Related
I am using Vapor for Swift backend. Following is the code i am working with.
drop.post("postTodo") { request in
var jsonContent: JSON?
if let contentType = request.headers["Content-Type"], contentType.contains("application/json"), let jsonData = request.json {
jsonContent = jsonData
print("Got JSON: \(jsonContent)")
}
guard let id = jsonContent?.node.object?["id"]?.string
else {
return JSON(["message": "Please include mandatory parameters"])
}
let tempId = Int(id)!
I am getting "id" as optional string for eg: Optional("123") for jsonContent?.node.object?["id"]?.string
When I try to convert it to int using Int(id)! i get back nil
If i try to do let tempId = Int(id!) it gives error.
But when i do the same thing in playground i get proper int value.
let id: String?
id = "1234"
let myInt = Int(id!)
Why Optional string to Int is not working properly in my Vapor app ?
Any idea.
If "id" is an optional string, then you probably don't want to be force unwrapping it with the "!".
The safest approach would be something like:
if let id = id
{
let myIdAsInt = Int(id)
}
The reason it "works" in the playground, is you are definitely assigning a non-nil value to the string (therefore you get away with the force unwrap).
String!might contain a string, or it might contain nil. It’s like a regular optional, but Swift lets you access the value directly without the unwrapping safety. If you try to do it, it means you know there’s a value there – but if you’re wrong your app will crash.
var optionalString: String? = "123"
// first check if it doesn't contain nil
if let str = optionalString {
// string to -> Int
if let id = Int(str) {
print(id) // work with id
}
} else {
// optionalString contains nil
}
what i found is in my iOS code i had a struct with optional properties coz of which when mapped to Dict gave object with optional values to keys.
If I make properties non optional and send it to vapor backend after it works fine.
So basically it was the case of using Optionals properly.
Quite often, you need to write code such as the following:
if someOptional != nil {
// do something with the unwrapped someOptional e.g.
someFunction(someOptional!)
}
This seems a bit verbose, and also I hear that using the ! force unwrap operator can be unsafe and best avoided. Is there a better way to handle this?
It is almost always unnecessary to check if an optional is not nil. Pretty much the only time you need to do this is if its nil-ness is the only thing you want to know about – you don’t care what’s in the value, just that it’s not nil.
Under most other circumstances, there is a bit of Swift shorthand that can more safely and concisely do the task inside the if for you.
Using the value if it isn’t nil
Instead of:
let s = "1"
let i = Int(s)
if i != nil {
print(i! + 1)
}
you can use if let:
if let i = Int(s) {
print(i + 1)
}
You can also use var:
if var i = Int(s) {
print(++i) // prints 2
}
but note that i will be a local copy - any changes to i will not affect the value inside the original optional.
You can unwrap multiple optionals within a single if let, and later ones can depend on earlier ones:
if let url = NSURL(string: urlString),
data = NSData(contentsOfURL: url),
image = UIImage(data: data)
{
let view = UIImageView(image: image)
// etc.
}
You can also add where clauses to the unwrapped values:
if let url = NSURL(string: urlString) where url.pathExtension == "png",
let data = NSData(contentsOfURL: url), image = UIImage(data: data)
{ etc. }
Replacing nil with a default
Instead of:
let j: Int
if i != nil {
j = i
}
else {
j = 0
}
or:
let j = i != nil ? i! : 0
you can use the nil-coalescing operator, ??:
// j will be the unwrapped value of i,
// or 0 if i is nil
let j = i ?? 0
Equating an optional with a non-optional
Instead of:
if i != nil && i! == 2 {
print("i is two and not nil")
}
you can check if optionals are equal to non-optional values:
if i == 2 {
print("i is two and not nil")
}
This also works with comparisons:
if i < 5 { }
nil is always equal to other nils, and is less than any non-nil value.
