I am getting the error message Cannot assign value of type 'String?' to type 'Int'
I have browsed through other questions like this but it still shows the error.
if sunscreenName.text != nil && reapplyTime.text != nil {
sunscreen = sunscreenName.text!
reApplyTime = reapplyTime.text
//Some sort of message such as Progress hud
}
Thanks in advance!
I got your problem, actually what happens here Swift is is type safe langauge
So what you are doing is is to store a String value in Int which will not happen automatically you need to convert it to Int
like this
Int(sunscreenName.text)
But there is a catch there not all string are convertible to Int type, fo e.g.
let name = "roshan"
if you try to convert it to Int it will give you a nil
let a = Int(name)
So its better you do a optional Binding here provided by Swift
if let sunValue = Int(sunscreenName.text),let reApplyValue = Int(reapplyTime.text) {
sunscreen = sunValue
reApplyTime = reApplyValue
}
I recommend reading through The Swift Programming Language to get a better understanding of Swift and its fundamental concepts, since this question is fairly basic.
You make several mistakes:
if sunscreenName.text != nil && reapplyTime.text != nil {
This is wrong. In Swift, if you plan to use the value later, you should use if let rather than comparing to nil. Comparing to nil leaves the values optional, but if let unwraps them. So, do this instead:
if let sunscreenText = sunscreenName.text, let reapplyText = reapplyTime.text {
Now you have the sunscreenText and reapplyText variables, which are typed String, not String? (i.e. they are not optional).
Now, there's these two lines.
sunscreen = sunscreenName.text!
reApplyTime = reapplyTime.text
You don't say which one is giving the error, but the issue is the same in either case. First, use our unwrapped sunscreenText and reapplyText variables instead of sunscreenName.text! and reapplyTime.text. Next, if one of these is meant to be an Int instead of a String, cast it. Swift is not like JavaScript, in that it won't automatically convert values from one type to another, so if something is a string and we need an integer, we have to convert it ourselves.
(assuming reapplyTime was the line that was giving the error:)
if let reapplyInt = Int(reapplyText) {
reapplyTime = reapplyInt
}
The reason we have to unwrap is because Int(String) can return nil if the string is something that can't be converted to an integer. Alternately, we could just provide a default value:
reapplyTime = Int(reapplyText) ?? 0 // sets to 0 if it can't parse the string as an integer
I have created my own class in Swift as below.
class Product: NSObject {
var product_id:Int?
var product_number:String?
var product_price:Float?
var product_descrption:String?
}
Now i am setting value in each property like this
let p=Product()
p.product_id=1
p.product_price=220.22
p.productdescrption="Some description"
p.product_number="W2_23_233"
But when i get the value from price then for price i get value like "Optional 220.22" But i don't get appended word "Optional" in description".So to resolve this i added "!" for unwrapping the value of float but i did not have to do this for String please tell why this is happening?
If you are printing any of these values should say Optional(...). If you are assigning the values to a label, that will not include the Optional(...), The reason that it shows Optional(...) when you print the value using print(), is just to show you its an optional. For safety, instead of using the !, try using if lets.
An example with your code,
if let id = p.product_id {
print(id) //Does not contain Optional()
}
You can also combine them, to do them all at one time. (Only do this if you don't want to print unless all values are non-nil)
if let id = p.product_id,
let price = p.product_price,
let description = p.productdescrption,
let productNumber = p.product_number {
//Enter code here that does something with these values
}
Note, if you aren't on swift 3, I believe you only have to write let on the first condition.
If you print any optional variable without unwrapping no matter what type it is, Optional will be appended to the variable's value.
print(p.product_price) will print Optional(220.220001)
print(p.product_descrption) will print Optional("Some description")
To print only value you need to unwrap the optional variables.
print(p.product_price!) will print 220.22
print(p.product_descrption!) will print Some description
This forced unwrapping will only work if the optionals does not contain nil. Otherwise it will give you a runtime error.
So to check for nil you can use if let statement.
No matter what type of variable. If you assign a value to an optional variable, It always enclosed with Optional(...)
Optional without forced unwrapping:
print("product_price = \(p.product_price) \n product_descrption = \(p.product_descrption)")
Output:
product_price = Optional(220.22)
product_descrption = Optional(Some description)
Optional with forced unwrapping:
print("product_price = \(p.product_price!) \n product_descrption = \(p.product_descrption!)")
