When I set firstThing to default nil this will work, without the default value of nil I get a error that there is a missing parameter when calling the function.
By typing Int? I thought it made it optional with a default value of nil, am I right? And if so, why doesn't it work without the = nil?
func test(firstThing: Int? = nil) {
if firstThing != nil {
print(firstThing!)
}
print("done")
}
test()
Optionals and default parameters are two different things.
An Optional is a variable that can be nil, that's it.
Default parameters use a default value when you omit that parameter, this default value is specified like this: func test(param: Int = 0)
If you specify a parameter that is an optional, you have to provide it, even if the value you want to pass is nil. If your function looks like this func test(param: Int?), you can't call it like this test(). Even though the parameter is optional, it doesn't have a default value.
You can also combine the two and have a parameter that takes an optional where nil is the default value, like this: func test(param: Int? = nil).
The default argument allows you to call the function without passing an argument. If you don't pass the argument, then the default argument is supplied. So using your code, this...
test()
...is exactly the same as this:
test(nil)
If you leave out the default argument like this...
func test(firstThing: Int?) {
if firstThing != nil {
print(firstThing!)
}
print("done")
}
...then you can no longer do this...
test()
If you do, you will get the "missing argument" error that you described. You must pass an argument every time, even if that argument is just nil:
test(nil) // this works
Swift is not like languages like JavaScript, where you can call a function without passing the parameters and it will still be called. So to call a function in Swift, you need to assign a value to its parameters.
Default values for parameters allow you to assign a value without specifying it when calling the function. That's why test() works when you specify a default value on test's declaration.
If you don't include that default value, you need to provide the value on the call: test(nil).
Also, and not directly related to this question, but probably worth to note, you are using the "C++" way of dealing with possibly null pointers, for dealing with possible nil optionals in Swift. The following code is safer (specially in multithreading software), and it allows you to avoid the forced unwrapping of the optional:
func test(firstThing: Int? = nil) {
if let firstThing = firstThing {
print(firstThing)
}
print("done")
}
test()
You are conflating Optional with having a default. An Optional accepts either a value or nil. Having a default permits the argument to be omitted in calling the function. An argument can have a default value with or without being of Optional type.
func someFunc(param1: String?,
param2: String = "default value",
param3: String? = "also has default value") {
print("param1 = \(param1)")
print("param2 = \(param2)")
print("param3 = \(param3)")
}
Example calls with output:
someFunc(param1: nil, param2: "specific value", param3: "also specific value")
param1 = nil
param2 = specific value
param3 = Optional("also specific value")
someFunc(param1: "has a value")
param1 = Optional("has a value")
param2 = default value
param3 = Optional("also has default value")
someFunc(param1: nil, param3: nil)
param1 = nil
param2 = default value
param3 = nil
To summarize:
Type with ? (e.g. String?) is an Optional may be nil or may contain an instance of
Type
Argument with default value may be omitted from a call to
function and the default value will be used
If both Optional and has default, then it may be omitted from function call OR may be included and can be provided with a nil value (e.g. param1: nil)
in case you need to use a bool param, you need just to assign the default value.
func test(WithFlag flag: Bool = false){.....}
then you can use without or with the param:
test() //here flag automatically has the default value: false
test(WithFlag: true) //here flag has the value: true
"Optional parameter" means "type of this parameter is optional". It does not mean "This parameter is optional and, therefore, can be ignored when you call the function".
The term "optional parameter" appears to be confusing. To clarify, it's more accurate to say "optional type parameter" instead of "optional parameter" as the word "optional" here is only meant to describe the type of parameter value and nothing else.
If you want to be able to call the func with or without the parameter you can create a second func of the same name which calls the other.
func test(firstThing: Int?) {
if firstThing != nil {
print(firstThing!)
}
print("done")
}
func test() {
test(firstThing: nil)
}
now you can call a function named test without or without the parameter.
// both work
test()
test(firstThing: 5)
Don't let the question mark fools you!
Optional is an enum which has two cases:
#frozen public enum Optional<Wrapped> : ExpressibleByNilLiteral {
/// The absence of a value.
///
/// In code, the absence of a value is typically written using the `nil`
/// literal rather than the explicit `.none` enumeration case.
case none
/// The presence of a value, stored as `Wrapped`.
case some(Wrapped)
}
code from the original compiled source of the Optional inside Xcode
When you are defining a function that accept some Type of arguments, you can pass a default value withe the same type.
in your case
the type of the firstThing is Optional<Int> (also known as Int?). So if you want the caller to the oportunity to ignore the paramter, you MUST do the job for the caller and pass a default value.
Usually we need the .none case of the optional so we can do:
func test(firstThing: Optional<Int> = .none) { ... }
This is exactly the same as:
func test(firstThing: Int? = nil) { ... }
Also!
Who seys that we the default value of an optional is a nil? maybe passing nil means that the function should remove something by updating it's value to 'nil'. So don't asume "the default value for optional is a nil"
It is little tricky when you try to combine optional parameter and default value for that parameter. Like this,
func test(param: Int? = nil)
These two are completely opposite ideas. When you have an optional type parameter but you also provide default value to it, it is no more an optional type now since it has a default value. Even if the default is nil, swift simply removes the optional binding without checking what the default value is.
So it is always better not to use nil as default value.
Default value doesn't mean default value of data type .Here default value mean value defined at the time of defining function.
we have to declare default value of variable while defining variable in function.
I have an if-else statement where I am checking if the value coming from user defaults is nil or not like this:
if defaults.bool(forKey: "abcd") != nil{
//Do something
}
else{
//do something else
}
But Xcode is giving me an error saying:
"Comparing non-optional value of type 'Bool' to 'nil' always returns true"
Can someone explain what's happening here and how to fix this?
bool(forKey:) returns a NON-optional, which cannot be nil. If the key is missing in the user defaults, the return value will be false.