Be careful! There can be gotchas here:
let a: Any = "hello"
let b: Any = "goodbye"
if (a as? Double) == (b as? Double) {
print("these will be equal because both nil...")
}
Calling a method (or reading a property) on an optional
Instead of:
let j: Int
if i != nil {
j = i.successor()
}
else {
// no reasonable action to take at this point
fatalError("no idea what to do now...")
}
you can use optional chaining, ?.:
let j = i?.successor()
Note, j will also now be optional, to account for the fatalError scenario. Later, you can use one of the other techniques in this answer to handle j’s optionality, but you can often defer actually unwrapping your optionals until much later, or sometimes not at all.
As the name implies, you can chain them, so you can write:
let j = s.toInt()?.successor()?.successor()
Optional chaining also works with subscripts:
let dictOfArrays: ["nine": [0,1,2,3,4,5,6,7]]
let sevenOfNine = dictOfArrays["nine"]?[7] // returns {Some 7}
and functions:
let dictOfFuncs: [String:(Int,Int)->Int] = [
"add":(+),
"subtract":(-)
]
dictOfFuncs["add"]?(1,1) // returns {Some 2}
Assigning to a property on an optional
Instead of:
if splitViewController != nil {
splitViewController!.delegate = self
}
you can assign through an optional chain:
splitViewController?.delegate = self
Only if splitViewController is non-nil will the assignment happen.
Using the value if it isn’t nil, or bailing (new in Swift 2.0)
Sometimes in a function, there’s a short bit of code you want to write to check an optional, and if it’s nil, exit the function early, otherwise keep going.
You might write this like this:
func f(s: String) {
let i = Int(s)
if i == nil { fatalError("Input must be a number") }
print(i! + 1)
}
or to avoid the force unwrap, like this:
func f(s: String) {
if let i = Int(s) {
print(i! + 1)
}
else {
fatalErrr("Input must be a number")
}
}
but it’s much nicer to keep the error-handling code at the top by the check. This can also lead to unpleasant nesting (the "pyramid of doom").
Instead you can use guard, which is like an if not let:
func f(s: String) {
guard let i = Int(s)
else { fatalError("Input must be a number") }
// i will be an non-optional Int
print(i+1)
}
The else part must exit the scope of the guarded value, e.g. a return or fatalError, to guarantee that the guarded value will be valid for the remainder of the scope.
guard isn’t limited to function scope. For example the following:
var a = ["0","1","foo","2"]
while !a.isEmpty {
guard let i = Int(a.removeLast())
else { continue }
print(i+1, appendNewline: false)
}
prints 321.
Looping over non-nil items in a sequence (new in Swift 2.0)
If you have a sequence of optionals, you can use for case let _? to iterate over all the non-optional elements:
let a = ["0","1","foo","2"]
for case let i? in a.map({ Int($0)}) {
print(i+1, appendNewline: false)
}
prints 321. This is using the pattern-matching syntax for an optional, which is a variable name followed by ?.
You can also use this pattern matching in switch statements:
func add(i: Int?, _ j: Int?) -> Int? {
switch (i,j) {
case (nil,nil), (_?,nil), (nil,_?):
return nil
case let (x?,y?):
return x + y
}
}
add(1,2) // 3
add(nil, 1) // nil
Looping until a function returns nil
Much like if let, you can also write while let and loop until nil:
while let line = readLine() {
print(line)
}
You can also write while var (similar caveats to if var apply).
where clauses also work here (and terminate the loop, rather than skipping):
while let line = readLine()
where !line.isEmpty {
print(line)
}
Passing an optional into a function that takes a non-optional and returns a result
Instead of:
let j: Int
if i != nil {
j = abs(i!)
}
else {
// no reasonable action to take at this point
fatalError("no idea what to do now...")
}
you can use optional’s map operator:
let j = i.map { abs($0) }
This is very similar to optional chaining, but for when you need to pass the non-optional value into the function as an argument. As with optional chaining, the result will be optional.
