().
This is a common Swift convention: "object.foo" edits it in place. "object.fooed" returns a mutated copy.
That said, in your case, I would recommend doing this using the existing Measurement API:
import Foundation
extension Measurement where UnitType == UnitLength {
func toFeetAndInches() -> (feet: Measurement<UnitLength>, inches: Measurement<UnitLength>) {
let numberOfWholeFeet = self.converted(to: .feet).value.rounded(.towardZero)
return (
feet: Measurement(value: numberOfWholeFeet, unit: UnitLength.feet),
inches: Measurement(value: self.value - numberOfWholeFeet, unit: UnitLength.feet).converted(to: .inches)
)
}
}
let heightInFeet = Measurement(value: 5.5, unit: UnitLength.feet)
let (feet, inches) = heightInFeet.toFeetAndInches()
let mf = MeasurementFormatter()
mf.unitOptions = .providedUnit // force the use of feet/inches, rather than the unit appropriate for the user's locale.
mf.unitStyle = .medium
print(mf.string(for: feet)!, mf.string(for: inches)!) // => "5 ft. 6 in."
If you look at the reference for the round function of Float type, you will see that it returns nothing. It just mutate the float you called this method on.
You can do
var feetRounded = heightInFeet
feetRounded.round(.down)
Please take a look at the documentation. The method
mutating func round(_ rule: FloatingPointRoundingRule)
has no return value (aka Void aka ())
If you need a result you have to use rounded(_:) which has a return value
func rounded(_ rule: FloatingPointRoundingRule) -> Float
round changes the value in place; you use it like this:
var heightInFeet: Float = 5.45
heightInFeet.round(.down)
You notice that no value is returned; there is no = in the second line. We do not need to set anything to the result of the round call, because it has no result.
If, as in your code, you accidentally do capture the "result", it is expressed as type (). So () is the "result" type of a method call that has no result.
When we accidentally write this:
var heightInFeet: Float = 5.45
let x = heightInFeet.round(.down)
we get this error: "Constant 'x' inferred to have type '()', which may be unexpected." That is just a fancy way of saying, "You've taken a method call that has no result and captured its 'result'. You probably didn't mean to do that!" And indeed, you didn't.
Let's say I have the following class:
class User: NSObject {
var name = "Fred"
var age = 24
var email = "fred#freddy.com"
var married = false
}
I want to be able to write a generic function that takes in a list of KeyPaths for a known class type, read the values and print to screen. The problem is, the I can't get the following code to compile as the type of the KeyPath's Value is not known, and will be different for each time. What do I have to do to make this work generically?
Consider the following:
struct KeyPathProperties<T> {
var name: String
var relatedKeyPaths: [KeyPath<T, Any>]
}
extension KeyPath where Root == User {
var properties: KeyPathProperties<Root> {
switch self {
case \Root.name:
return KeyPathProperties(name: "name", relatedKeyPaths: [\Root.age, \Root.email])
default:
fatalError("Unknown key path")
}
}
}
This line fails to compile:
return KeyPathProperties(name: "name", relatedKeyPaths: [\Root.age, \Root.email])
with this error:
Cannot convert value of type 'KeyPath<User, Int>' to expected element type 'KeyPath<User, Any>'
This is what I wish to be able to do, for instance:
let myUser = User()
var keyPathProps = KeyPathProperties(name: "name", relatedKeyPaths: [\User.age, \User.email])
for keyPath in props.relatedKeyPaths {
print("Value: \(myUser[keyPath: keyPath])")
}
The above won't compile of course. Essentially I want to store keyPaths in an array at runtime, so I can generically at some point in time get values out of the User. I need to know if I can re-write the above in some way where the compiler can safely and correctly determine the type of the keyPath's value at runtime.
This is a conceptual use case for a much more complex architectural issue I'm trying to solve with hopefully less code.
MORE INFORMATION:
At runtime I wish to keep track of the properties that get modified - these properties are held in a modifiedProps array in each object / instance. At some point at runtime, I wish to be able to enumerate over this array of KeyPaths and print their values like so:
for modifiedKeyPath in self.modifiedProps {
print ("\(self[keyPath: modifiedKeyPath])"
}
In short - I need to be able to capture the generic type of the KeyPath within KeyPathProperties. How do I achieve this?
SIDE NOTE: I can already easily achieve this by using Swift 3 style string based KeyPaths (by adding #objc to the class properties). I can store an array of keyPaths as strings and later do:
let someKeyPath = #keyPath(User.email)
...
myUser.value(forKeyPath: someKeyPath)
I just cannot do this with Swift 4 KeyPaths generically.
