How to check which class an item is in Swift 2? - ios

I have a function that is passed an item, the item could be either one of two object classes. I'm trying to test the item for a variable that only exists in one of the classes to identify which of the two classes it is. I'm not sure what the correct way to do this is. I'm trying:
let theItem = item as! ObjectOne
if let objectOneUniqueVariable = theItem.uniqueVariable {
print("It's an instance of ObjectOne")
//Do more things
} else {
theItem = item as! ObjectTwo
print("It's an instance of ObjectTwo")
//Do more things
}
The uniqueVariable is a sting - I'm currently getting the error:
Initialiser for conditional binding must have Optional type, not 'String'
I can add what ever I need to Object One in order for this test to work. Any ideas?

Swift has an elegant solution: Switch on the class type.
switch item {
case let object1 as ObjectOne:
print("It's an instance of ObjectOne")
//Do more things
case let object2 as ObjectTwo:
print("It's an instance of ObjectTwo")
//Do more things
default:
println("something else")
}
In the minimal case if you only cared about getting the value if it's the type of object1, a minimalist way of getting an optional is:
let optionalValue = (item as? ObjectOne)?.theMember
You could then unwrap it, if/when you need it.

Related

why I have to unwrap value before I use

A block is defined like below
// Declare block ( optional )
typealias sorting = (([Schedule], [String]) -> [Schedule])?
var sortSchedule: sorting = { (schedules, sortDescription) in
var array = [Schedule]()
for string in sortDescription
{
for (index, schedule) in schedules.enumerate()
{
if string == schedule.startTime
{
array.append(schedule)
break
}
}
}
return array
}
At some points, I am invoking a block by doing
let allSchedules = sortSchedule?(result, sortDescription())
for schedule in allSchedules // Xcode complains at here
{
..........
}
Im using ? because I want to make sure that if the block exists, then do something. However, Xcode complains for the for loop
value of optional type [Schedule]? not upwrapped, did you mean to use '!' or '?'?
Im not sure why because the return type of a block is an array which can have 0 or more than one items.
Does anyone know why xcode is complaining.
You are use ? in line let allSchedules = sortSchedule?(result, sortDescription()) not "for sure that if the block exists", but just for note, that you understand that it can be nil. Behind scene allSchedules have type Array<Schedule>?. And you can not use for in cycle for nil. You better use optional binding:
if let allSchedules = sortSchedule?(result, sortDescription())
{
for schedule in allSchedules
{
//..........
}
}

How do I find the index of an item in an array of [Any]()?

var items = [Any]()
I want to find the index of an object User (in this case, a class).
items.indexOf(myUserInstance) throws an error (Cannot convert value of type 'User' to expected argument type '#noescape(Any) throws -> Bool)
items.indexOf(myUserInstance as? Any) throws an error too.
Is there a simple extension that will allow me to use indexOf? Right now, I'm looping every item, type-casting, and checking if the ID property equals...
can try something like
items.indexOf { x in
if let u = x as? User {
if(u === user){
return true;
}
}
return false
}
might be a more elegant way of writing this, but thats all i got so far
How about items.indexOf { $0 as? User === user } ?
This will return an optional containing either the first index the User object is within the array, or nil (.None) if the user is not within the collection.
If your User class is deriver from NSObject and it overrides the isEqual: method, then you can use the == operator instead of === to also find objects that are equal in contents; == will call the isEqual: method behind the scenes instead of comparing the pointers, as oposed to ===.
// we can retrive index like this
(items as! NSArray).indexOfObject(user)

How can I use the OR operator with optional binding?

I have a piece of code that I would like to execute if a variable can be casted as one of two types.
if let myOptional = variableToCast as! firstTypeToTryToCastAs ||
let myOptional = variableToCast as! secondTypeToTryToCastAs {
//Execute some code
}
However, Swift (as of version 2.0) apparently does not allow this. I'm looking for a way to do this without having to create two separate if blocks.
My code only uses the super type, therefore the code dealing with both types are the same. However, I can't cast it to the super type because I don't want to code to execute if variableToCast is one of the many other possible types that are also derived from the same super type.
Based on your comment:
My code only uses the super type, therefore the code dealing with both
types are the same. However, I can't cast it to the super type because
I don't want to code to execute if variableToCast is one of the many
other possible types that are also derived from the same super type.
In your optional binding if let cast to the super type and then limit it to the classes you are interested in with a where clause:
class SuperType {
}
class firstTypeToTryToCastAs: SuperType {
}
class secondTypeToTryToCastAs: SuperType {
}
class thirdTypeToTryToCastAs: SuperType {
}
var variableToCast: AnyObject = firstTypeToTryToCastAs()
if let myOptional = variableToCast as? SuperType where myOptional is firstTypeToTryToCastAs || myOptional is secondTypeToTryToCastAs {
print("this works")
} else {
print("not the type we are looking for")
}
I'm looking for a way to do this without having to create two separate if blocks
Why?
You can't use the same code in the if block anyway because myOptional will be of a different type. I would go with
if let myOptional = variableToCast as! firstTypeToTryToCastAs
{
// code dealing with a firstTypeToTryToCastAs
}
else if let myOptional = variableToCast as! secondTypeToTryToCastAs
{
// code dealing with a secondTypeToTryToCastAs
}
On the other hand if your two types have a common super type, cast it to that super type.
To add onto JeremyP's answer, if you don't actually care what the value of myOptional is, but just want to execute some code if it is one of those two types, you could do this instead
if variableToCast is firstTypeToTryToCastAs || variableToCast is secondTypeToTryToCastAs {
//Code for both cases
}

Swift if let evaluates successfully on Optional(nil)

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).

Swift: Testing optionals for 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
}

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