I have a NSObject Subclass. Say CityWalks
class CityWalks{
var totalCount:Int?
}
How do I use this property further? Should I check the nil coalescing every time this value is accessed.
example:
let aObject =
say in one fucntion (function1()) , I need to access this value, then it would like (aObject!.totalCount ?? 0)
func function1(){
...Some Access code for the object....
(aObject!.totalCount ?? 0)
}
Similarly in every other function(function2()) , I will have to write the same code.
func function2(){
...Some Access code for the object....
(aObject!.totalCount ?? 0)
}
So, what could be a better approach for such field, considering this property might receive a value from server or might not.
If you have a default value for this property just assign this value as default value.
class YourClass {
var totalCount = 0
}
I'd recommend you avoid using an optional value if it's possible. Because optional values its a first place when you can get an error.
As stated in the comments and the other answer using an optional is not really optimal in your case. It seems like you might as well use a default value of 0.
However, to clarify, you have to check the value when unwrapping the optional.
Sometimes it's possible to pass an optional to UIElement etc and then you don't really need to do anything with them
There are pretty ways of checking for nil in optional values built into swift so you can build pretty neat code even though you work with optional.
Look in to guard let and if let if you want to know more about unwrapping values safely.
if let
if let totalWalks = aObject?.totalCount {
//operate on totalWalks
}
guard
guard let totalWalks = aObject?.totalCount else { return }
//operate on totalWalks
There are also cases where you will want to call a function on an optional value and in this case you can do so with ?
aObject?.doSomething()
Any return values this function might have will now be wrapped in an optional and you might have to unwrap them as well with an if let or guard
When working with optionals you should try to avoid forcing the unwrap with ! as even though you at the moment know that the value is not null that might after a change in the code not be true anymore.
Related
I tried for a long time to turn the text into an Int but it did not work. I tried it like this:
(AnzahlString is a textfield)
var AnzahlAInt = 0
if let AnzahlAString = AnzahlString.text {
let AnzahlAInt = Int(AnzahlAString)
}
But then I always get the error:
Value of optional type 'Int?' must be unwrapped to a value of type 'Int'
Then I added a ! at the end of Int(AnzahlAString)! so I don't get a error, but now when I press on the button, the app crashes. It was predictable, but how can I change this now to an Int without the !?
At first glance, it looks like you have two things to check for:
is AnzahlString.text present, and
does it represent an Int
The first check is in fact not necessary, since .text will never return nil, even though it's marked as Optional. This means you can safely force-unwrap it.
The second check is easily done by using the ?? operator:
let AnzahlAInt = Int(AnzahlString.text!) ?? 0
PS, just as a stylistic hint: variable names in Swift ususally start with a lowercase letter, names starting with capital letters are used for types.
PPS: your code as written shadows AnzahlAInt - the value of your var is never changed.
The reason why the resulting Int is optional, is that parsing might or might not succeed. For example, if you try to parse the string "Fluffy Bunnies" into an Int, there is no reasonable Int that can be returned, therefore the result of parsing that string will be nil.
Furthermore, if you force the parser by using !, you're telling Swift that you know for sure that the string you pass will always result in a valid Int, and when it doesn't, the app crashes.
You need to handle the situation in which the parse result is nil. For example:
if let AnzahlAIntResult = Int(AnzahlAString) {
// We only get here if the parse was successful and we have an Int.
// AnzahlAIntResult is now an Int, so it can be assigned to AnzahlAInt.
AnzahlAInt = AnzahlAIntResult
}
You did a good job so far but missed out one thing.
This line tries to convert the String into an Int. However this can fail, since your String can be something like this "dfhuse".
let AnzahlAInt = Int(AnzahlAString)
This is why the result of Int(AnzahlAString) is an Optional (Int?). To use it as an real Int, you have to unwrap it.
First solution is the !, however, every time this does fail your app crashes. Not a good Idea to use so.
