How closure captures values in Swift? - ios

I am running the below code -
class Element {
var name: String
init(name: String) {
self.name = name
}
deinit {
print("Element is deinitializing...")
}
}
var element: Element? = Element(name: "Silver")
var closure = {
print(element?.name ?? "default value")
}
print(isKnownUniquelyReferenced(&element))
element?.name = "Gold"
element = nil
closure()
and it prints -
true
Element is deinitializing...
default value
In the above, isn't the closure captures element strongly? How the element is getting nil inside the closure?

From Swift Programming Guide - Closures
A closure can capture constants and variables from the surrounding context in which it’s defined. The closure can then refer to and modify the values of those constants and variables from within its body, even if the original scope that defined the constants and variables no longer exists.
A closure captures variables, not the contents of variables. When we are talking about local variables in a function (which are normally allocated on stack), it makes sure they are accessible even when the function exits and other local variables are deallocated, therefore we can do things like this:
func myFunc() {
var array: [Int] = []
DispatchQueue.main.async {
// executed when myFunc has already returned!
array.append(10)
}
}
Your example is similar. The closure captures the variable. This is a variable on module level, therefore its scope always exists. When you reassign its value, it will affect the value read inside the closure.
Or, in other words, the closure will be equivalent to:
var closure = {
print(CurrentModule.element?.name ?? "default value")
}
where CurrentModule is the name of your main module (which is usually the name of your project).
To prevent this behavior and capture the value of the variable instead, we can use closure capture list. Unfortunately, the official documentation does not properly explain what exactly is a capture list. Basically, using a capture list you declare variables local to the closure using values that are available when the closure is created.
For example:
var closure = { [capturedElement = element] in
print(capturedElement?.name ?? "default value")
}
This will create a new variable capturedElement inside the closure, with the current value of variable element.
Of course, usually we just write:
var closure = { [element] in
print(element?.name ?? "default value")
}
which is a shorthand for [element = element].

Related

Using init as a Closure

Recently I saw the following code line in a book (about CoreData)
return modelURLs(in: modelName).compactMap(NSManagedObjectModel.init)
I know what the code does but the question is: Why and how does it work?
There should be a closure as the argument of the compactMap function but there's only a "NSManagedObjectModel.init" in NORMAL parenthesis. What's the secret about it? What is it doing there? I would understand it if there's a static/class property called init which returns a closure but I don't think there is.
Unfortunately the book doesn't say more about this line of code. I would like to have further readings from the apple docs but I can't find anything. When I make a google search about "init in closures" then I don't get helpful results.
So you guys are my last hope :)
By the way: the function modelURLs(in: modelName) returns an Array of URLs but that's not really important here.
When using closures different syntax can be used as in the below example that converts an int array to a string array
let array = [1, 2, 3]
The following calls to compactMap will all correctly convert the array and generate the same result
let out1 = array.compactMap({return String($0)})
let out2 = array.compactMap({String($0)})
let out3 = array.compactMap {String($0)}
let out4 = array.compactMap(String.init)
When there are two init methods that takes the same number and types of argument then you must add the full signature for the init method to use. Consider this simple example struct
struct TwoTimesInt: CustomStringConvertible {
let value: Int
let twiceTheValue: Int
var description: String {
return "\(value) - \(twiceTheValue)"
}
init(value: Int) {
self.value = value
self.twiceTheValue = 2 * value
}
}
With only 1 init method we can do
let out5 = array.compactMap(TwoTimesInt.init)
But if we add a second init method
init(twiceTheValue: Int) {
self.value = twiceTheValue / 2
self.twiceTheValue = twiceTheValue
}
Then we need to give the full signature of the init method to use
let out6 = array.compactMap( TwoTimesInt.init(value:) )
Another thing worth mentioning when it comes to which method is selected is to look at the full signature of the init method including if it returns an optional value or not. So for example if we change the signature of the second init method to return an optional value
init?(twiceTheValue: Int) {
self.value = twiceTheValue / 2
self.twiceTheValue = twiceTheValue
}
then compactMap will favour this init since it expects a closure that returns an optional value, so if we remove the argument name in the call
let out7 = array.compactMap(TwoTimesInt.init)
will use the second init while the map function on the other hand will use the first init method if called the same way.
let out8 = array.map(TwoTimesInt.init)

Swift how variable declaration works

Can some one please explain to me how it works
var count: Int?
count = 1
if let count = count {
//do something.
}
Why there is no compiler error at if let count = count as we already created a variable named count as var count: Int?. How come two variables with same name possible?
Swift treated as both variables as a different one. The "count" that declared first can be treated as global one while the constant "count" is only available inside if condition, so it can't access outside if condition.
Optional Variable: It can contain a value or a Nil value. Nil represents the absence of a value or nothing
var count: Int? // Optional Variable
Here variable count is global variable.
Optional Binding: It is the way by which we try to retrieve a values from a chain of optional variable.
if let count = count {
//do something.
}
Here constant count value is available only with in the scope. It cannot be accessed outside the scope.
This is called optional binding
if let constantName = someOptional {
statements
}
someOptional is checked to see if it's nil or has data. If it's nil, the if-statement just doesn't get executed. If there's data, the data gets unwrapped and assigned to constantName for the scope of the if-statement. Then the code inside the braces is executed.

