When i use the following code and have nameTextField be "Jeffrey" (or any other name)
#IBAction func helloWorldAction(nameTextField: UITextField) {
nameLabel.text = "Hello, \(nameTextField.text)"
}
nameLabel displays... Hello, Optional("Jeffrey")
But, when I change the previous code to include a "!" like this:
#IBAction func helloWorldAction(nameTextField: UITextField) {
nameLabel.text = "Hello, \(nameTextField.text!)"
}
The code works as expected and nameLabel displays.... Hello, Jeffrey
Why is the "!" required, in the video tutorial I used to create this simple program he did not use the "!" and the program worked as expected.
Another alternative is to use the null coalescing operator within the interpolated string for prettier text without the need for if let.
nameLabel.text = "Hello, \(nameTextField.text ?? "")"
It's less readable in this case, but if there were a lot of strings it might be preferable.
Optionals must be unwrapped. You must check for it or force unwrap as you do. Imagine the optional as a box where you put a value. Before you can access it, you need to put it out.
if let name = nameTextField.text {
nameLabel.text = "Hello, \(name)"
}
Here's a handy extension to unwrap Any? to String.
Set a default value for nil values.
extension String {
init(_ any: Any?) {
self = any == nil ? "My Default Value" : "\(any!)"
}
}
// Example
let count: Int? = 3
let index: Int? = nil
String(count)
String(index)
// Output
// 3
// My Default Value
You can also use optional map.
This is where I learned of how to use it.
Basically, map will take an optional and return a value if there's a value and return nil if there's no value.
It think this makes more sense in code, so here's the code I found useful:
func getUserAge() -> Int? {
return 38
}
let age = getUserAge()
let ageString = age.map { "Your age is \($0)" }
print(ageString ?? "We don't know your age.")
I guess this may not be super helpful in the case where you're passing in an optional string, (nil coalescing works just fine in that case), but this is super helpful for when you need to use some value that isn't a string.
It's even more useful when you want to run logic on the given value since map is a closure and you can use $0 multiple times.
Related
I am trying to print the value of strVal after lblTxt in the label when the text field is empty but it not accepting the value, instead just printing the value of lblTxt. I know it works when the left hand side variable if of optional, but i am taking value from the textfield how should I make that optional? I am very new to this.
The type of textFieldTest.text is an optional String, or String?. There are several ways to use optionals:
var text4:String? = textFieldTest.text // assign it to another optional variable
var text5 = textFieldTest.text // the same, the compiler will infer that text5 is also an optional string
textFieldTest.text?.append(" hello") // will only call `append(_:)` on `text` if it's not nil, otherwise this will be ignored
let text6:String = textFieldTest.text ?? "It was nil!" // use the coalescing operator
let text7:String = textFieldTest.text! // force-unwrap it into a String; if it's nil, your app will crash
if let text8 = textFieldTest.text {
// this code block will ONLY execute if textField.text was not nil
// in that case, text8 will be of type String (non-optional)
// so it shouldn't be treated as an optional
}
Your code unwraps the optional, and runs a code block only in case the optional was not nil. Then, in the same code block, you still want to test for nil, but again, if it were nil, the code block would not execute in the first place. And you even perform the test on a non-optional type which is not meaningful.
As your code does not seem to require textFieldTest.text to contain a value (actually, on the contrary: you want to explicitly also handle the scenario where it doesn't), you must avoid making your code block only execute in case textFieldTest.text unwraps to a String.
So simply do not unwrap it:
#IBaction func stringButtonPressed(_ sender: Any) {
labelTest.text = "\(lblTxt) \(textFieldTest.text ?? strval!)"
labelTest.backgroundColor = UIColor.yellow
...
}
Or you can use the unwrap logic to do the same in a more verbose way:
#IBaction func stringButtonPressed(_ sender: Any) {
if let text3 = textFieldTest.text {
labelTest.text = "\(lblTxt) \(text3)"
} else {
labelTest.text = "\(lblTxt) \(strval!)"
}
labelTest.backgroundColor = UIColor.yellow
...
}
In any case, be aware that the block after if let is conditional.
If you declare a variable with if let you basically saying Execute this code blocks when its not nil so text3 ?? strval! in that line strval will not gonna execute because text3 isn't nil
If text3 is nil , code blocks in brackets won't execute.
I'm new to swift and I don't really understand how to use optional value correctly.