The error tells you what your misconception is:
Cannot convert value of type 'KeyPath<User, Int>'
to expected element type 'KeyPath<User, Any>'
You seem to think that you can use a KeyPath<User, Int> where a KeyPath<User, Any> is expected, ostensibly on the grounds that an Int is an Any. But that's not true. These are generic types, and generic types are not covariant — that is, there is no substitution principle for generics based on their parameterized types. The two types are effectively unrelated.
If you need an array of key paths regardless of their parameterized types, you would need an array of PartialKeyPath or AnyKeyPath. It seems that in your use case the root object is the same throughout, so presumably you want PartialKeyPath.
I need to convert an Integer value to String. I made a variable with an Integer value and then I have printed that using print. What is the different between the below approaches?
var word = "Count is "
var count = 100
print(word+String(describing: count)); // Count is 100
print(word+String(count)); // Count is 100
Your question is actually unnecessary because if all you want to do here is print, you can just do it directly:
print("Count is", count) // Count is 100
That's because print takes a variadic parameter and inserts space as the default separator.
However, let's answer the question anyway.
Answer
It's the difference between coercion and representation.
Coercion. Certain types can be changed into certain other types. You can change a Double to an Int and an Int to a Double. You can change an Int to a String and a String (maybe) to an Int. This is possible because the second type has an initializer whose parameter is the first type. This is something you would do in your actual program, like this:
let sum : Int = x + y
self.myLabel.text = "Your total is \(String(sum))"
Representation. For debugging purposes, all types can benefit from being representable as a string. Suppose you have a Person type. You can't change a Person to a String and vice versa — that makes no sense — but you would surely like to be able to print a Person to the console to see if you are getting the right results. This is possible because the Person type itself supplies a printable description. This is something you would do for debugging purposes:
let p = Person(first:"Matt", last:"Neuburg")
print("p is \(String(describing:p))")
Comments
Comment 1. This distinction is fairly new in Swift. It used to be that String(...) was used to express both notions. But the powers that be realized that that was a confusing conflation of the two distinct mechanisms. So nowadays, an attempt to write String(p) will fail, whereas earlier it would have succeeded. String has no Person initializer, so String(p) is forbidden; you now have to say explicitly that you are describing, not coercing.
Comment 2. The need to print a description is so obviously common that you do not have to pass thru String(describing:) if all you want to do is log the object itself. You can write print(p), because this is a shorthand for print(String(describing:p)). Similarly, string interpolation calls String(describing:) for you, so you can write print("p is \(p)"). And, as I said at the outset, print takes a variadic parameter and inserts space as the default separator, so you can write print("p is", p).
Reading the docs might help!
Here's an excerpt from String.init(describing:)
Use this initializer to convert an instance of any type to its
preferred representation as a String instance. The initializer creates
the string representation of instance in one of the following ways,
depending on its protocol conformance:
If instance conforms to the TextOutputStreamable protocol, the result is obtained by calling instance.write(to: s) on an empty string
s.
If instance conforms to the CustomStringConvertible protocol, the result is instance.description.
If instance conforms to the CustomDebugStringConvertible protocol, the result is instance.debugDescription.
An unspecified result is supplied automatically by the Swift standard library.
Int conforms to CustomStringConvertible, so String(describing: someInt) is the same as someInt.description.
Now let's look at String(someInt), this calls this initializer:
public init<T>(_ value: T) where T : LosslessStringConvertible
Int also conforms to LosslessStringConvertible. In fact, LosslessStringConvertible inherits from CustomStringConvertible. LosslessStringConvertible specifies only one requirement - an initializer that takes a string.
The docs for init<T>(_:) says this:
Creates an instance from the description of a given LosslessStringConvertible instance.
Even though description is not in a monospace font, I am quite sure (though not 100%) that this refers to the description property of CustomStringConvertible.
Assuming that init<T>(_:) indeed returns the description, String(describing: someInt) always returns the same value as String(someInt).
Note that this does not hold true for all types. It only holds true for types that conforms to LosslessStringConvertible but not TextOutputStreamable. If you have a type that conforms to both, results can be different:
struct A: TextOutputStreamable, LosslessStringConvertible {
init?(_ description: String) { }
init() { }
var description: String {
return "foo"
}
func write<Target>(to target: inout Target) where Target : TextOutputStream {
target.write("bar")
}
}
String(describing: A()) // bar
String(A()) // foo
I know about the ampersand as a bit operation but sometimes I see it in front of variable names. What does putting an & in front of variables do?
