I'm newbie in Xcode (Swift 3).
Why this code works:
let someUrl = URL(string: "https://www.apple.com")
But this one returns error:
let someConstant = "https://www.apple.com"
let someUrl = URL(string: someConstant)
Cannot use instance member 'someConstant' within property initializer; property initializers run before 'self' is available
the error is not exactly in your code and not with the object URL. but it depends where you placed the code.
you placed it at Place A where it becomes static variables. and thats why you can not access your someConstant at this place from an other instance variable. if you place your code snippet in a function methodName ( Place B) your code snippets work. for your case it is enough to place just let someUrl = URL(string: someConstant) at Place B
class a {
// Place A: instance constant here
func methodName(){
// Place B
}
}
see also: more results at stackoverflow
Related
So I am teaching myself Swift and I get optionals when I am declaring them, so for example:
var thisString:String?
I would have to either force unwrap thisString, or use
if let anotherString = thisString {
or use
guard if let another string = thisString else { return }
or nil coalesce it
let anotherString = thisString ?? "Something else"
But where I am getting hung up is there are times I create something that I don't think it an optional but the compiler does.
For example, why is the URL an optional here?
let myURL = URL(string: "https://itunes.apple.com/search?term=U2")
var myRequest = URLRequest(url: myURL!)
I didn't declare it as an optional and it clearly has a value. So why does the compiler see this as an optional? If I didn't force unwrap it I would get this error:
Value of optional type 'URL?' not unwrapped; did you mean to use '!' or '?'?
Is there some standard that I am missing? I've gone over both the docs and Swift Programming Book from Big Nerd and I still don't seem to get this part.
Thanks for the assist.
But where I am getting hung up is there are times I create something that I don't think it an optional but the compiler does.
But what you think is not what matters. What matters is what the API call you are making actually returns. You find that out by reading the docs.
So why does the compiler see this as an optional
Well, let's read the docs. You are calling URL(string:). That is the same as URL's init(string:). Look at the docs:
https://developer.apple.com/documentation/foundation/url/1779737-init
The declaration is
init?(string: String)
See the question mark? That means, "This call returns an Optional." So you should expect an Optional here.
The compiler can't determine if the url that you define in the string is valid or not.
Suppose instead of:
let myURL = URL(string: "https://itunes.apple.com/search?term=U2")
You miss typed the myURL definition as:
let myURL = URL(string: "https:/itunes.apple.com/search?term=U2")
The string contains a malformed URL, so the program would crash the moment you went to define myURL.
let myURL = URL(string: "https://itunes.apple.com/search?term=U2")
Here you are creating an url from string.That string might not be a valid string.All the strings are not valid url. So you are getting an optional because if that string can be turned in to a valid url then that url will be returned else nil will be returned. See the Apple documentation here.The initializer what you are using is a failable initializer itself.
init?(string: String)
#Keshav is correct, to get a better idea, hold the option button and click on the 'string' part of the init function for the URL class. You will see in the swift reference the init declaration is init?(string: String). This means that a optional is returned. Any function can return a optional, you can have func returnMyString(_ myString: String) -> String {} or func returnMyString(_ myString: String) -> String? {}. Both of those functions are pretty much the same, except the second one returns a optional.
URL has optional initializers. For example you can have
class A {
init?() {
return nil //You will get Optional<A>.none
}
}
A() === Optional<A>.none //true
which kind of implies that initialization failed. Such initializers wrap returned object into Optional. In Swift nil == Optional<Any>.none so you can speak of them interchangeably.
For example if you will attempt to construct a URL with something that is not an actual url, it will return nil.
let notAURL = URL(string: "{") //notAURL will be nil
On a side note: I believe optional initializers are a very poor design choice since they don't communicate anything about what went wrong inside the init. If init is fallible, it should throw. I don't understand why Swift designers allow optional initializers and I see why it births a lot of confusion.
I try to convert my code to swift 3 an I have spent hours on the following error:
Type 'Any' has no subscript members
Here's was my original code:
let data: AnyObject = user.object(forKey: "profilePicture")![0]
I looked at the answers here but I'm still stuck. (I do programming as a hobby, I'm not a pro :/)
I've try that:
let object = object.object(forKey: "profilePicture") as? NSDictionary
let data: AnyObject = object![0] as AnyObject
But now I get this error:
Variable used within its own initial value
Second issue: Use always a different variable name as the method name, basically use more descriptive names than object anyway.
First issue: Tell the compiler the type of the value for profilePicture, apparently an array.
if let profilePictures = user["profilePicture"] as? [[String:Any]], !profilePictures.isEmpty {
let data = profilePictures[0]
}
However, the array might contain Data objects, if so use
if let profilePictures = user["profilePicture"] as? [Data], !profilePictures.isEmpty {
let data = profilePictures[0]
}
Or – what the key implies – the value for profilePicture is a single object, who knows (but you ...)
And finally, as always, don't use NSArray / NSDictionary in Swift.
I just used XCode 8 and let it convert my existing project. Now I face the error that there is no init function for new URL without parameters.
class fileObj : NSObject, NSCopying, Comparable {
var URL = NSURL() // initial
...
The new code looks like:
class fileObj : NSObject, NSCopying, Comparable {
var myUrl = Foundation.URL() // initial
...
How should I init the new URL var?
it makes absolutely zero sense to do so, but currently (Swift3, Xcode Version 8.0 (8A218a)) it is working and gives you a totally blank URL object with no purpose at all, as you just asked for.
var myURL: URL = NSURLComponents().url!
