I declare an implicitly unwrapped optional as:
var numberOfRows: Int!
and initialize it in init:
numberOfRows = 25
Later I need to decrement it by one so I write:
numberOfRows--
but this doesn't compile. The error message says the decrement operator can't be applied to an implicitly unwrapped optional. With a little experimentation I find that the following compiles without error:
numberOfRows!--
I would like to understand this. What is the explanation for what seems like the extra '!'?
Implicitly unwrapped optional is a type on its own, and is different from the type it that wraps. Some operators on optionals and implicitly unwrapped optionals are pre-defined for you out of the box by the language, but for the rest you have to define them yourself.
In this particular case an operator postfix func --(inout value: Int!) -> Int! is just not defined. If you want to use postfix -- operator on Int! just the same way you use it on Int then you will have to define one.
E.g. something like:
postfix func --<T: SignedIntegerType>(inout value: T!) -> T! {
guard let _value = value else { return nil }
value = _value - 1
return _value
}
If we look at what the optional type is, we will see that this is enum like:
enum Optional<T> {
case Some(T)
case None
}
And it can be Some Type like Int for example or None and in this case it's have nil value.
When you make this:
var numberOfRows: Int!
you directly is indicated by the ! that this is not Int type but this is the enum Optional<Int> type. At moment of creation it's will be Some<Int> if equal it value but with this ! you have that it is enum Optional<Int> and in some next moment it will be the None. That's why you have to use ! second time when make this:
numberOfRows!--
Your nomberOfRows value is Optional<Int> type and it may be Int or nil and you have to directly indicate that this is Int type to make -- action.
Related
Consider the following example where foo is explicitly defined as optional:
let foo: Int? = 10
let bar = foo.map { $0 * 2 }
This compiles and works as expected. Now consider a subsequent example using optional chaining:
let foo: [Int]? = [0, 1, 2]
let bar = foo?.count.map { $0 * 2 }
This fails to compile with the following message:
error: value of type 'Int' has no member 'map'
Why does the compiler see foo?.count as an Int and not an Int?? This defies the Swift Language Book:
To reflect the fact that optional chaining can be called on a nil value, the result of an optional chaining call is always an optional value, even if the property, method, or subscript you are querying returns a non-optional value.
This is an operator precedence issue. You need to add parentheses to change the order:
let bar = (foo?.count).map { $0 * 2 }
Consider the case of a Collection like String rather than an Int:
let foo: String? = "ABC"
let bar = foo?.uppercased().map(\.isLowercase) // [false, false, false]
In this case, it's sensible that the .map applies to the String rather than to the Optional. If the precedence were the other way, then you'd need a ?. at every step (or a lot of parentheses :D)
The type of count unwraps to Int if available. Int does not implement the function map. Optional chaining simply fails when it finds nil. The compiler is trying to tell you if it does not fail the operation map cannot happen. The result is bar which is indeed an optional, because if the operation fails( because something unwraps to nil) the entire chain must return nil.
I have a class where I have a property like this:
var residenceId: Int!
When building with Xcode 10 and Swift 4.2 there are no issues. However, after installing Xcode 11 Beta and converting to Swift 5 I get the following warning:
Implicitly unwrapped property 'residenceId' declared here
Later in the class I also have this function:
func jsonDictionary() -> [String : Any] {
return ["residenceId": residenceId]
}
Here I get the warning
Coercion of implicitly unwrappable value of type 'Int?' to 'Any' does not unwrap optional
Are we no longer allowed to use implicitly unwrapped optional?
EDIT:
After some more looking into it, I have come to believe that "Implicitly unwrapped property 'residenceId' declared here" is not actually a warning, but rather some info (it is in a grey label rather than usual yellow) to help me understand why I am getting the second warning.
And to clarify, my question is if we are no longer able to use the '!' symbol on properties to define an implicitly unwrapped property (obviously only if we are sure it will not be nil) in order to later avoid explicitly unwrapping it (and thus simplifying the code).
Since Swift 4, ImplicitlyUnwrappedOptional or ! as we knew it, became Optional.
