This is one thing I do not understand in Swift - ios

Consider these lines:
I create a NSButton based class with this:
typealias onClickHandler = (NSTextfieldSuper)->Void
var onClick: onClickHandler?
When the user clicks on an instance of that button, I do this:
if (self.onClick != nil) {
onClick?(self)
}
I use that button later, from another class, with this:
let button = SuperButton()
button.onClick = { (textField: NSTextfieldSuper)->Void in
}
I am not sure if this is the correct syntax. I would like to process the button sent from the first closure on the parent class, where the button is created.
This was the only form I was able to type this without Xcode complaining. If this is correct, what is the purpose of this ->Void there? What could this possibly returning?
I just want to process that button sent.
By the way, as a bonus, I have to initialize several buttons with this, all running the same function. It would be nice to do something like
func doSomething () {
}
and then
let button = SuperButton()
button.onClick = doSomething
any ideas?

This was the only form I was able to type this without Xcode complaining. If this is correct, what is the purpose of this ->Void there? What could this possibly returning?
It is the same as in your typealias, in Swift a function type has the form:
(parameter definitions) -> return type
and functions which return nothing have a return type of Void (similar to C). The full form off a closure expression is:
{ (parameter definitions) ->return typeinbody}
Without any inference this expression provides the full type of the closure, and the -> Void Return type in your example specifies that your closure returns nothing. In your assignment this full type will be checked at compile time to conform to the type of onClick.
Now Swift will infer lots of stuff and there are various shorthands available for closure expressions, you will find that Swift accepts:
button.onClick = { textField in }
as well here with both the argument and return types of the closure being inferred.
By the way, as a bonus, [...] any ideas?
Just make the types match:
func doSomething(textField : NSTextfieldSuper) { }
button.onClick = doSomething
Unlike in (Objective-)C functions and closures (blocks in C) are interchangeable (as they are in plenty of other languages, C is the oddfellow here)
HTH