The best solution would be Optional Binding, as you already used to get the text of your text field.
if let AnzahlAString = AnzahlString.text {
if let safeInt = Int(AnzahlAString) {
// You can use safeInt as a real Int
} else {
print("Converting your String to an Int failed badly!")
}
}
Hope this helps you. Feel free to ask again if something is unclear.
For unwrapping you can also use guard like this
Simple and easy
guard let AnzahlAInt = Int(AnzahlString.text!) else {
return
}
print(AnzahlAInt)
I have codes like below
var currentDocumentPart: DocumentPart
var documentPartArray: [DocumentPart]
let currentDocumentPartIndex = documentPartArray.indexOf(currentDocumentPart)
let previousDocumentPart = documentPartArray[currentDocumentPartIndex! - 1]
I want to implement if let statement to check whether previousDocumentPart is empty, but it doesn't let me since previousDocumentPart's type is not optional. How should I change? or is there other way to check if it is nil besides if let?
And due to my implementation in code it seems like it is hard to change the type of currentDocumentPart and documentPartArray to optional.
The type of previousDocumentPart is inferred from the value assigned to it which in this case is DocumentPart (not an optional, hence no nil value allowed).
If you want to be able to check for nil values change the definition of documentPartArray to :
var documentPartArray: [DocumentPart?]
This question already has answers here:
Why would I use if and let together, instead of just checking if the original variable is nil? (Swift)
(2 answers)
Closed 7 years ago.
Apple has this segment of code on one of their sample projects:
let existingImage = cache.objectForKey(documentIdentifier) as? UIImage
if let existingImage = existingImage where cleanThumbnailDocumentIDs.contains(documentIdentifier) {
return existingImage
}
why is apple using this if let? Isn't more logical to simply use
if cleanThumbnailDocumentIDs.contains(documentIdentifier) {
return existingImage!
}
???!!
If you use
let existingImage = cache.objectForKey(documentIdentifier) as? UIImage
if let existingImage = existingImage where cleanThumbnailDocumentIDs.contains(documentIdentifier) {
return existingImage
}
This will make sure that if existingImage == nil,it will not
execute return existingImage.
Besides,if let also unwrap existingImage from UIImage? to
UIImage
As Abhinav mentioned above, Apple introduced a new type called optional type with Swift.
What does optional mean?
Short and Sweet, "Optional types are types, which can contain a value of a particular data type or nil".
You can read more about optionals and their advantages here : swift-optionals-made-simple
Now whenever you want to make use of value present in an optional type, first you need to check what it contains i.e. does it contains a proper value or it contains nil. This process is called optional unwrapping.
Now there are two types of unwrapping,
Forced unwrapping : If you're sure that an optional will have an value all the time, you can then unwrap the value present in the optional type using "!" mark. This is force unwrapping.
The one more way is to use if let expression, this is safe unwrapping, here you'll check in your program that, if optional has a value you will do something with it; if it doesn't contain value you'd do something else. A simple example is this (You can test this in play ground:
func printUnwrappedOptional (opt:String?) {
if let optionalValue = opt { //here we try to assign opt value to optionalValue constant, if assignment is successful control enters if block
println(optionalValue) // This will be executed only if optionalValue had some value
}
else {
println("nil")
}}
var str1:String? = "Hello World" //Declaring an optional type of string and assigning it with a value
var str2:String? //Declaring an optional type of string and not assigning any value, it defaults to nil
printUnwrappedOptional(str1) // prints "Hello World"
printUnwrappedOptional(str2) // prints "nil"
Hope this clears your question, read through the link given above it'll be more clear to you. Hope this helps. :)
Edit: In Swift 2.0, Apple introduced "guard" statements, once you're good with optionals go through this link, guard statement in swift 2. This is another way to deal with optionals.
Using if let, makes sure that the object (existingImage) is not nil, and it unwraps it automatically, so you are sure inside the if that the condition is true, and the object is not nil, and you can use it without unwrap it !