This is nice when you want an optional anyway. For example, reduce1 is like reduce, but uses the first value as the seed, returning an optional in case the array is empty. You might write it like this (using the guard keyword from earlier):
extension Array {
func reduce1(combine: (T,T)->T)->T? {
guard let head = self.first
else { return nil }
return dropFirst(self).reduce(head, combine: combine)
}
}
[1,2,3].reduce1(+) // returns 6
But instead you could map the .first property, and return that:
extension Array {
func reduce1(combine: (T,T)->T)->T? {
return self.first.map {
dropFirst(self).reduce($0, combine: combine)
}
}
}
Passing an optional into a function that takes an optional and returns a result, avoiding annoying double-optionals
Sometimes, you want something similar to map, but the function you want to call itself returns an optional. For example:
// an array of arrays
let arr = [[1,2,3],[4,5,6]]
// .first returns an optional of the first element of the array
// (optional because the array could be empty, in which case it's nil)
let fst = arr.first // fst is now [Int]?, an optional array of ints
// now, if we want to find the index of the value 2, we could use map and find
let idx = fst.map { find($0, 2) }
But now idx is of type Int??, a double-optional. Instead, you can use flatMap, which “flattens” the result into a single optional:
let idx = fst.flatMap { find($0, 2) }
// idx will be of type Int?
// and not Int?? unlike if `map` was used
I think you should go back to the Swift programming book and learn what these things are for. ! is used when you are absolutely sure that the optional isn't nil. Since you declared that you are absolutely sure, it crashes if you're wrong. Which is entirely intentional. It is "unsafe and best avoided" in the sense that asserts in your code are "unsafe and best avoided". For example:
if someOptional != nil {
someFunction(someOptional!)
}
The ! is absolutely safe. Unless there is a big blunder in your code, like writing by mistake (I hope you spot the bug)
if someOptional != nil {
someFunction(SomeOptional!)
}
in which case your app may crash, you investigate why it crashes, and you fix the bug - which is exactly what the crash is there for. One goal in Swift is that obviously your app should work correctly, but since Swift cannot enforce this, it enforces that your app either works correctly or crashes if possible, so bugs get removed before the app ships.
You there is one way. It is called Optional Chaining. From documentation:
Optional chaining is a process for querying and calling properties,
methods, and subscripts on an optional that might currently be nil. If
the optional contains a value, the property, method, or subscript call
succeeds; if the optional is nil, the property, method, or subscript
call returns nil. Multiple queries can be chained together, and the
entire chain fails gracefully if any link in the chain is nil.
Here is some example
class Person {
var residence: Residence?
}
class Residence {
var numberOfRooms = 1
}
let john = Person()
if let roomCount = john.residence?.numberOfRooms {
println("John's residence has \(roomCount) room(s).")
} else {
println("Unable to retrieve the number of rooms.")
}
// prints "Unable to retrieve the number of rooms."
You can check the full article here.
We can use optional binding.
var x:Int?
if let y = x {
// x was not nil, and its value is now stored in y
}
else {
// x was nil
}
After lot of thinking and researching i have came up with the easiest way to unwrap an optional :
Create a new Swift File and name it UnwrapOperator.swift
Paste the following code in the file :
import Foundation
import UIKit
protocol OptionalType { init() }
extension String: OptionalType {}
extension Int: OptionalType {}
extension Int64: OptionalType {}
extension Float: OptionalType {}
extension Double: OptionalType {}
extension CGFloat: OptionalType {}
extension Bool: OptionalType {}
extension UIImage : OptionalType {}
extension IndexPath : OptionalType {}
extension NSNumber : OptionalType {}
extension Date : OptionalType {}
extension UIViewController : OptionalType {}
postfix operator *?
postfix func *?<T: OptionalType>( lhs: T?) -> T {
guard let validLhs = lhs else { return T() }
return validLhs
}
prefix operator /
prefix func /<T: OptionalType>( rhs: T?) -> T {
guard let validRhs = rhs else { return T() }
return validRhs
}
Now the above code has created 2 operator [One prefix and one postfix].
