I am having troubles while converting optional string to int.
println("str_VAR = \(str_VAR)")
println(str_VAR.toInt())
Result is
str_VAR = Optional(100)
nil
And i want it to be
str_VAR = Optional(100)
100
At the time of writing, the other answers on this page used old Swift syntax. This is an update.
Convert Optional String to Int: String? -> Int
let optionalString: String? = "100"
if let string = optionalString, let myInt = Int(string) {
print("Int : \(myInt)")
}
This converts the string "100" into the integer 100 and prints the output. If optionalString were nil, hello, or 3.5, nothing would be printed.
Also consider using a guard statement.
You can unwrap it this way:
if let yourStr = str_VAR?.toInt() {
println("str_VAR = \(yourStr)") //"str_VAR = 100"
println(yourStr) //"100"
}
Refer THIS for more info.
When to 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. Let’s have a look:
if let yourStr = str_VAR?.toInt() {
println("str_VAR = \(yourStr)")
println(yourStr)
}else {
//show an alert for something else
}
The if let structure unwraps str_VAR?.toInt() (i.e. checks if there’s a value stored and takes that value) and stores its value in the yourStr constant. You can use yourStr inside the first branch of the if. Notice that inside the if you don’t need to use ? or ! anymore. It’s important to realise thatyourStr is actually of type Int that’s not an Optional type so you can use its value directly.
Try this:
if let i = str_VAR?.toInt() {
println("\(i)")
}
Related
I make a query that returns a NSNumber. I then attempt to cast the NSNumber to String. For some reason, it always prints/ compares as an optional...but when I check the variables type it says string...Why is it optional? I need it to be a string!
let whoseTurn = selectedGame?["currentSender"]
let whoseTurnAsString: String = String(describing: whoseTurn)
if let whoseTurn = selectedGame?["currentSender"] as? NSNumber {
let whoseTurnAsString = "\(whoseTurn)"
print(whoseTurnAsString)
}
This is the right way to do optional chaining and make sure you are not forcing an optional
whoseTurn is an optional wrapping your NSNumber. You aren't "casting" it here to a string, you are making a string that "describes" it, and in this case that description includes the fact that whoseTurn is an optional... If you don't want that you'll need to unwrap it,
let whoseTurnAsString: String = String(describing: whoseTurn!)
(note the ! at the end)
This line of code let whoseTurn = selectedGame?["currentSender"] will return an optional.
This line of code let whoseTurnAsString: String = String(describing: whoseTurn) will return a String describing that optional value, which will be a string like this: Optional(5) or Optional(6). It describes the value saying that it is an optional.
So you need to unwrap that optional in order to get the wrapped value, you can force unwrap selectedGame like this:
let whoseTurn = selectedGame!["currentSender"] and then use the normal String initializer like this: String(whoseTurn).
Or, preferably, safely unwrap it like this:
if let whoseTurn = selectedGame?["currentSender"] {
let whoseTurnAsString = String(whoseTurn)
}
String can be optional also '?' or '!' indicates optional, check documentation on optionals.
I have following simple code after last line sum swift gives error
and var sum=input*2 should be changed to var sum=input!*2
I am not sure why since i didn't declare variable a as optional.
Does swift make input as optional ? Thanks
Var a="2"
let input = int(a)
var sum=input*2
Casting a string to an int returns and Optional because it could fail.
For example, let result = Int("foo") will return nil, because "foo" is not a valid Int.
What if you did
Var a = "This is most defiantly not a number and even if it were its too long to fit within an int 123457993849038409238490ff9f-09-0f9-09f dd0d0066646464646464349023849038490328 I'm a teapot".
let input = int(a)
Do you think that could be converted to an int? Now do you understand why its an optional?
What happens if the String can't be converted to an Int? Such as Int("A"). It becomes an optional because the compiler can't know for sure that the String you are passing in can become an Int.
guard let intVal = Int(a) else { return }
or
if let intVal = Int(a) {
//you have a valid Int here
}
is the way to go about handling this situation
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 an "if let" statement that is being executed, despite the "let" part being nil.
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]! {
println(leftInc)
let valueString : String = formatter.stringFromNumber(NSNumber(double: leftInc!))!
self.leftIncisorTextField?.text = valueString
self.btnLeftIncisor.associatedLabel?.text = valueString
}
// self.analysis.inputs is a Dictionary<String, Double?>
The inputs dictionary holds information entered by the user - either a number, or nil if they haven't entered anything in the matching field yet.
Under the previous version of Swift, the code was written as this:
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]?? {
and worked correctly.
I saw a similar question here, but in that instance the problem seemed to be the result of using Any?, which is not the case here.
Swift 2.2
In your if let you define another optional, that's why nil is a legitimate case. if let is intended mainly to extract (maybe) non optional value from an optional.
You might try:
if let leftInc : Double = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!].flatMap ({$0}) {
// leftInc is not an optional in this scope
...
}
Anyway I'd consider to not do it as a one liner but take advantage of guard case. Just in order to enhance readability. And avoid bang operator (!).
The if-let is for unwrapping optionals. You are allowing nil values by setting the type to an optional Double.
The if statement should be:
if let leftInc = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!] as? Double{
...
}
This will attempt to get an object out of inputs, if that fails it returns nil and skips it. If it does return something it will attempt to convert it to a Double. If that fails it skips the if statement as well.
if inputs is a dictionary like [Something:Double] then you don't need the last as? Double as indexing the dictionary will return a Double?
I recommend reading the swift book on optional chaining.
You could break it down further -
if let optionalDouble = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!], leftInc = optionalDouble {
....
}
as your dictionary has optional values - this way of writing it might make it clearer what's going on
if let k = dict["someKey"]{}, dict["someKey"] will be an object of type Any
this can bypass a nill
So do a typecast to get it correct like if let k = dict["someKey"] as! String {}
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.