The situation is: Firstly, I have a model class to store some values of its properties and send the request to server to get values. code:
import Foundation
class User: NSObject {
var name: String?
func getInfo(updateUI: () -> ()) {
let manager = AFHTTPSessionManager()
manager.POST(URLString, parameters: nil, success: { (task: NSURLSessionDataTask, responseObject: AnyObject?) in
let dic = responseObject as? NSDictionary
self.name = dic?.objectForKey("name")
updateUI()
}, failure: { (task: NSURLSessionDataTask?, error: NSError) in
log.obj(error)
}）
}
Secondly, I want to use this model in a ViewController to get values from server and update UI. code:
class UserViewController: UIViewController {
#IBOutlet var nameLabel: UILabel!
var user = User()
override func viewDidLoad() {
super.viewDidLoad()
user.getInfo({
nameLabel.text = user.name!
})
}
}
I think it's a dangerous way to handle this work because there is nothing can ensure that I have a certain name when I want to put it on the label. It works pretty well in most cases but it can crash when fail to get a right value from server(for example, there is no value for key "name").
I have two ideas about questions above:
Give a default value for the properties like: var name = ""
Test the value before unwarp it like:
if let foo = user.name {
nameLabel.text = foo
}
I want to choose the first way because:
I think showing a default string on the label is much more acceptable than crash because fail to unwarp a optional value.
When I have a lot more values to use, the code will be very long and diffcult to read.
Did I choose a right one? Or both of them didn't understand the usage of optional value and there is a better way? Someone can help me?
You can safely unwrap an optional or give it a default value without declaring a new variable like so:
nameLabel.text = user.name ?? ""
This is called the nil coalescing operator and will try to unwrap an optional, and if the unwrap is unsuccessful will give it the default value that you have placed on the rhs of the operator.
Your instincts are correct that force unwrapping optional is ill-advised. You are also correct that providing a non-nil default value defeats the purpose of an optional.
if let foo = user.name binding is a sound way of doing this, but if all you're doing is assigning a value, you can spare yourself the curly braces by using the nil coalescing operator, as pbush25 states:
nameLabel.text = user.name ?? "Default Label Text"
This means "unwrap user.name and if its value is not nil, use it. Otherwise, use the following value:" You can even chain them together if, for example, you wanted to use user's ID as the fallback label text, but the ID might also not exist:
nameLabel.text = user.name ?? user.id ?? "No name or ID".
Even if you want to crash in the case that the optional value is nil, it would be better to assert or use a preconditionFailure() than to simply force unwrap the optional so that you can at least provide an error message.
Dears
I have this case where chatId is a property of type Int
let StringMessage = String(self.listingChat?.messages.last?.chatId)
When I debug I find that StringMessage is returning Optional(15) Which means it is unwrapped. But at the same time XCode does not allow me to put any bangs (!) to unwrap it. So I am stuck with Unwrapped Variable. I know its noob question but it I really cant get it. Your help is appreciated.
Thank you
It depends on what you want the default value to be.
Assuming you want the default value to be an empty string (""), You could create a function or a method to handle it.
func stringFromChatId(chatId: Int?) -> String {
if let chatId = chatId {
return String(chatId)
} else {
return ""
}
}
let stringMessage = stringFromChatId(self.listingChat?.messages.last?.chatId)
Or you could handle it with a closure.
let stringMessage = { $0 != nil ? String($0!) : "" }(self.listingChat?.messages.last?.chatId)
If you don't mind crashing if self.listingChat?.messages.last?.chatId is nil, then you should be able to directly unwrap it.
let StringMessage = String((self.listingChat?.messages.last?.chatId)!)
or with a closure
let stringMessage = { String($0!) }(self.listingChat?.messages.last?.chatId)
Update
Assuming chatId is an Int and not an Optional<Int> (AKA Int?) I missed the most obvious unwrap answer. Sorry, I was tired last night.
let StringMessage = String(self.listingChat!.messages.last!.chatId)
Force unwrap all the optionals along the way.
Optionals have a very nice method called map (unrelated to map for Arrays) which returns nil if the variable is nil, otherwise it calls a function on the (non-nil) value. Combined with a guard-let, you get very concise code. (I've changed the case of stringMessage because variables should begin with a lower-case letter.)
guard let stringMessage = self.listingChat?.messages.last?.chatId.map { String($0) } else {
// Do failure
}
// Success. stringMessage is of type String, not String?