At the time of unwrapping you can use either of these operator before or after the optionals
The explanation is simple, the operators returns the constructor value if they get nil in variable else the contained value inside the variable.
Below is the example of usage :
var a_optional : String? = "abc"
var b_optional : Int? = 123
// before the usage of Operators
print(a_optional) --> Optional("abc")
print(b_optional) --> Optional(123)
// Prefix Operator Usage
print(/a_optional) --> "abc"
print(/b_optional) --> 123
// Postfix Operator Usage
print(a_optional*?) --> "abc"
print(b_optional*?) --> 123
Below is the example when variable contains nil :
var a_optional : String? = nil
var b_optional : Int? = nil
// before the usage of Operators
print(a_optional) --> nil
print(b_optional) --> nil
// Prefix Operator Usage
print(/a_optional) --> ""
print(/b_optional) --> 0
// Postfix Operator Usage
print(a_optional*?) --> ""
print(b_optional*?) --> 0
Now it is your choice which operator you use, both serve the same purpose.
I am building a game in Xcode and I'm storing the details for the level in text files e.g. Level1.txt, Level2.txt etc.
I read in the data from a text file and store it in a Dictionary.
When I try to assign the values from the dictionary to the global variables, it doesn't work.
Text File Contents (Level1.txt)
LevelNum:1
weaponPickupRate:10.0
weaponPickupAmount:50.0
monsterMinSpeed:10.0
monsterMaxSpeed:15.0
monsterRate:1.0
totalMonsters:10.0
goldPerMonster:10
Global Variables
var settings = [String: Any]()
var monsterMaxSpeed = 0.0
Function For Obtaining Level Details
func GenerateLevel(levelNumber: Int) {
fileName = "level\(levelNumber).txt"
levelPath = "\(NSBundle.mainBundle().resourcePath!)/\(fileName)"
var err: NSError? = NSError()
let s = String(contentsOfFile: levelPath, encoding: NSUTF8StringEncoding, error: &err)
if let content = s {
var array = content.componentsSeparatedByString("\n")
for a in array {
var v = a.componentsSeparatedByString(":")
settings[v[0]] = v[1]
}
}
println(settings) // A
var e = settings["monsterMaxSpeed"]
println(e) // B
monsterMaxSpeed = settings["monsterMaxSpeed"] // C
}
Println(setting) (A) - prints:
[monsterRate: 1.0, monsterMinSpeed: 10.0, weaponPickupRate: 10.0, weaponPickupAmount: 50.0, goldPerMonster: 10, totalMonsters: 10.0, LevelNum: 1, monsterMaxSpeed: 15.0]
Println(e) (B) prints:
Optional("15.0")
This Line Does not work
it shows up an error and doesn't allow me to build my project. The Error given is:
'(String, Any)' is not convertible to 'Double'
monsterMaxSpeed = settings["monsterMaxSpeed"]
Please can someone help and advise me what I need to do?
Thanks,
Ryann
You have two problems here.
First, fetching from a [String:Any] dictionary by key does not return an Any. It returns an Any? i.e. an optional that may or may not contain an Any. This is because that key may not be present in the dictionary.
You need to test if the value is non-nil and unwrap the value if it is:
if let speed = settings["monsterMaxSpeed"] {
monsterMaxSpeed = speed
}
else {
// handle there being no speed setting in your file
// by reporting an error or similar
}
Or, if you’re happy with just using a default, you can use the nil coalescing operator:
// if the key is present, us the unwrapped value, if not use 0.0
monsterMaxSpeed = settings["monsterMaxSpeed"] ?? 0.0
Second, you’ve declared monsterMaxSpeed as a Double not an Any. So once you resolve your optional unwrapping problem you’ll get a second problem. You need to convert the Any to a Double using as?. The ? in as? is important – if the value is not a double (suppose there was a rogue character in the entry in your file), you will get a nil back. Again, you would need to test for this and handle the error.