Check:
let a: ImplicitlyUnwrappedOptional<Int> = 1
will spit out the error:
'ImplicitlyUnwrappedOptional' has been renamed to 'Optional'
So instead if we do:
let a: Int! = 1
print(type(of: a)) //will print "Optional<Int>"
It's still Optional<Int> but indicates to the compiler that it can be implicitly unwrapped.
Implicit Unwrapping is Part of a Declaration.
...
consider ! to be a synonym for ? with the addition that it adds a flag on the declaration letting the compiler know that the declared value can be implicitly unwrapped.
Ref: Reimplementation of Implicitly Unwrapped Optionals
Now getting to the main question:
If you do:
let a: Int! = 1
let b: Any = a
print(type(of: b)) //prints "Optional<Int>"
It will give the following warning:
Expression implicitly coerced from 'Int?' to 'Any'
or as per Xcode 11
Coercion of implicitly unwrappable value of type 'Int?' to 'Any' does not unwrap optional
Note here that we tried to get a non-optional Any out of an Int? which means we were basically expecting an Int but just by specifying Any it won't also unwrap the Optional.
It will remain an Optional, and this is the meaning behind that warning.
Solutions:
To handle this warning gracefully, we can do any of the following:
let a: Int! = 1
let b: Any? = a
type(of: b) //Optional<Any>.Type
let c: Any! = a
type(of: c) //Optional<Any>.Type
let d: Any = a!
type(of: d) //Int.Type
EDIT: (based on comment)
! instead of ? have any practical difference for the programmer?
! tells the compiler that it can be implicitly unwrapped so it can help ease in the need for optional chaining.
Example:
With ?
class A {
var n: Int? = 1
}
class B {
var a: A? = A()
}
let b: B? = B()
print(b?.a?.n)
/*
but the following won't compile as compiler
will not implicitly unwrap optionals
print(b.a.n)
*/
With !
class A {
var n: Int! = 1
}
class B {
var a: A! = A()
}
let b: B! = B()
print(b.a.n) //compiler will implicitly unwrap `!` similar to print(b!.a!.n)
//also... you can still safely unwrap if you want
print(b?.a?.n)
b.a.n and b?.a?.n will both give an Optional<Int> at the end
Ofcourse if b or a is nil then b.a.n will crash because it's implicitly unwrapping b and a to get to n which is Optional<Int>
To get Int instead of Optional<Int>, you would do b.a.n! and so in your case you would do: print(residenceId!) to get Int
I hope I made sense
I have a custom operator ?= that is used to simply assign optional values to non-optionals.
var a: String
let b: String?
a = b ?? a
// is simplified to
a ?= b
This works well, but I don't want people to be able to use this operator when b is not optional. It's not going to cause any problems but it looks bad and I don't like it.
My function signature is currently:
func ?=<T> (originalValue: inout T, optionalNewValue: T?)
How could I make this cause a compiler warning or error when optionalNewValue is not actually an optional?
As other have said in the comments, the ?= operator might not be the best solution to your problems. Now if you're keen to that operator, or adds value to your code base, then there is a workaround that would allow you to reject non optional values on the right hand side of the operator.
You can use an OptionalCompatible protocol for the right hand side of the operator:
protocol OptionalCompatible {
associatedtype Wrapped
var optionalValue: Wrapped? { get }
}
extension Optional: OptionalCompatible {
var optionalValue: Wrapped? { return self }
}
func ?=<T, U> (originalValue: inout T, optionalNewValue: U) where U: OptionalCompatible, U.Wrapped == T {
originalValue = optionalNewValue.optionalValue ?? originalValue
}
Using a custom protocol will disable the auto-promotion of non-optionals to optionals, allowing only optionals to be used.
There is a caveat, though: in theory the conformance OptionalCompatible can be added to any type.
How could I make this cause a compiler warning or error when optionalNewValue is not actually an optional?
You cannot. It is a built-in fact of Swift that a nonOptional of type T can always be assigned or passed where an Optional<T> is expected.
I have the following variable:
var npill : String!
It's an Int value, but I can't set it as Int because of:
npillIn: fieldNumeroPillole.text!,
How can I convert this var to a Int var? I have tried the following:
var number1: Int = (npill! as NSString).intValue
By the above code I receive the following error:
cannot use instance member 'npill' within property initializer, property initializers run before "self" is aviable
If I then set:
var number1: Int = (self.npill! as NSString).intValue
The error it outputs is as follows:
Value of type '(NSObject) -> () -> Farmaco' has no member 'npill'
If anyone knows how I should be converting it properly, please let me know.