If you want trinary logic here (nil/true/false) use object(forKey:) looking for an NSNumber, and if present, take its boolValue.
As
defaults.bool(forKey: "abcd")
will return false by default check Docs , so it will never be optional
The Boolean value associated with the specified key. If the specified key doesn‘t exist, this method returns false.
The func bool(forKey: "abcd") returns Bool type not optional.
Which means you cant compare it to bool, what you can do is simply:
if defaults.bool(forKey: "abcd") {
//Do something
} else {
//do something else
}
Now if the key exists and has true value it will get into the if statement, if it does not exists or is false it will go to the else.
If you have any doubts you can read about the func in the following Apple developer link: Apple:bool(forKey:)
Objective-c property in swift. If you're using some objective c property in swift and it says something like "Comparing non-optional value of type 'XYZ' to 'nil' always returns true" you have to make that objective c property to "_Nullable" so that property may not be optional anymore. Like #property (strong,nonatomic) NSString *_Nullable someString;
defaults.bool(forKey: "abcd") != nil
The first part, defaults.bool(forKey: "abcd"), returns a non-optional boolean. We know that because bool(forKey:) returns Bool, not Bool?. Therefore, you'll always get a Bool value, i.e. either true or false, never nil. Note the documentation:
If the specified key doesn‘t exist, this method returns false.
"Comparing non-optional value of type 'Bool' to 'nil' always returns true"
The compiler is simply pointing out that it knows that defaults.bool(forKey: "abcd") can't be nil, and since you're comparing it to nil, you're probably making a mistake. Your else block will never execute.
Can someone explain what's happening here and how to fix this?
It depends on what you mean for the code to do. If you want to take different actions depending on whether the value is true or false, then compare it to one of those values. If you want to get an optional value back, use object(forKey:) (which returns an optional) instead.
i solved like this:
if(Userdefaults.standart.bool(forkey: "blablabool"){
}
This works..
When you call this if its null it returns false.
This question already has answers here:
Swift optionals - warning on conditional cast from 'x' to 'x' always succeeds
(2 answers)
If let clause when getting from NSUserDefaults()
(2 answers)
Closed 5 years ago.
if let seen: Bool = defaults.bool(forKey: UtilitiesKeys.mainTutorialSeen) {
return seen
}
return false
if i do this swift shows me an issue:
Conditional cast from 'Bool' to 'Bool' always succeeds, Non-optional
expression of type 'Bool' used in a check for optionals.
Since there might not be a value for my key, I don't want to force unwrap it. Now obviously it's working like this, but how do I safely unwrap the value without having swift complaining?
I have tried to use guard as well...
if let seen = UserDefaults.standard.object(forKey: UtilitiesKeys.mainTutorialSeen)) as? Bool {
return seen
}else {
return false
}
Apple document says:
/*!
-boolForKey: is equivalent to -objectForKey:, except that it converts the returned value to a BOOL. If the value is an NSNumber, NO
will be returned if the value is 0, YES otherwise. If the value is an
NSString, values of "YES" or "1" will return YES, and values of "NO",
"0", or any other string will return NO. If the value is absent or
can't be converted to a BOOL, NO will be returned.
open func bool(forKey defaultName: String) -> Bool
Initializer for conditional binding must have Optional type, not 'Bool'
hope this help you.
As suggested in doc:
-boolForKey: is equivalent to -objectForKey:, except that it converts the returned value to a BOOL. If the value is an NSNumber, NO will be returned if the value is 0, YES otherwise. If the value is an NSString, values of "YES" or "1" will return YES, and values of "NO", "0", or any other string will return NO. If the value is absent or can't be converted to a BOOL, NO will be returned.
open func bool(forKey defaultName: String) -> Bool
It is not an optional anymore. So you don't need to cast it with if let.
You can directly use:
let seen = defaults.bool(forKey: UtilitiesKeys.mainTutorialSeen)
Why does the following always print "prefix" which is incorrect but print "no prefix" when String is not optional or implicitly unwrapped optional?
var value:String! = "aaa" // Same incorrect behavior on Optional String as well.
if value?.hasPrefix("bbb") {
NSLog("prefix")
}
else {
NSLog("no prefix")
}
The if statement is checking if the statement returns a value or nil, not if it returns true or false. You can use another if statement to check the value of hasPrefix().
var value:String! = "aaa" // Same incorrect behavior on Optional String as well.
if let hasPrefix = value?.hasPrefix("bbb") {
if hasPrefix{
NSLog("prefix")
}
else {
NSLog("no prefix")
}
}
else {
NSLog("nil value")
}
This is related (not sure if duplicate) of: How to check if Optional is not nil and property is true in one expression?
Summarizing it a bit for your question:
When you use Optional Chaining, the returned value is always an Optional. That means value?.hasPrefix("bbb") returns Bool? and not Bool. Thus, if value is not nil, value?.hasPrefix("bbb") will always not be nil, and so it will go into the first case, no matter if its true or false.
This is explained in the Swift book section on Optional Chaining. It makes sense because when you use Optional Chaining, the return value always has a chance to return nil, and that doesn't depend on the final value in the "chain".
You if statement just gives you an optional result because you are calling the "hasPrefix()" on an optional value.
This wouldn't even work in the newer versions of swift and would give the error :
Value of optional type 'Bool?' not unwrapped; did you mean to use '!'
or '?'?
Refer this for documentation on the same.
Following would give you the required result:
if let val = value, val.hasPrefix("bbb") {
NSLog("prefix")
} else {
NSLog("no prefix")
}
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