With Swift, Apple has introduced a new concept/type - Optional Type. I think you better go through Apple Documentation.
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.
existingImage is an optional (as? UIImage) and therefor needs to be unwrapped before used, otherwise there would be a compiler error. What you are doing is called forced unwrapping via !. Your program will crash, if existingImage == nil and is therefor only viable, if you are absolutely sure, that existingImage can't be nil
if let and optional types is more help where is there is changes to get nil values to void crashes and unwanted code executions.
In Swift 2.0,
guard
will help us lot where our intention is clear not to execute the rest of the code if that particular condition is not satisfied
In The Swift Programming Language (book of Apple) I've read that you can create optional variables in 2 ways: using a question mark (?) or by using an exclamation mark (!).
The difference is that when you get the value of an optional with (?) you have to use an exclamation mark every time you want the value:
var str: String? = "Question mark?"
println(str!) // Exclamation mark needed
str = nil
While with an (!) you can get it without a suffix:
var str: String! = "Exclamation mark!"
println(str) // No suffix needed
str = nil
What is the difference and why are there 2 ways if there is no difference at all?
The real benefit to using implicitly unwrapped optionals (declared with the !) is related to class initialisation when two classes point to each other and you need to avoid a strong-reference cycle. For example:
Class A <-> Class B
Class A's init routine needs to create (and own) class B, and B needs a weak reference back to A:
class A {
let instanceOfB: B!
init() {
self.instanceOfB = B(instanceOfA: self)
}
}
class B {
unowned let instanceOfA: A
init(instanceOfA: A) {
self.instanceOfA = instanceOfA
}
}
Now,
Class B needs a reference to class A to be initialised.
Class A can only pass self to class B's initialiser once it's fully initialised.
For Class A to be considered as initialised before Class B is created, the property instanceOfB must therefore be optional.
However, once A's been created it would be annoying to have to access instanceOfB using instanceOfB! since we know that there has to be a B
To avoid this, instanceOfB is declared as an implicity unwrapped optional (instanceOfB!), and we can access it using just instanceOfB. (Furthermore, I suspect that the compiler can optimise the access differently too).
An example of this is given on pages 464 to 466 of the book.
Summary:
Use ? if the value can become nil in the future, so that you test for this.
Use ! if it really shouldn't become nil in the future, but it needs to be nil initially.
You should go beyond the syntactic sugar.
There are two completely different polymorphic types. The syntactic sugar just uses either one or the other of these types.
When you write Foo? as a type you really have Optional<Foo>, while when you write Foo! you really have ImplicitlyUnwrappedOptional<Foo>.
These are two different types, and they are different from Foo as well.
The values you create with ? are plain optional values as you mentioned, you should access it via optional binding (if let unwrappedValue = myOptionalValue) or by using the exclamation point syntax myOptionalValue!.doSomething().
The values you create with ! are called implicitly unwrapped optionals. With them, you don't need to manually unwrap before using them. When you do val myOptionalValue!.doSomething().
The value will be automatically unwrapped for you when you use myOptionalValue directly, be careful with this though, because accessing an implicitly unwrapped value when there is actually no value in it (when it is nil) will result in a runtime error.
? (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 : String? // declared variable with default nil value
println(defaultNil) >> nil
var canBeNil : String? = "test"
println(canBeNil) >> optional(test)
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 : String! = "test"
print(canNotBeNil) >> "test"
var cantBeNil : String = "test"
cantBeNil = nil // can't do this as it's not optional and show a compile time error
For more detail, refer a document by Apple Developer Commitee, in detail
The String! kind is called an implicitly unwrapped optional:
Sometimes it’s clear from a program’s structure that an optional will always have a value, after that value is first set. In these cases, it’s useful to remove the need to check and unwrap the optional’s value every time it’s accessed, because it can be safely assumed to have a value all of the time.
