I'm trying to ON meter with AVAudioRecorder, but it give error as
Cann't assign to the result of expression
Declaration:
var recorder : AVAudioRecorder?
..
Definition
self.recorder?.meteringEnabled = true // Error here
You need to unwrap the self.recorder variable like so:
self.recorder!.meteringEnabled = true
Optional chaining does not support the setting of property values. It is only used for querying (see: getting) properties, and calling methods.
So code should be
if let recorder = self.recorder?{
recorder.meteringEnabled = true
}
This will only set the properties on the object if it has a value. Since your ivar is most likely defined as var player: AVAudioPlayer?, the compiler doesn't actually know whether it will have a value.
As mentioned in the language guide, under the Calling Properties Through Optional Chaining section:
You cannot, however, set a property’s value through optional chaining.
Related
GKTurnBasedParticipant has a property: player, which has class GKPlayer. The declaration is written as:
var player: GKPlayer { get }
Yet the API documentation says,
The value of this property may be nil if this slot in the match has
not yet been filled by an actual player.
So if it can be nil, why isn't the declaration:
var player: GKPlayer?
What am I not understanding here? Is it something with the getter? This is actually important because the player should be nil for automatched games where a second player has yet to join.
When I do conditional unwrapping (after migrating to Swift 4.2)
if let onlineGKPlayer = participant.player {
this now is a compiler error:
Initializer for conditional binding must have Optional type, not 'GKPlayer'
https://developer.apple.com/documentation/gamekit/gkturnbasedparticipant/1521037-player
Not sure when it was updated, but the API documentation now lists it as an Optional:
var player: GKPlayer? { get }
It seems you had the right idea
I have a NSObject Subclass. Say CityWalks
class CityWalks{
var totalCount:Int?
}
How do I use this property further? Should I check the nil coalescing every time this value is accessed.
example:
let aObject =
say in one fucntion (function1()) , I need to access this value, then it would like (aObject!.totalCount ?? 0)
func function1(){
...Some Access code for the object....
(aObject!.totalCount ?? 0)
}
Similarly in every other function(function2()) , I will have to write the same code.
func function2(){
...Some Access code for the object....
(aObject!.totalCount ?? 0)
}
So, what could be a better approach for such field, considering this property might receive a value from server or might not.
If you have a default value for this property just assign this value as default value.
class YourClass {
var totalCount = 0
}
I'd recommend you avoid using an optional value if it's possible. Because optional values its a first place when you can get an error.
As stated in the comments and the other answer using an optional is not really optimal in your case. It seems like you might as well use a default value of 0.
However, to clarify, you have to check the value when unwrapping the optional.
Sometimes it's possible to pass an optional to UIElement etc and then you don't really need to do anything with them
There are pretty ways of checking for nil in optional values built into swift so you can build pretty neat code even though you work with optional.
Look in to guard let and if let if you want to know more about unwrapping values safely.
if let
if let totalWalks = aObject?.totalCount {
//operate on totalWalks
}
guard
guard let totalWalks = aObject?.totalCount else { return }
//operate on totalWalks
There are also cases where you will want to call a function on an optional value and in this case you can do so with ?
aObject?.doSomething()
Any return values this function might have will now be wrapped in an optional and you might have to unwrap them as well with an if let or guard
When working with optionals you should try to avoid forcing the unwrap with ! as even though you at the moment know that the value is not null that might after a change in the code not be true anymore.
In The Swift Programming Language (book of Apple) I've read that you can create optional variables in 2 ways: using a question mark (?) or by using an exclamation mark (!).
The difference is that when you get the value of an optional with (?) you have to use an exclamation mark every time you want the value:
var str: String? = "Question mark?"
println(str!) // Exclamation mark needed
str = nil
While with an (!) you can get it without a suffix:
var str: String! = "Exclamation mark!"
println(str) // No suffix needed
str = nil
What is the difference and why are there 2 ways if there is no difference at all?
The real benefit to using implicitly unwrapped optionals (declared with the !) is related to class initialisation when two classes point to each other and you need to avoid a strong-reference cycle. For example:
Class A <-> Class B
Class A's init routine needs to create (and own) class B, and B needs a weak reference back to A:
class A {
let instanceOfB: B!
init() {
self.instanceOfB = B(instanceOfA: self)
}
}
class B {
unowned let instanceOfA: A
init(instanceOfA: A) {
self.instanceOfA = instanceOfA
}
}
Now,
Class B needs a reference to class A to be initialised.
Class A can only pass self to class B's initialiser once it's fully initialised.
For Class A to be considered as initialised before Class B is created, the property instanceOfB must therefore be optional.
However, once A's been created it would be annoying to have to access instanceOfB using instanceOfB! since we know that there has to be a B
To avoid this, instanceOfB is declared as an implicity unwrapped optional (instanceOfB!), and we can access it using just instanceOfB. (Furthermore, I suspect that the compiler can optimise the access differently too).
An example of this is given on pages 464 to 466 of the book.
Summary:
Use ? if the value can become nil in the future, so that you test for this.
Use ! if it really shouldn't become nil in the future, but it needs to be nil initially.
You should go beyond the syntactic sugar.
There are two completely different polymorphic types. The syntactic sugar just uses either one or the other of these types.
When you write Foo? as a type you really have Optional<Foo>, while when you write Foo! you really have ImplicitlyUnwrappedOptional<Foo>.
These are two different types, and they are different from Foo as well.
The values you create with ? are plain optional values as you mentioned, you should access it via optional binding (if let unwrappedValue = myOptionalValue) or by using the exclamation point syntax myOptionalValue!.doSomething().