Output:
product_price = 220.22
product_descrption = Some description
I am using a keychain wrapper written in Swift.
When value is nil, this wrapper I'm using automatically saves data as "nil" instead of saving as an actual nil value.
For ex, ["last_name" : "nil"].
In one of my label in my iOS app, I'm returning last_name value.
It bothers me that a text in label is saying "nil" when there supposed to be nothing.
I remember in Swift that there is a syntax like A ?? B which puts B instead if A is not valid. But I cannot remember it right now.
In my app, I want to something like below:
if the value I'm looking for is nil, then input "" instead of "nil"
If keychain["last_name"] == nil ?? ""
I know this is a horrible explanation but this is all I could come up with.
?? is the nil coalescing operator. It is used for unwrapping optionals — a ?? b is shorthand for a != nil ? a! : b.
You probably want to use the ternary operator instead. You could even combine them like this:
let lastName = keychain["last_name"] == "nil" ? "" : (keychain["last_name"] ?? "")
That said, it might be easier to just filter out these responses immediately after getting the keychain instead of checking every time:
var keychain = ["first_name" : "aaron", "last_name" : "nil"]
keychain.forEach { if $0.1 == "nil" { keychain.removeValueForKey($0.0) } }
print(keychain) //["first_name": "aaron"]
The $0 is a shorthand argument name. Here's the long form version of the same code:
keychain.forEach { (key, value) -> () in
if value == "nil" {
keychain.removeValueForKey(key)
}
}
I have a custom object called Field. I basically use it to define a single field in a form.
class Field {
var name: String
var value: Any?
// initializers here...
}
When the user submits the form, I validate each of the Field objects to make sure they contain valid values. Some fields aren't required so I sometimes deliberately set nil to the value property like this:
field.value = nil
This seems to pose a problem when I use an if-let to determine whether a field is nil or not.
if let value = field.value {
// The field has a value, ignore it...
} else {
// Add field.name to the missing fields array. Later, show the
// missing fields in a dialog.
}
I set breakpoints in the above if-else and when field.value has been deliberately set to nil, it goes through the if-let block, not the else. However, for the fields whose field.value I left uninitialized and unassigned, the program goes to the else block.
I tried printing out field.value and value inside the if-let block:
if let value = field.value {
NSLog("field.value: \(field.value), value: \(value)")
}
And this is what I get:
field.value: Optional(nil), value: nil
So I thought that maybe with optionals, it's one thing to be uninitialized and another to have the value of nil. But even adding another if inside the if-let won't make the compiler happy:
if let value = field.value {
if value == nil { // Cannot invoke '==' with an argument list of type '(Any, NilLiteralConvertible)'
}
}
How do I get around this? I just want to check if the field.value is nil.
I believe this is because Any? allows any value and Optional.None is being interpreted as just another value, since Optional is an enum!
AnyObject? should be unable to do this since it only can contain Optional.Some([any class object]), which does not allow for the case Optional.Some(Optional) with the value Optional.None.
This is deeply confusing to even talk about. The point is: try AnyObject? instead of Any? and see if that works.
More to the point, one of Matt's comment mentions that the reason he wants to use Any is for a selection that could be either a field for text input or a field intended to select a Core Data object.
The Swifty thing to do in this case is to use an enum with associated values, basically the same thing as a tagged/discriminated union. Here's how to declare, assign and use such an enum:
enum DataSelection {
case CoreDataObject(NSManagedObject)
case StringField(String)
}
var selection : DataSelection?
selection = .CoreDataObject(someManagedObject)
if let sel = selection { // if there's a selection at all
switch sel {
case .CoreDataObject(let coreDataObj):
// do what you will with coreDataObj
case .StringField(let string):
// do what you will with string
}
}
Using an enum like this, there's no need to worry about which things could be hiding inside that Any?. There are two cases and they are documented. And of course, the selection variable can be an optional without any worries.
There's a tip to replace my Any? type with an enum but I couldn't get this error out of my head. Changing my approach doesn't change the fact that something is wrong with my current one and I had to figure out how I arrived at an Optional(nil) output.