These kinds of optionals are defined as implicitly unwrapped optionals. You write an implicitly unwrapped optional by placing an exclamation mark (String!) rather than a question mark (String?) after the type that you want to make optional.
in the optional chaining section you find the answer:
example class:
class Person {
var residence: Residence?
}
class Residence {
var numberOfRooms = 1
}
If you try to access the numberOfRooms property of this person’s residence, by placing an exclamation mark after residence to force the unwrapping of its value, you trigger a runtime error, because there is no residence value to unwrap:
let roomCount = john.residence!.numberOfRooms
// this triggers a runtime error
The code above succeeds when john.residence has a non-nil value and will set roomCount to an Int value containing the appropriate number of rooms. However, this code always triggers a runtime error when residence is nil, as illustrated above.
Optional chaining provides an alternative way to access the value of numberOfRooms. To use optional chaining, use a question mark in place of the exclamation mark:
if let roomCount = john.residence?.numberOfRooms {
println("John's residence has \(roomCount) room(s).")
} else {
println("Unable to retrieve the number of rooms.")
}
// prints "Unable to retrieve the number of rooms."
Well mentioned by #tarmes above. Noticed another usage of implicit optional:
Lets say I have an optional Int:
let firstInt: Int? = 9
And I'm trying to use optional pattern matching and use this optional Int like this:
if case let myFirstInt? = firstInt where myFirstInt > 1 {
print("Valid")
} else {
print("Invalid")
}
Notice that I'm using implicit optional with local parameter myFirstInt making it safe for nil condition linked with optional firstInt. If now, I make firstInt as nil, it will execute else condition. If, instead, I use force-unwrap with firstInt that would lead to crash, something like this:
ok so I am trying to return nil if a certain type is passed into my function. In this case im passing in an instance of my class "BlogPost" and a type within this blogpost. I also have an array called "types" and I have assigned the variable Videos to the last index of that array. If this type is passed into my function I would like to return nil (so assuming im going to need an optional here for returning a possible nil) this is what I have so far :-
so all in all I need to pass in an instance of my blog post but always return nil if a certain type is passed in. Hope this makes sense
Update:
The types array is defined as follows:
let types : [String] = ["technology", "Fashion", "Animals"]
this is the array I am referring to in the function. Basically if that last entry of the array is entered into the function I need to return nil
sure this is blogpost it does actually have an empty string for type
great so im getting there what Ive done now is change the blogpost.type to choose one at random. So now if the specfic type is chosen from this array how would I do that still getting an error. This is what I have updated to
so now all I need to do is access the 2 type in that array and if I do access it return nil. Any thoughts on that? so to drag it on thanks
I don't think you can. You can create failable initialisers which does what you need but you cannot use it with normal function.
The best solution for you would be return optional Int or String and when you call the function just check the result for nil and do what you need to do, otherwise ignore it:
func randomViews(blog : BlogPost.Type) -> Int? {
case 10:
return nil
case 10, 20 :
return 0
default:
random
}
if (randomViews(parameter) == nil) {
//function returned nil
}
You have displayed error because you compare optional blog to Videos, you have to unwrap it first, for example if you are sure the blog has always have a value use:
if blog! == Videos
if not sure is safer to use:
if let blg = blog {
if blg == Videos {
}
else {
// blog has not have a value
}
You are passing blog as a BlogPost.Type parameter. That is not correct. You should have either just passed it the String parameter, or you could pass it the BlogPost itself:
func randomViews(blog: BlogPost) {
let videos = types[2]
if blog.type == videos {
// do whatever you want
}
// carry on
}
Unrelated to your question at hand, but notice that I use let instead of var when defining videos. Always use let if the value will not (and cannot) change.
Also note that I use lowercase letter v in videos, because Cocoa naming conventions dictate that variables generally start with lowercase letters, whereas types, classes, structs, and enums generally start with uppercase letters.