The values you create with ! are called implicitly unwrapped optionals. With them, you don't need to manually unwrap before using them. When you do val myOptionalValue!.doSomething().
The value will be automatically unwrapped for you when you use myOptionalValue directly, be careful with this though, because accessing an implicitly unwrapped value when there is actually no value in it (when it is nil) will result in a runtime error.
? (Optional) indicates your variable may contain a nil value while ! (unwrapper) indicates your variable must have a memory (or value) when it is used (tried to get a value from it) at runtime.
The main difference is that optional chaining fails gracefully when the optional is nil, whereas forced unwrapping triggers a runtime error when the optional is nil.
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 nonoptional value. You can use this optional return value to check whether the optional chaining call was successful (the returned optional contains a value), or did not succeed due to a nil value in the chain (the returned optional value is nil).
Specifically, the result of an optional chaining call is of the same type as the expected return value, but wrapped in an optional. A property that normally returns an Int will return an Int? when accessed through optional chaining.
var defaultNil : String? // declared variable with default nil value
println(defaultNil) >> nil
var canBeNil : String? = "test"
println(canBeNil) >> optional(test)
canBeNil = nil
println(canBeNil) >> nil
println(canBeNil!) >> // Here nil optional variable is being unwrapped using ! mark (symbol), that will show runtime error. Because a nil optional is being tried to get value using unwrapper
var canNotBeNil : String! = "test"
print(canNotBeNil) >> "test"
var cantBeNil : String = "test"
cantBeNil = nil // can't do this as it's not optional and show a compile time error
For more detail, refer a document by Apple Developer Commitee, in detail
The String! kind is called an implicitly unwrapped optional:
Sometimes it’s clear from a program’s structure that an optional will always have a value, after that value is first set. In these cases, it’s useful to remove the need to check and unwrap the optional’s value every time it’s accessed, because it can be safely assumed to have a value all of the time.
These kinds of optionals are defined as implicitly unwrapped optionals. You write an implicitly unwrapped optional by placing an exclamation mark (String!) rather than a question mark (String?) after the type that you want to make optional.
in the optional chaining section you find the answer:
example class:
class Person {
var residence: Residence?
}
class Residence {
var numberOfRooms = 1
}
If you try to access the numberOfRooms property of this person’s residence, by placing an exclamation mark after residence to force the unwrapping of its value, you trigger a runtime error, because there is no residence value to unwrap:
let roomCount = john.residence!.numberOfRooms
// this triggers a runtime error
The code above succeeds when john.residence has a non-nil value and will set roomCount to an Int value containing the appropriate number of rooms. However, this code always triggers a runtime error when residence is nil, as illustrated above.
Optional chaining provides an alternative way to access the value of numberOfRooms. To use optional chaining, use a question mark in place of the exclamation mark:
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."
Well mentioned by #tarmes above. Noticed another usage of implicit optional:
Lets say I have an optional Int:
let firstInt: Int? = 9
And I'm trying to use optional pattern matching and use this optional Int like this:
if case let myFirstInt? = firstInt where myFirstInt > 1 {
print("Valid")
} else {
print("Invalid")
}
Notice that I'm using implicit optional with local parameter myFirstInt making it safe for nil condition linked with optional firstInt. If now, I make firstInt as nil, it will execute else condition. If, instead, I use force-unwrap with firstInt that would lead to crash, something like this:
As I read the swift guide it says: The value of a constant doesn’t need to be known at compile time, but you must assign it a value exactly once.
I tried in REPL but without any luck:
let aConst;
aConst=23;
So how to declare a constant without setting initial value?
Example
let myConstant = getSomeValueFromMethod()
This is what it means that the value doesn't have to be known at compile time...
You can't declare a constant, then assign it at the global scope. If you have to do something like this, use a variable instead. When a constant is declared at global scope, it must be initialized with a value.
Docs
You can't just declare a constant without assigning it some sort of value. The compiler needs to know that the constant will have some sort of value. Consider the following example where a constant "isTablet" is computed based on variables that aren't known until runtime.
let isTablet = {
if (UIDevice.currentDevice().userInterfaceIdiom == .Pad) {
return true
} else {
return false
}
}()
Another example beside Suthan's:
var someVar: NSString
...
someVar = "Some String"
let someUnknownConstant = someVar
I'm trying to set following property of AVAudioPlayer, but it gives an error.
Error:
Cannot assign result of this property
Code:
self.player = AVAudioPlayer(contentsOfURL: url, error: &error)
self.player?.numberOfLoops = 0 // Error here
self.player?.delegate = self // Error here
if(!self.player?.prepareToPlay()){... } // This seems to good.
Fully confused? Is there any workaround behind this?
Optional chaining does not support the setting of property values. It is only used for querying (see: getting) properties, and calling methods.
Your code should instead read:
if let myPlayer = self.player? {
myPlayer.numberOfLoops = 0
myPlayer.delegate = self
}
This will only set the properties on the object if it has a value. Since your ivar is most likely defined as var player: AVAudioPlayer?, the compiler doesn't actually know whether it will have a value.
As mentioned in the language guide, under the Calling Properties Through Optional Chaining section:
You cannot, however, set a property’s value through optional chaining.
What I do is declare the type of my class's player property as AVAudioPlayer!. I can do this because I happen to know that this class will always have an audio player. This means, in turn, that I don't have to use ? or ! ever again; I can just say self.player and everything just works.
If you use this approach and you are ever in doubt about whether you really have a player, simply say if self.player - it will be false if no player is there, and now you can avoid talking to self.player as if it were an actual AVAudioPlayer (which would crash your app).