I was able to reproduce the error by writing the following view controller in a new single-view project. Notice the init signature.
import UIKit
class Field {
var name: String = "Name"
var value: Any?
init(_ name: String, _ value: Any) {
self.name = name
self.value = value
}
}
class AppState {
var currentValue: Field?
}
class ViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
let f = Field("Amount", AppState().currentValue)
NSLog("\(f.value)")
}
}
In short, I was passing a nil value (AppState().currentValue) to an initializer that accepts Any, and assigns it to a property whose type is Any?. The funny thing here is if I directly passed nil instead, the compiler will complain:
let f = Field("Amount", nil) // Type 'Any' does not conform to protocol 'NilLiteralConvertible'
It seems that somewhere along the way, Swift wraps the nil value of AppState().currentValue in an optional, hence Optional(nil).
I'm using Xcode 6 Beta 4. I have this weird situation where I cannot figure out how to appropriately test for optionals.
If I have an optional xyz, is the correct way to test:
if (xyz) // Do something
or
if (xyz != nil) // Do something
The documents say to do it the first way, but I've found that sometimes, the second way is required, and doesn't generate a compiler error, but other times, the second way generates a compiler error.
My specific example is using the GData XML parser bridged to swift:
let xml = GDataXMLDocument(
XMLString: responseBody,
options: 0,
error: &xmlError);
if (xmlError != nil)
Here, if I just did:
if xmlError
it would always return true. However, if I do:
if (xmlError != nil)
then it works (as how it works in Objective-C).
Is there something with the GData XML and the way it treats optionals that I am missing?
In Xcode Beta 5, they no longer let you do:
var xyz : NSString?
if xyz {
// Do something using `xyz`.
}
This produces an error:
does not conform to protocol 'BooleanType.Protocol'
You have to use one of these forms:
if xyz != nil {
// Do something using `xyz`.
}
if let xy = xyz {
// Do something using `xy`.
}
To add to the other answers, instead of assigning to a differently named variable inside of an if condition:
var a: Int? = 5
if let b = a {
// do something
}
you can reuse the same variable name like this:
var a: Int? = 5
if let a = a {
// do something
}
This might help you avoid running out of creative variable names...
This takes advantage of variable shadowing that is supported in Swift.
Swift 3.0, 4.0
There are mainly two ways of checking optional for nil. Here are examples with comparison between them
1. if let
if let is the most basic way to check optional for nil. Other conditions can be appended to this nil check, separated by comma. The variable must not be nil to move for the next condition. If only nil check is required, remove extra conditions in the following code.
Other than that, if x is not nil, the if closure will be executed and x_val will be available inside. Otherwise the else closure is triggered.
if let x_val = x, x_val > 5 {
//x_val available on this scope
} else {
}
2. guard let
guard let can do similar things. It's main purpose is to make it logically more reasonable. It's like saying Make sure the variable is not nil, otherwise stop the function. guard let can also do extra condition checking as if let.
The differences are that the unwrapped value will be available on same scope as guard let, as shown in the comment below. This also leads to the point that in else closure, the program has to exit the current scope, by return, break, etc.
guard let x_val = x, x_val > 5 else {
return
}
//x_val available on this scope
One of the most direct ways to use optionals is the following:
Assuming xyz is of optional type, like Int? for example.
if let possXYZ = xyz {
// do something with possXYZ (the unwrapped value of xyz)
} else {
// do something now that we know xyz is .None
}
This way you can both test if xyz contains a value and if so, immediately work with that value.
With regards to your compiler error, the type UInt8 is not optional (note no '?') and therefore cannot be converted to nil. Make sure the variable you're working with is an optional before you treat it like one.
From swift programming guide
If Statements and Forced Unwrapping
You can use an if statement to find out whether an optional contains a
value. If an optional does have a value, it evaluates to true; if it
has no value at all, it evaluates to false.
So the best way to do this is
// swift > 3
if xyz != nil {}
and if you are using the xyz in if statement.Than you can unwrap xyz in if statement in constant variable .So you do not need to unwrap every place in if statement where xyz is used.
if let yourConstant = xyz {
//use youtConstant you do not need to unwrap `xyz`
}
This convention is suggested by apple and it will be followed by devlopers.