It works as an inout to make the variable an in-out parameter. In-out means in fact passing value by reference, not by value. And it requires not only to accept value by reference, by also to pass it by reference, so pass it with & - foo(&myVar) instead of just foo(myVar)
As you see you can use that in error handing in Swift where you have to create an error reference and pass it to the function using & the function will populate the error value if an error occur or pass the variable back as it was before
Why do we use it? Sometimes a function already returns other values and just returning another one (like an error) would be confusing, so we pass it as an inout. Other times we want the values to be populated by the function so we don't have to iterate over lots of return values, since the function already did it for us - among other possible uses.
It means that it is an in-out variable. You can do something directly with that variable. It is passed by address, not as a copy.
For example:
var temp = 10
func add(inout a: Int){
a++
}
add(inout:&temp)
temp // 11
There's another function of the ampersand in the Swift language that hasn't been mentioned yet. Take the following example:
protocol Foo {}
protocol Bar {}
func myMethod(myVar: Foo & Bar) {
// Do something
}
Here the ampersand syntax is stating that myVar conforms to both the Foo and Bar protocol.
As another use case, consider the following:
func myMethod() -> UIViewController & UITableViewDataSource {
// Do something
}
Here we're saying that the method returns a class instance (of UIViewController) that conforms to a certain protocol (UITableViewDataSource). This is rendered somewhat obsolete with Swift 5.1's Opaque Types but you may see this syntax in pre-Swift 5.1 code from time to time.
If you put & before a variable in a function, that means this variable is inout variable.
#Icaro already described what it means, I will just give an example to illustrate the difference between inout variables and in variables:
func majec(inout xValue:Int, var yValue:Int) {
xValue = 100
yValue = 200
}
var xValue = 33
var yValue = 33
majec(&xValue, yValue: yValue)
xValue //100
yValue //33
As noted in other answers, you use prefix & to pass a value to an inout parameter of a method or function call, as documented under Functions > Function Argument Labels and Parameter Names > In-Out Parameters in The Swift Programming Language. But there's more to it than that.
You can, in practice, think about Swift inout parameters and passing values to them as being similar to C or C++ pass-by-address or pass-by-reference. In fact, the compiler will optimize many uses of inout parameters down to roughly the same mechanics (especially when you're calling imported C or ObjC APIs that deal in pointers). However, those are just optimizations — at a semantic level, inout really doesn't pass addresses around, which frees the compiler to make this language construct more flexible and powerful.
For example, here's a struct that uses a common strategy for validating access to one of its properties:
struct Point {
private var _x: Int
var x: Int {
get {
print("get x: \(_x)")
return _x
}
set {
print("set x: \(newValue)")
_x = newValue
}
}
// ... same for y ...
init(x: Int, y: Int) { self._x = x; self._y = y }
}
(In "real" code, the getter and setter for x could do things like enforcing minimum/maximum values. Or x could do other computed-property tricks, like talking to a SQL database under the hood. Here we just instrument the call and get/set the underlying private property.)
Now, what happens when we pass x to an inout parameter?
func plusOne(num: inout Int) {
num += 1
}
var pt = Point(x: 0, y: 1)
plusOne(num: &pt.x)
// prints:
// get x: 0
// set x: 1
So, even though x is a computed property, passing it "by reference" using an inout parameter works the same as you'd expect it to if x were a stored property or a local variable.
This means that you can pass all sorts of things "by reference" that you couldn't even consider in C/C++/ObjC. For example, consider the standard library swap function, that takes any two... "things" and switches their values:
var a = 1, b = 2
swap(&a, &b)
print(a, b) // -> 2 1
var dict = [ "Malcolm": "Captain", "Kaylee": "Mechanic" ]
swap(&dict["Malcolm"], &dict["Kaylee"])
print(dict) // -> ["Kaylee": "Captain", "Malcolm": "Mechanic"], fanfic ahoy
let window1 = NSWindow()
let window2 = NSWindow()
window1.title = "window 1"
window2.title = "window 2"
var windows = [window1, window2]
swap(&windows[0], &windows[1])
print(windows.map { $0.title }) // -> ["window 2", "window 1"]
The the way inout works also lets you do fun stuff like using the += operator on nested call chains:
window.frame.origin.x += 10
... which is a whole lot simpler than decomposing a CGRect just to construct a new one with a different x coordinate.
This more nuanced version of the inout behavior, called "call by value result", and the ways it can optimize down to C-style "pass by address" behavior, is covered under Declarations > Functions > In-Out Parameters in The Swift Programming Language.
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