Your observation is indeed correct, there is no empty initialiser because this would be an invalid URL, therefore they decided to disallow that.
What I recommend you doing is not initialising the variable at first and making it an optional (URL?). Later in the code, you'll be able to initialise it.
I was trying to do something like this (it is a contrived example for demonstration purposes only):
class Test {
let hello = "hello"
let world = "world"
let phrase: String {
return self.hello + self.world
}
}
but you can't use let for computed properties in Swift. Is there a way to do this without having to write an init() method? Thanks!
The reason let doesn't work on a read-only calculated property is because it's used to state that the property's actual value will never change after being set – not that the property is read-only. As the Apple docs say (emphasis mine):
You must declare computed properties — including read-only computed
properties — as variable properties with the var keyword, because their
value is not fixed. The let keyword is only used for constant
properties, to indicate that their values cannot be changed once they
are set as part of instance initialization.
You therefore need to use var in order to reflect the fact that a calculated property's value could change at any time, as you're creating it on the fly when accessing it. Although in your code, this can't happen – as your hello and world properties are let constants themselves. However, Swift is unable to infer this, so you still have to use var.
For example:
class Test {
let hello = "hello"
let world = "world"
var phrase: String {
return self.hello + self.world
}
}
(This doesn't change the readability of the property – as because you haven't provided it with a setter, it's still read-only)
However in your case, you might want to consider using a lazy property instead, as your hello and world properties are constants. A lazy property is created when it's first accessed, and keeps its value for the rest of its lifetime – meaning you won't have to keep on concatenating two constants together every time you access it.
For example:
class Test {
let hello = "hello"
let world = "world"
lazy var phrase: String = {
return self.hello + self.world
}()
}
Another characteristic of let properties is that their value should always be known before initialisation. Because the value of a lazy property might not be known before then, you also need to define it as a var.
If you're still adamant on wanting a let property for this, then as far as I can see, you have two options.
The first is the neatest (although you've said you don't want to do it) – you can assign your phrase property in the initialiser. As long as you do this before the super.init call, you don't have to deal with optionals. For example:
class Test {
let hello = "hello"
let world = "world"
let phrase: String
init() {
phrase = hello+world
}
}
You simply cannot do it inline, as self at that scope refers to the static class, not an instance of the class. Therefore you cannot access the instance members, and have to use init() or a lazy/calculated property.
The second option is pretty hacky – you can mirror your hello and world properties at class level, so you can therefore access them inline in your phrase declaration. For example:
class Test {
static let hello = "hello"
static let world = "world"
// for some reason, Swift has trouble inferring the type
// of the static mirrored versions of these properties
let hello:String = Test.hello
let world:String = Test.world
let phrase = hello+world
}
If you don't actually need your hello or world properties as instance properties, then you can just make them static – which will solve your problem.
Yes to make it work as computed properties, replace let to var.
Like,
class Test {
let hello = "hello"
let world = "world"
var phrase: String {
return self.hello + self.world
}
}
This way you can use it without init()
My class has a property of type NSURL that is initialized from a string. The string is known at compile time.
For the class to operate appropriately, it must be set to its intended value at initialization (not later), so there is no point in defining it as an optional (implicitly unwrapped or otherwise):
class TestClass: NSObject {
private let myURL:NSURL
...
Assuming that NSURL(string:) (which returns NSURL?) will never fail if passed a valid URL string that is known at compile time, I can do something like this:
override init() {
myURL = NSURL(string: "http://www.google.com")!
super.init()
}
However, I somehow don't feel comfortable around the forced unwrapping and would like to guard the URL initialization somehow. If I try this:
guard myURL = NSURL(string: "http://www.google.com") else {
fatalError()
}
Value of optional type 'NSURL?' not unwrapped; did you mean to use '!'
or '?'?
(Note: there's no way to add a ! or ? anywhere the code above that will fix the error. Conditional unwrapping only happens with guard let... guard var..., and myURL is already defined)
I understand why this fails: Even a successful call to NSURL(string:) is returning the (valid) NSURL wrapped inside an optional NSURL?, so I still need to unwrap it somehow before assigning to myURL (which is non-optional, hence not compatible for assignment as-is).
I can get around this by using an intermediate variable:
guard let theURL = NSURL(string: "http://www.google.com") else {
fatalError()
}
myURL = theURL
...but this is obviously not elegant at all.
What should I do?
Update Another approach, that doesn't use guard, would be to use a switch, as optionals map to the Optional enum:
init?() {
switch URL(string: "http://www.google.com") {
case .none:
myURL = NSURL()
return nil
case let .some(url):
myURL = url
}
}
although you'd still get a url local variable.
Original answer
You can declare your initializer as a failable one and return nil in case the url string parsing fails, instead of throwing a fatal error. This will make it more clear to clients your the class that the initializer might fail at some point. You still won't get rid of the guard, though.
init?() {
guard let url = URL(string: "http:www.google.com") else {
// need to set a dummy value due to a limitation of the Swift compiler
myURL = URL()
return nil
}
myURL = url
}
This add a little complexity on the caller side, as it will need to check if the object creation succeeded, but it's the recommended pattern in case the object initializer can fail constructing the object. You'd also need to give up the NSObject inheritance as you cannot override the init with a failable version (init?).
You can find out more details about failable initializers on the Swift blog, Apple's documentation, or this SO question.