I think:
let StringMessage = String(self.listingChat?.messages.last?.chatId)!
I have a variable
var a: [AnyObject? -> Void]
and I am adding data in to it by append method. Now I want to check if the variable is nil or not. I tried using [] but not working and also tried "", this also not working, can anyone tell what is the meaning of this variable and how to check if it is nil.
As far as I understand, var a is an Array of functions that take an optional Object of any type, and return void. So these functions's parameter IS optional, but the Array itself isn't : it cannot be nil, or it would be declared [AnyObject? -> Void]? , no?
EDIT : if, nevertheless, you declared this a as an optional (but WHY would you do that ?) - adding a ? - you check an optional existence with if let :
if let b = a {
// a not nil, do some stuff
} else {
// a is null
}
If you just want to check if the array is empty, use isEmpty method from Swift Array
Update: Xcode 7.3 Swift 2.2
If you want to check if a variable is nil you should use if let to unwrap if for you. There is no need to create a second var.
let str = "123"
var a = Int(str)
if let a = a {
print(a)
}
Or
if let a = Int(str) {
print(a)
}
In Swift, nil is not a pointer—it is the absence of a value of a certain type. Optionals of any type can be set to nil, not just object types.
So, You can check it with below code:
let possibleNumber = "123"
let convertedNumber = possibleNumber.toInt()
if convertedNumber != nil {
println("convertedNumber contains some integer value.")
}
// prints "convertedNumber contains some integer value."
Please refer this about nil for more information.
In Swift 3.0
if let imageURL = dictObj["list_image"] as? String {
print(imageURL)
}
You can use if let. if let is a special structure in Swift that allows you to check if an Optional holds a value, and in case it does – do something with the unwrapped value.
var a:Int=0
if let b=a{
println(a)
} else {
println("Value - nil")
}
But for Strings you can also use .isEmpty() If you have initialized it to "".
var str:String=""
if !str.isEmpty(){
println(str)
}
For me none of the above solutions worked when I was using an AVFoundation object.
I would get Type 'AVCaptureDeviceInput does not conform to protocol 'BooleanType' when I tried if (audioDeviceInput) and I would get Binary operator '!=' cannot be applied to operands of type 'AVCaptureDeviceInput' and 'nil'.
Solution in my situation
if (audioDeviceInput.isEqual(nil))
nil is a pointer like any other and can be referenced as such, which is why this works.
I am trying with these lines of code
class Student {
var name: String
var age: Int?
init(name: String) {
self.name = name
}
func description() -> String {
return age != nil ? "\(name) is \(age) years old." : "\(name) hides his age."
}
}
var me = Student(name: "Daniel")
println(me.description())
me.age = 18
println(me.description())
Above code produces as follow
Daniel hides his age.
Daniel is Optional(18) years old.
My question is why there is Optional (18) there, how can I remove the optional and just printing
Daniel is 18 years old.
You have to understand what an Optional really is. Many Swift beginners think var age: Int? means that age is an Int which may or may not have a value. But it means that age is an Optional which may or may not hold an Int.
Inside your description() function you don't print the Int, but instead you print the Optional. If you want to print the Int you have to unwrap the Optional. You can use "optional binding" to unwrap an Optional:
if let a = age {
// a is an Int
}
If you are sure that the Optional holds an object, you can use "forced unwrapping":
let a = age!
Or in your example, since you already have a test for nil in the description function, you can just change it to:
func description() -> String {
return age != nil ? "\(name) is \(age!) years old." : "\(name) hides his age."
}
To remove it, there are three methods you could employ.
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.
The Implicitly Unwrapped Optional
if let unwrappedAge = age {
// continue in here
}
Note that the unwrapped type is now Int, rather than Int?.
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!
For testing/debugging purposes I often want to output optionals as strings without always having to test for nil values, so I created a custom operator.
I improved things even further after reading this answer in another question.
fileprivate protocol _Optional {
func unwrappedString() -> String
}
extension Optional: _Optional {
fileprivate func unwrappedString() -> String {
switch self {
case .some(let wrapped as _Optional): return wrapped.unwrappedString()
case .some(let wrapped): return String(describing: wrapped)
case .none: return String(describing: self)
}
}
}
postfix operator ~? { }
public postfix func ~? <X> (x: X?) -> String {
return x.unwrappedString
}
Obviously the operator (and its attributes) can be tweaked to your liking, or you could make it a function instead. Anyway, this enables you to write simple code like this:
var d: Double? = 12.34
print(d) // Optional(12.34)
print(d~?) // 12.34
d = nil
print(d~?) // nil
Integrating the other guy's protocol idea made it so this even works with nested optionals, which often occur when using optional chaining. For example:
let i: Int??? = 5
print(i) // Optional(Optional(Optional(5)))
print("i: \(i~?)") // i: 5
Update
Simply use me.age ?? "Unknown age!". It works in 3.0.2.