Related

Swift generic class, inheritance and covariance

I'm faced with the problem of using generic class and inheritance.
Brief description of the problem:
I have a base class called BookPageDataSource and two inherited classes (ReadingBookPageDataSource and StarsBookPageDataSource) with different implementations.
Also, I have a generic class BookPageViewController that contains the generic parameter of this data source and two inherited classes (ReadingBookPageViewController and StarsBookPageViewController) from this class.
I need to write a method the return parameter of which is BookPageViewController<DataSource>.
// Data Sources
class BookPageDataSource { }
class ReadingBookPageDataSource: BookPageDataSource { }
class StarsBookPageDataSource: BookPageDataSource { }
// Controllers
class BookPageViewController<DataSource: BookPageDataSource>: UIViewController {
let dataSource: DataSource
init(dataSource: DataSource) {
self.dataSource = dataSource
super.init(nibName: nil, bundle: nil)
}
required init?(coder aDecoder: NSCoder) {
return nil
}
}
final class ReadingBookPageViewController: BookPageViewController<ReadingBookPageDataSource> { }
final class StarsBookPageViewController: BookPageViewController<StarsBookPageDataSource> { }
// Communication
class Pager {
func currentPageController<DataSource>(at index: Int) -> BookPageViewController<DataSource> {
// for example
if index == 0 {
// How to remove the cast from the line below?
return readingPageController() as! BookPageViewController<DataSource>
}
return starsPageController() as! BookPageViewController<DataSource>
}
private func readingPageController() -> ReadingBookPageViewController {
return ReadingBookPageViewController(dataSource: ReadingBookPageDataSource())
}
private func starsPageController() -> StarsBookPageViewController {
return StarsBookPageViewController(dataSource: StarsBookPageDataSource())
}
}
The method currentPageController always crashes, because the DataSource is always equals to BookPageDataSource, not to ReadingBookPageDataSource or StarsBookPageDataSource.
Conceptual Discussion
Your concept for the architecture is flawed and this is leading to your issue.
Simple Generics Example
Here's a very simple example of a generic function, which just returns the value you give it:
func echo <T> (_ value: T) -> T { return value }
Because this function is generic, there is ambiguity about the type that it uses. What is T? Swift is a type-safe language, which means that ultimately there is not allowed to be any ambiguity about type whatsoever. So why is this echo function allowed? The answer is that when I actually use this function somewhere, the ambiguity about the type will be removed. For example:
let myValue = echo(7) // myValue is now of type Int and has the value 7
In the act of using this generic function I have removed the ambiguity by passing it an Int, and therefore the compiler has no uncertainty about the types involved.
Your Function
func currentPageController <DataSource> (at index: Int) -> BookPageViewController<DataSource>
Your function only uses the generic parameter DataSource in the return type, not in the input - how is the compiler supposed figure out what DataSource is?* I assume this is how you imagined using your function:
let pager = Pager()
let controller = pager.currentPageController(at: 0)
But now, what is the type of controller? What can you expect to be able to do with it? It seems that you're hoping that controller will take on the correct type based on the value that you pass in (0), but this is not how it works. The generic parameter is determined based on the type of the input, not the value of the input. You're hoping that passing in 0 will yield one return type, while 1 will yield a different one - but this is forbidden in Swift. Both 0 and 1 are of type Int, and the type is all that can matter.
As is usually the case with Swift, it is not the language/compiler that is preventing you from doing something. It is that you haven't yet logically formulated what is even is that you want, and the compiler is just informing you of the fact that what you've written so far doesn't make sense.
Solutions
Let's move on to giving you a solution though.
UIViewController Functionality
Presumably there is something that you wanted to use controller for. What is it that you actually need? If you just want to push it onto a navigation controller then you don't need it to be a BookPageViewController. You only need it to be a UIViewController to use that functionality, so your function can become this:
func currentPageController (at index: Int) -> UIViewController {
if index == 0 {
return readingPageController()
}
return starsPageController()
}
And you can push the controller that it returns onto a navigation stack.
Custom Functionality (Non-Generic)
If, however, you need to use some functionality which is specific to a BookPageViewController then it depends what it is you want to do. If there is a method on BookPageViewController like this:
func doSomething (input: Int) -> String
which doesn't make use of the generic parameter DataSource then probably you'll want to separate out that function into its own protocol/superclass which isn't generic. For example:
protocol DoesSomething {
func doSomething (input: Int) -> String
}
and then have BookPageViewController conform to it:
extension BookPageViewController: DoesSomething {
func doSomething (input: Int) -> String {
return "put your implementation here"
}
}
Now the return type of your function can be this non-generic protocol:
func currentPageController (at index: Int) -> DoesSomething {
if index == 0 {
return readingPageController()
}
return starsPageController()
}
and you can use it like this:
let pager = Pager()
let controller = pager.currentPageController(at: 0)
let retrievedValue = controller.doSomething(input: 7)
Of course, if the return type is no longer a UIViewController of any sort then you probably want to consider renaming the function and the related variables.
Custom Functionality (Generic)
The other option is that you can't separate out the functionality you need into a non-generic protocol/superclass because this functionality makes use of the generic parameter DataSource. A basic example is:
extension BookPageViewController {
func setDataSource (_ newValue: DataSource) {
self.dataSource = newValue
}
}
So in this case you really do need the return type of your function to be BookPageViewController<DataSource>. What do you do? Well, if what you really want is to use the setDataSource(_:) method defined above then you must have a DataSource object that you plan to pass in as an argument, right? If this is the case then we're making progress. Previously, you only had some Int value which you were passing into your function and the problem was that you couldn't specify your generic return type with that. But if you already have a BookPageDataSource value then it is at least logically possible for you to use this to specialize your
function.
What you say you want, however, is to just use an Int to get the controller at that index, regardless of what the DataSource type is. But if you don't care what the DataSource is of the returned BookPageViewController then how can you expect to set its DataSource to something else using the setDataSource(_:) method?
You see, the problem solves itself. The only reason you would need the return type of your function to be generic is if the subsequent functionality you need to make use of uses that generic type, but if this is the case then the controller you get back can't have just any old DataSource (you just wanted whichever one corresponds to the index you provide) - you need it to have exactly the type of DataSource which you plan to pass in when you use it, otherwise you're giving it the wrong type.
So the ultimate answer to your question is that, in the way that you were conceiving of it, there is no possible use for the function you were trying to construct. What's very cool about the way Swift is architected is that the compiler is actually able to figure out that logical flaw and prevent you from building your code until you've re-conceptualized it.
Footnote:
* It is possible to have a generic function which only uses the generic parameter in the return type and not in the input, but this won't help you here.

How to discriminate same method selector on Swift 3.0? [duplicate]