Although you must still either explicitly compare an optional with nil or use optional binding to additionally extract its value (i.e. optionals are not implicitly converted into Boolean values), it's worth noting that Swift 2 has added the guard statement to help avoid the pyramid of doom when working with multiple optional values.
In other words, your options now include explicitly checking for nil:
if xyz != nil {
// Do something with xyz
}
Optional binding:
if let xyz = xyz {
// Do something with xyz
// (Note that we can reuse the same variable name)
}
And guard statements:
guard let xyz = xyz else {
// Handle failure and then exit this code block
// e.g. by calling return, break, continue, or throw
return
}
// Do something with xyz, which is now guaranteed to be non-nil
Note how ordinary optional binding can lead to greater indentation when there is more than one optional value:
if let abc = abc {
if let xyz = xyz {
// Do something with abc and xyz
}
}
You can avoid this nesting with guard statements:
guard let abc = abc else {
// Handle failure and then exit this code block
return
}
guard let xyz = xyz else {
// Handle failure and then exit this code block
return
}
// Do something with abc and xyz
Swift 5 Protocol Extension
Here is an approach using protocol extension so that you can easily inline an optional nil check:
import Foundation
public extension Optional {
var isNil: Bool {
guard case Optional.none = self else {
return false
}
return true
}
var isSome: Bool {
return !self.isNil
}
}
Usage
var myValue: String?
if myValue.isNil {
// do something
}
if myValue.isSome {
// do something
}
One option that hasn't specifically been covered is using Swift's ignored value syntax:
if let _ = xyz {
// something that should only happen if xyz is not nil
}
I like this since checking for nil feels out of place in a modern language like Swift. I think the reason it feels out of place is that nil is basically a sentinel value. We've done away with sentinels pretty much everywhere else in modern programming so nil feels like it should go too.
Instead of if, ternary operator might come handy when you want to get a value based on whether something is nil:
func f(x: String?) -> String {
return x == nil ? "empty" : "non-empty"
}
Another approach besides using if or guard statements to do the optional binding is to extend Optional with:
extension Optional {
func ifValue(_ valueHandler: (Wrapped) -> Void) {
switch self {
case .some(let wrapped): valueHandler(wrapped)
default: break
}
}
}
ifValue receives a closure and calls it with the value as an argument when the optional is not nil. It is used this way:
var helloString: String? = "Hello, World!"
helloString.ifValue {
print($0) // prints "Hello, World!"
}
helloString = nil
helloString.ifValue {
print($0) // This code never runs
}
You should probably use an if or guard however as those are the most conventional (thus familiar) approaches used by Swift programmers.
Optional
Also you can use Nil-Coalescing Operator
The nil-coalescing operator (a ?? b) unwraps an optional a if it contains a value, or returns a default value b if a is nil. The expression a is always of an optional type. The expression b must match the type that is stored inside a.
let value = optionalValue ?? defaultValue
If optionalValue is nil, it automatically assigns value to defaultValue
Now you can do in swift the following thing which allows you to regain a little bit of the objective-c if nil else
if textfieldDate.text?.isEmpty ?? true {
}
var xyz : NSDictionary?
// case 1:
xyz = ["1":"one"]
// case 2: (empty dictionary)
xyz = NSDictionary()
// case 3: do nothing
if xyz { NSLog("xyz is not nil.") }
else { NSLog("xyz is nil.") }
This test worked as expected in all cases.
BTW, you do not need the brackets ().
If you have conditional and would like to unwrap and compare, how about taking advantage of the short-circuit evaluation of compound boolean expression as in
if xyz != nil && xyz! == "some non-nil value" {
}
Granted, this is not as readable as some of the other suggested posts, but gets the job done and somewhat succinct than the other suggested solutions.
If someone is also try to find to work with dictionaries and try to work with Optional(nil).
let example : [Int:Double?] = [2: 0.5]
let test = example[0]
You will end up with the type Double??.
To continue on your code, just use coalescing to get around it.
let example : [Int:Double?] = [2: 0.5]
let test = example[0] ?? nil
Now you just have Double?
This is totally logical, but I searched the wrong thing, maybe it helps someone else.
Since Swift 5.7:
if let xyz {
// Do something using `xyz` (`xyz` is not optional here)
} else {
// `xyz` was nil
}