Old Answer
Without force unwrapping (no mach signal/crash if nil) another nice way of doing this would be:
(result["ip"] ?? "unavailable").description.
result["ip"] ?? "unavailable" should have work too, but it doesn't, not in 2.2 at least
Of course, replace "unavailable" with whatever suits you: "nil", "not found" etc
To unwrap optional use age! instead of age. Currently your are printing optional value that could be nil. Thats why it wrapped with Optional.
In swift Optional is something which can be nil in some cases. If you are 100% sure that a variable will have some value always and will not return nil the add ! with the variable to force unwrap it.
In other case if you are not much sure of value then add an if let block or guard to make sure that value exists otherwise it can result in a crash.
For if let block :
if let abc = any_variable {
// do anything you want with 'abc' variable no need to force unwrap now.
}
For guard statement :
guard is a conditional structure to return control if condition is not met.
I prefer to use guard over if let block in many situations as it allows us to return the function if a particular value does not exist.
Like when there is a function where a variable is integral to exist, we can check for it in guard statement and return of it does not exist.
i-e;
guard let abc = any_variable else { return }
We if variable exists the we can use 'abc' in the function outside guard scope.
age is optional type: Optional<Int> so if you compare it to nil it returns false every time if it has a value or if it hasn't. You need to unwrap the optional to get the value.
In your example you don't know is it contains any value so you can use this instead:
if let myAge = age {
// there is a value and it's currently undraped and is stored in a constant
}
else {
// no value
}
I did this to print the value of string (property) from another view controller.
ViewController.swift
var testString:NSString = "I am iOS Developer"
SecondViewController.swift
var obj:ViewController? = ViewController(nibName: "ViewController", bundle: nil)
print("The Value of String is \(obj!.testString)")
Result :
The Value of String is I am iOS Developer
Check out the guard statement:
for student in class {
guard let age = student.age else {
continue
}
// do something with age
}
When having a default value:
print("\(name) is \(age ?? 0) years old")
or when the name is optional:
print("\(name ?? "unknown") is \(age) years old")
I was getting the Optional("String") in my tableview cells.
The first answer is great. And helped me figure it out. Here is what I did, to help the rookies out there like me.
Since I am creating an array in my custom object, I know that it will always have items in the first position, so I can force unwrap it into another variable. Then use that variable to print, or in my case, set to the tableview cell text.
let description = workout.listOfStrings.first!
cell.textLabel?.text = description
Seems so simple now, but took me a while to figure out.
This is not the exact answer to this question, but one reason for this kind of issue.
In my case,
I was not able to remove Optional from a String with "if let" and "guard let".
So use AnyObject instead of Any to remove optional from a string in swift.
Please refer link for the answer.
https://stackoverflow.com/a/51356716/8334818
If you just want to get rid of strings like Optional(xxx) and instead get xxx or nil when you print some values somewhere (like logs), you can add the following extension to your code:
extension Optional {
var orNil: String {
if self == nil {
return "nil"
}
return "\(self!)"
}
}
Then the following code:
var x: Int?
print("x is \(x.orNil)")
x = 10
print("x is \(x.orNil)")
will give you:
x is nil
x is 10
PS. Property naming (orNil) is obviously not the best, but I can't come up with something more clear.
With the following code you can print it or print some default value. That's what XCode generally recommend I think
var someString: String?
print("Some string is \(someString ?? String("Some default"))")
If you are printing some optional which is not directly printable but has a 'to-printable' type method, such as UUID, you can do something like this:
print("value is: \(myOptionalUUID?.uuidString ?? "nil")")
eg
let uuid1 : UUID? = nil
let uuid2 : UUID? = UUID.init()
print("uuid1: \(uuid1?.uuidString ?? "nil")")
print("uuid2: \(uuid2?.uuidString ?? "nil")")
-->
uuid1: nil
uuid2: 0576137D-C6E6-4804-848E-7B4011B40C11