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
Related
Converting a String to Int returns an optional value but converting a Double to Int does not return an optional value. Why is that? I wanted to check if a double value is bigger than maximum Int value, but because converting function does not return an optional value, I am not be able to check by using optional binding.
var stringNumber: String = "555"
var intValue = Int(stringNumber) // returns optional(555)
var doubleNumber: Double = 555
var fromDoubleToInt = Int(doubleNumber) // returns 555
So if I try to convert a double number bigger than maximum Integer, it crashes instead of returning nil.
var doubleNumber: Double = 55555555555555555555
var fromDoubleToInt = Int(doubleNumber) // Crashes here
I know that there's another way to check if a double number is bigger than maximum Integer value, but I'm curious as why it's happening this way.
If we consider that for most doubles, a conversion to Int simply means dropping the decimal part:
let pieInt = Int(3.14159) // 3
Then the only case in which the Int(Double) constructor returns nil is in the case of an overflow.
With strings, converting to Int returns an optional, because generally, strings, such as "Hello world!" cannot be represented as an Int in a way that universally makes sense. So we return nil in the case that the string cannot be represented as an integer. This includes, by the way, values that can be perfectly represented as doubles or floats:
Consider:
let iPi = Int("3.14159")
let dPi = Double("3.14159")
In this case, iPi is nil while dPi is 3.14159. Why? Because "3.14159" doesn't have a valid Int representation.
But meanwhile, when we use the Int constructor which takes a Double and returns non-optional, we get a value.
So, if that constructor is changed to return an optional, why would it return 3 for 3.14159 instead of nil? 3.14159 can't be represented as an integer.
But if you want a method that returns an optional Int, returning nil when the Double would overflow, you can just write that method.
extension Double {
func toInt() -> Int? {
let minInt = Double(Int.min)
let maxInt = Double(Int.max)
guard case minInt ... maxInt = self else {
return nil
}
return Int(self)
}
}
let a = 3.14159.toInt() // returns 3
let b = 555555555555555555555.5.toInt() // returns nil
Failable initializers and methods with Optional return types are designed for scenarios where you, the programmer, can't know whether a parameter value will cause failure, or where verifying that an operation will succeed is equivalent to performing the operation:
let intFromString = Int(someString)
let valueFromDict = dict[someKey]
Parsing an integer from a string requires checking the string for numeric/non-numeric characters, so the check is the same as the work. Likewise, checking a dictionary for the existence of a key is the same as looking up the value for the key.
By contrast, certain operations are things where you, the programmer, need to verify upfront that your parameters or preconditions meet expectations:
let foo = someArray[index]
let bar = UInt32(someUInt64)
let baz: UInt = someUInt - anotherUInt
You can — and in most cases should — test at runtime whether index < someArray.count and someUInt64 < UInt32.max and someUInt > anotherUInt. These assumptions are fundamental to working with those kinds of types. On the one hand, you really want to design around them from the start. On the other, you don't want every bit of math you do to be peppered with Optional unwrapping — that's why we have types whose axioms are stated upfront.
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 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)")
}
It should be easy but I can only find the reverse conversion.
How can I convert Int32 to Int in Swift?
Unless the problem is different?
I have a value stored in Core Data and I want to return it as an Int.
Here is the code I am using, which does not work:
func myNumber () -> Int {
var myUnit:NSManagedObject
myUnit=self.getObject(“EntityName”) // This is working.
return Int(myUnit.valueForKey(“theNUMBER”)?.intValue!)
}
Am I missing something or isn't this ridiculously easy?
let number1: Int32 = 10
let number2 = Int(number1)
The error is your ? after valueForKey.
Int initializer doesnt accept optionals.
By doing myUnit.valueForKey(“theNUMBER”)?.intValue! gives you an optional value and the ! at the end doesnt help it.
Just replace with this:
return Int(myUnit.valueForKey(“theNUMBER”)!.intValue)
But you could also do like this if you want it to be fail safe:
return myUnit.valueForKey(“theNUMBER”)?.integerValue ?? 0
And to shorten you function you can do this:
func myNumber() -> Int {
let myUnit = self.getObject("EntityName") as! NSManagedObject
return myUnit.valueForKey("theNUMBER")?.integerValue ?? 0
}
Swift 4.0 producing "Cannot invoke initializer for type 'Int' with an argument list of type '(() -> Int32)"
let number1: Int32 = 10
let number2 = Int(number1)
Simply do this
Int("\(Int32value)")
I'm unable to understand why swift is making things difficult.
Sometimes "?" make things twisted ,
adding "!" to Int32 and then convert it to int works
let number1 = someInt32!
let number2 = Int(number1)
I am attempting to use a hash that contains a nullable int as a value. The below is the code.
var facs:[(pk:Int,phone:Int?,name:String)] = []
var phone: AnyObject? = fac["phone"]!
var phoneLong:Int?;
if(phone == nil){
phoneLong = nil
}
else{
phoneLong = phone as? Int
}
var id = fac["id"]! as Int
var name = fac["name"]! as String
facs.append(pk:id, phone:phoneLong, name:name)
However, I get a compile error on the facs.append line that states Type 'T' does not conform to protocol 'IntegerLiteralConvertible'. I've tried a few variations, and the only way I can get rid of the error is to make phone a non-nullable int, which is not what I need. Thoughts?
It looks like the append method doesn't correctly detect the parameters as a tuple - just make that explicit by assigning the tuple to a variable:
let params = (pk:id, phone:phoneLong, name:name)
facs.append(params)