[NOTE This question was originally formulated under Swift 2.2. It has been revised for Swift 4, involving two important language changes: the first method parameter external is no longer automatically suppressed, and a selector must be explicitly exposed to Objective-C.]
Let's say I have these two methods in my class:
#objc func test() {}
#objc func test(_ sender:AnyObject?) {}
Now I want to use Swift 2.2's new #selector syntax to make a selector corresponding to the first of these methods, func test(). How do I do it? When I try this:
let selector = #selector(test) // error
... I get an error, "Ambiguous use of test()." But if I say this:
let selector = #selector(test(_:)) // ok, but...
... the error goes away, but I'm now referring to the wrong method, the one with a parameter. I want to refer to the one without any parameter. How do I do it?
[Note: the example is not artificial. NSObject has both Objective-C copy and copy: instance methods, Swift copy() and copy(sender:AnyObject?); so the problem can easily arise in real life.]
[NOTE This answer was originally formulated under Swift 2.2. It has been revised for Swift 4, involving two important language changes: the first method parameter external is no longer automatically suppressed, and a selector must be explicitly exposed to Objective-C.]
You can work around this problem by casting your function reference to the correct method signature:
let selector = #selector(test as () -> Void)
(However, in my opinion, you should not have to do this. I regard this situation as a bug, revealing that Swift's syntax for referring to functions is inadequate. I filed a bug report, but to no avail.)
Just to summarize the new #selector syntax:
The purpose of this syntax is to prevent the all-too-common runtime crashes (typically "unrecognized selector") that can arise when supplying a selector as a literal string. #selector() takes a function reference, and the compiler will check that the function really exists and will resolve the reference to an Objective-C selector for you. Thus, you can't readily make any mistake.
(EDIT: Okay, yes you can. You can be a complete lunkhead and set the target to an instance that doesn't implement the action message specified by the #selector. The compiler won't stop you and you'll crash just like in the good old days. Sigh...)
A function reference can appear in any of three forms:
The bare name of the function. This is sufficient if the function is unambiguous. Thus, for example:
#objc func test(_ sender:AnyObject?) {}
func makeSelector() {
let selector = #selector(test)
}
There is only one test method, so this #selector refers to it even though it takes a parameter and the #selector doesn't mention the parameter. The resolved Objective-C selector, behind the scenes, will still correctly be "test:" (with the colon, indicating a parameter).
The name of the function along with the rest of its signature. For example:
func test() {}
func test(_ sender:AnyObject?) {}
func makeSelector() {
let selector = #selector(test(_:))
}
We have two test methods, so we need to differentiate; the notation test(_:) resolves to the second one, the one with a parameter.
The name of the function with or without the rest of its signature, plus a cast to show the types of the parameters. Thus:
#objc func test(_ integer:Int) {}
#nonobjc func test(_ string:String) {}
func makeSelector() {
let selector1 = #selector(test as (Int) -> Void)
// or:
let selector2 = #selector(test(_:) as (Int) -> Void)
}
Here, we have overloaded test(_:). The overloading cannot be exposed to Objective-C, because Objective-C doesn't permit overloading, so only one of them is exposed, and we can form a selector only for the one that is exposed, because selectors are an Objective-C feature. But we must still disambiguate as far as Swift is concerned, and the cast does that.
(It is this linguistic feature that is used — misused, in my opinion — as the basis of the answer above.)
Also, you might have to help Swift resolve the function reference by telling it what class the function is in:
If the class is the same as this one, or up the superclass chain from this one, no further resolution is usually needed (as shown in the examples above); optionally, you can say self, with dot-notation (e.g. #selector(self.test), and in some situations you might have to do so.
Otherwise, you use either a reference to an instance for which the method is implemented, with dot-notation, as in this real-life example (self.mp is an MPMusicPlayerController):
let pause = UIBarButtonItem(barButtonSystemItem: .pause,
target: self.mp, action: #selector(self.mp.pause))
...or you can use the name of the class, with dot-notation:
class ClassA : NSObject {
#objc func test() {}
}
class ClassB {
func makeSelector() {
let selector = #selector(ClassA.test)
}
}
(This seems a curious notation, because it looks like you're saying test is a class method rather than an instance method, but it will be correctly resolved to a selector nonetheless, which is all that matters.)
I want to add a missing disambiguation: accessing an instance method from outside the class.
class Foo {
#objc func test() {}
#objc func test(_ sender: AnyObject?) {}
}
From the class' perspective the full signature of the test() method is (Foo) -> () -> Void, which you will need to specify in order to get the Selector.
#selector(Foo.test as (Foo) -> () -> Void)
#selector(Foo.test(_:))
Alternatively you can refer to an instance's Selectors as shown in the original answer.
let foo = Foo()
#selector(foo.test as () -> Void)
#selector(foo.test(_:))
In my case (Xcode 11.3.1) the error was only when using lldb while debugging. When running it works properly.

Getter for object property needs to return UnsafeMutablePointer<T>?

I'm working in Swift and one of the protocols I'm using needs to return an UnsafeMutablePointer<T> of a particular object.
I have something like this:
#objc var myProperty:UnsafeMutablePointer<someObject>
{
get
{
// I call a class function here to get a 'someObject'
// return object which I need to pass back a pointer to it.
return UnsafeMutablePointer<someObject>
}
}
The problem is that Xcode doesn't like this. It complains that '>' is not a unary operator.
I've also tried removing the UnsafeMutablePointer<> and use an & in front of someObject but it complains that the & is to be used immediately in a list of arguments for a function.
I suppose I just can't find the right syntax for this? Any help would be appreciated.
If someObject has the type SomeClass, then you need to update your declaration like this:
#objc var myProperty:UnsafeMutablePointer<SomeClass>
{
get
{
return UnsafeMutablePointer<SomeClass>(unsafeAddressOf(someObject))
}
}
The generic argument needs to be the type of the returned data, and you also need to intialize a specialized UnsafeMutablePointer with the memory address of the desired object.