Update
Thank you to #Hamish for pointing out what the OP was asking
So the problem seems to be this
import Foundation
class Foo {
var npill : String!
var number1: Int = (npill! as NSString).intValue
}
error: cannot use instance member 'npill' within property initializer; property initializers run before 'self' is available
var number1: Int = (npill! as NSString).intValue
^
What's going on here?
You are using a property to populate another property, and this is not allowed.
Solution
However you can easily fix the problem postponing the initialisation of number1. Infact if you make number1 lazy, it will be populated only when used.
class Foo {
var npill : String!
lazy var number1: Int = { return Int(self.npill!)! }()
}
Warning: Of course this code will crash if npill is still nil when number1 is used.
Old version
You can simply write
let npill: String! = "34"
if let npill = npill, let num = Int(npill) {
print(num) // <-- here you have your Int
}
(As #Hamish pointed out in a comment below, I misunderstood what the OP was really asking about. I'll leave my answer, however, as some curiosa and insights regarding ! type annotation, which may be relevant for future readers of this question)
For any type of String optionals, their values needs to be unwrapped prior to using the failable init?(_ text: String) initializer or Int.
In your example, the variable npill is an optional, as you've annotated its type with the ! specifier (which should be used with care). Quoting from the implemented evolution proposal SE-0054 [emphasis mine]
Appending ! to the type of a Swift declaration will give it optional
type and annotate the declaration with an attribute stating that it
may be implicitly unwrapped when used.
Hence, it's entirely legal to use npill directly with the init?(_ text: String) initializer of Int, as it will be unwrapped (without any safety check for nil content!) on-the-fly upon use.
// UNSAFE example!
var npill: String! = "42"
if let npillInt = Int(npill) {
/* ^^^^^^^^ ^^^^^- since 'npill' has a type annotated with
| '!', it will be unsafely unwrapped at
| this point
\
the optional binding here safely unwraps the return from
the failable Int initializer, but has nothing to do with
the unwrapping of 'npill' */
print(npillInt) // 42
}
// why unsafe? consider
npill = nil
if let npillInt = Int(npill) { // runtime exception!
// ...
}
Generally you should avoid using the ! annotation, however, unless you are entirely certain that the content of the resulting optional variable will never ever be nil.
Leaving aside the cons of even using the ! annotation: you may implement a safe version of the unsafe example above, by overriding the unsafe implicit unwrapping with safe explicit unwrapping techniques. For a given optional variable declared using the ! annotation, we may still apply safe means to unwrap it, e.g. optional binding or using the nil coalescing operator. #appzYourLife has already showed one perfectly valid and safe way to handle the unwrapping and attempted type conversion of npill using optional binding, so I'll simply include another example using the nil coalescing operator instead:
// "safe" example (STILL: why use the `!` annotation?)
var npill: String! = "42"
if let npillInt = Int(npill ?? "x") {
/* ^^^^^ if 'npill' is 'nil', the Int initializer will
be given the value "x", which itself will lead
it to fail, which is safe here as we intend to
bind the result of the initialization to 'npillInt' */
print(npillInt) // 42
}
npill = nil
if let npillInt = Int(npill ?? "x") {
// ... doesnt enter
}
The consensus of the examples above is that if we're even slightly uncertain whether npill can ever be nil or not, we need to treat it as if it was just an optional not type annotated with ! (i.e. String?); overriding the default unsafe unwrapping with safe means when working with the variable. In such a case, why would we even want to use the ! typ annotation at all, as it only brings fragility/danger to our application?
In Swift 3, when one has optional function parameters, what is the difference between functions with:
func doThis(num: Int!)
and
func doThat(num: Int?)
Both declare that num is an optional Int.
If you do Int! it can be implicitly unwrapped inside of your function.
That means you can use it in places where a plain (non-optional) Int is required. In that case, it will fail if it happens to be nil.
With a "proper" Int? the compiler will not let you use num where an Int is required and forces you to include a check/guard first.
The Int! construct is mostly there for interoperability with Objective-C code where it is not clear if a reference type is optional or not. You should avoid it in new code.