Happily, you can do this all in one go:
monsterMaxSpeed = (settings["monsterMaxSpeed"] as? Double) ?? 0.0
(it’s probably the confluence of the two of these issues that’s causing you to get a particularly enigmatic error – the error relates to the other version of Dictionary.subscript which takes an index, not a key, and returns a key/value pair, which isn’t optional, because indices should only address entries that are definitely in the dictionary)
I am doing what I believe to be a very simple task. I'm trying to get a value out of a dictionary if the key exists. I am doing this for a couple keys in the dictionary and then creating an object if they all exist (basically decoding a JSON object). I am new to the language but this seems to me like it should work, yet doesn't:
class func fromDict(d: [String : AnyObject]!) -> Todo? {
let title = d["title"]? as? String
// etc...
}
It gives me the error: Operand of postfix ? should have optional type; type is (String, AnyObject)
HOWEVER, if I do this, it works:
class func fromDict(d: [String : AnyObject]!) -> Todo? {
let maybeTitle = d["title"]?
let title = maybeTitle as? String
// etc...
}
It appears to be basic substitution but I may be missing some nuance of the language. Could anyone shed some light on this?
The recommended pattern is
if let maybeTitle = d["title"] as? String {
// do something with maybeTitle
}
else {
// abort object creation
}
It is possibly really a question of nuance. The form array[subscript]? is ambiguous because it could mean that the whole dictionary (<String:AnyObject>) is optional while you probably mean the result (String). In the above pattern, you leverage the fact that Dictionary is designed to assume that accessing some key results in an optional type.
After experimenting, and noticing that the ? after as is just as ambiguous, more, here is my solution:
var dictionary = ["one":"1", "two":"2"]
// or var dictionary = ["one":1, "two":2]
var message = ""
if let three = dictionary["three"] as Any? {
message = "\(three)"
}
else {
message = "No three available."
}
message // "No three available."
This would work with all non-object Swift objects, including Swift Strings, numbers etc. Thanks to Viktor for reminding me that String is not an object in Swift. +
If you know the type of the values you can substitute Any? with the appropriate optional type, like String?
There are a few of things going on here.
1) The ? in d["title"]? is not correct usage. If you're trying to unwrap d["title"] then use a ! but be careful because this will crash if title is not a valid key in your dictionary. (The ? is used for optional chaining like if you were trying to call a method on an optional variable or access a property. In that case, the access would just do nothing if the optional were nil). It doesn't appear that you're trying to unwrap d["title"] so leave off the ?. A dictionary access always returns an optional value because the key might not exist.
2) If you were to fix that:
let maybeTitle = d["title"] as? String
The error message changes to: error: '(String, AnyObject)' is not convertible to 'String'
The problem here is that a String is not an object. You need to cast to NSString.
let maybeTitle = d["title"] as? NSString
This will result in maybeTitle being an NSString?. If d["title"] doesn't exist or if the type is really NSNumber instead of NSString, then the optional will have a value of nil but the app won't crash.
3) Your statement:
let title = maybeTitle as? String
does not unwrap the optional variable as you would like. The correct form is:
if let title = maybeTitle as? String {
// title is unwrapped and now has type String
}
So putting that all together:
if let title = d["title"] as? NSString {
// If we get here we know "title" is a valid key in the dictionary, and
// we got the type right. title has now been unwrapped and is ready to use
}
title will have the type NSString which is what is stored in the dictionary since it holds objects. You can do most everything with NSString that you can do with String, but if you need title to be a String you can do this:
if var title:String = d["title"] as? NSString {
title += " by Poe"
}
and if your dictionary has NSNumbers as well:
if var age:Int = d["age"] as? NSNumber {
age += 1
}
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
}