Cannot convert value of type 'Option' to expected argument type '#noescape (Option) throws -> Bool'

I have a class called Option. This is a Realm object so it's a subclass of Realm's own base class Object.
In a view controller I have an array property that holds a bunch of Option objects.
private var options = [Option]()
In a method down the view controller, I need to check if a certain Option object is contained within the aforementioned options array.
Previously (Swift 1.2), I have the checking logic like this.
func itemSelected(selection: Object) {
let option = selection as! Option
if !contains(options, option) {
// act accordingly
}
}
Now I'm converting the project to Swift 2 (I have updated the Realm's version to Swift 2 version as well). I updated the code to this.
func itemSelected(selection: Object) {
let option = selection as! Option
if !options.contains(option) {
// act accordingly
}
}
But now I'm getting the following compile error!
Cannot convert value of type 'Option' to expected argument type '#noescape (Option) throws -> Bool'
I can't figure out why. Any ideas?
This is because the contains function now expects a closure rather than an element on all non-equatable types. All you need to do is change it to the following
if !(options.contains{$0==option}) {
// act accordingly
}
What this does is it passes a closure in to the function, which returns true only if that closure satisfies any of its elements. $0 stands for the current element in the array that the contains function is testing against, and it returns true if that element is equal to the one that you are looking for.
While the first answer, that indicates this issue occurs because the contains method needs to operate on an Equatable type, is true, that's only half the story. The Realm Object class inherits NSObject, which conforms to Equatable (thus this should work without a closure). For more discussion on this, you can refer to this issue on the Realm GitHub page: https://github.com/realm/realm-cocoa/issues/2519. The Realm developers indicate that they believe this is a bug in Swift.
Ultimately, the suggested workaround is to re-declare the conformance to Equatable and Hashable, like so (this is copied verbatim from GitHub user bdash's comment on the previously posted issue):
public class A: Object, Equatable, Hashable {
}
public func ==(lhs: A, rhs: A) -> Bool {
return lhs.isEqual(rhs)
}
You'd replace the all instances of type A with type Option in that sample.
I've tested this solution, and it works for me in XCode 7.2.1, using Swift version 2.1.1.

Generics type constraint vs inheritance

Is there a difference between these two function declarations?
func doSomething<T: UIViewController>(controller: T) {...}
vs.
func doSomething(controller: UIViewController) {...}
In Type Constraint Syntax section of the Apples Swift programming language book, there's this code sample:
func​ ​someFunction​<​T​: ​SomeClass​, ​U​: ​SomeProtocol​>(​someT​: ​T​, ​someU​: ​U​) {
// function body goes here
}
with this description:
The hypothetical function above has two type parameters. The first type parameter, T, has a type constraint that requires T to be a subclass of SomeClass. ...
So in which cases is it better to use generic function described above?
They are different, but in the way you are using them, they amount to pretty much exactly the same result.
The difference is that when you call the generic version, the compiler sets T statically to be whatever type is passed in as the argument. When calling methods on that argument, this makes almost no difference – either way the calls on its methods will be dynamically dispatched, and you can't touch any parts of T that aren't guaranteed to be available from the constraint.
But suppose you made a change to this method to not just take an argument, but also return one, of the same type:
// T here will take the type of whatever is going in/out of the function
// be that UIViewController or a subtype of it
func doSomethingGenerically<T: UIViewController>(controller: T) -> T {
// some logic that results in a new controller being returned
}
// here the return type is fixed to be UIViewController
func doSomethingViaBaseClass(controller: UIViewController) -> UIViewController {
// some logic that results in a new controller being returned
}
Now, suppose you had a subclass of UIViewController that you were passing in, like so:
let subClass: MyUIViewController = ...
let controller1 = doSomethingGenerically(subClass)
let controller2 = doSomethingViaBaseClass(subClass)
Here, the type of the variable controller1 will be MyUIViewController, because that is what was passed in to the function so that is what T is. But the type of the variable controller2 will be UIViewController because that is the fixed type that doSomethingViaBaseClass returns.
Note, this doesn't mean the object they reference will be different - that depends on what the body of the function implements. It's just the type of the variables referring to it that will change.
There are other subtle differences but this is the main one to know about. In the case of structs, however, there are more differences worth noting. As it happens I wrote an article about them yesterday that might help.

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