Swift how to "pass by value" of a object

I am quite new in Swift. And I create a class(for example):
class Fraction{
var a: Int
init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
}
and I also build a in other class function:
class func A_plusplus(f:Fraction){
f.a++
}
Then in the executive class I write:
var object = Fraction(a:10)
print("before run func = " + object.toString())
XXXclass.A_plusplus(object)
print("after ran func =" + object.toString() )
So the console output is
before run func = 10; after ran func =11
The question is how can I just send a copy of the "object" to keep its value which equal to 10
And if functions are always pass-by-reference, why we still need the keyword: "inout"
what does difference between A_plusplus(&object)//[if I make the parameter to be a inout parameter] and A_plusplus(object)
Universally, I don't want to use struct. Although this will solve my
problem exactly, I do pass-by-value rarely.So I don't want program's
copying processes slow my user's phone down :(
And It seems conforming the NSCopying protocol is a good option.But
I don't know how to implement the function:
func copyWithZone(zone:
NSZone)-> AnyObject? correctly
If your class is subclass of NSObject,better to use NSCopying
class Fraction:NSObject,NSCopying{
var a:Int
var b:NSString?
required init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
func copyWithZone(zone: NSZone) -> AnyObject {
let theCopy=self.dynamicType.init(a: self.a)
theCopy.b = self.b?.copy() as? NSString
return theCopy
}
}
class XXXclass{
class func A_plusplus(f:Fraction){
f.a++
f.b = "after"
}
}
var object = Fraction(a:10)
object.b = "before"
print("before run func = " + object.toString())
print(object.b!) //“Before”
XXXclass.A_plusplus(object.copy() as! Fraction)
print("after ran func =" + object.toString() )
print(object.b!)//“Before”
If it is just a common swift class,You have to create a copy method
class Fraction{
var a: Int
init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
func copy()->Fraction{
return Fraction(a: self.a)
}
}
class XXXclass{
class func A_plusplus(f:Fraction){
f.a++
}
}
var object = Fraction(a:10)
print("before run func = " + object.toString())
XXXclass.A_plusplus(object.copy())
print("after ran func =" + object.toString() )
To make it clear,you have to know that there are mainly two types in swift
Reference types. Like Class instance,function type
Value types,Like struct and others(Not class instance or function type)
If you pass in a Reference types,you pass in the copy of Reference,it still point to the original object.
If you pass in a Copy type,you pass in the copy of value,so it has nothing to do with the original value
Let us talk about inout,if you use it,it pass in the same object or value.It has effect on Value type
func add(inout input:Int){
input++
}
var a = 10
print(a)//10
add(&a)
print(a)//11
Swift has a new concept so called "struct"
You can define Fraction as struct (Not class)
And
struct Fraction{
...
}
var object = Fraction(a:10)
var object1 = object //then struct in swift is value type, so object1 is copy of object (not reference)
And if you use struct then try to use inout in A_plusplus function
Hope this will help you.
how can I just send a copy of the "object" to keep its value which equal to 10
In Swift classes and functions are always passed by reference. Structs, enums and primitive types are passed by value. See this answer.
You can't pass an object by value. You would have to manually copy it before passing it by reference (if that's what you really want).
Another way is to turn your class into a struct, since it would then be passed by value. However, keep in mind there a few other differences between classes and structs, and it might not necessarily be what you want.
And if functions are always pass-by-reference, why we still need the keyword: "inout"
According to the swift documentation, inout is used when
you want a function to modify a parameter’s value, and you want those changes to persist after the function call has ended, define that parameter as an in-out parameter instead.
So in practice with inout you can pass a value type (such as struct or primitive) by reference. You shouldn't really use this very often. Swift provides tuples, that could be used instead.
what does difference between A_plusplus(&object)//[if I make the parameter to be a inout parameter] and A_plusplus(object)
There is no difference for your A_plusplus function. In that function you don't modify the parameter f itself, you modify the f.a property.
The following example shows the effect of using inout when passing a class object. Both functions are the same, differing only in its parameter definition.
class Person {
var name: String
init(name: String) { self.name = name }
}
var me = Person(name: "Lennon") // Must be var to be passed as inout
// Normal object by reference with a var
func normalCall(var p: Person) {
// We sure are able to update p's properties,
// and they will be reflected back to me
p.name = "McCartney"
// Now p points to a new object different from me,
// changes won't be reflected back to me
p = Person(name: "Ringo")
}
// Inout object reference by value
func inoutCall(inout p: Person) {
// We still can update p's properties,
p.name = "McCartney"
// p is an alias to me, updates made will persist to me
p = Person(name: "Ringo")
}
print("\(me.name)") //--> Lennon
normalCall(me)
print("\(me.name)") //--> McCartney
inoutCall(&me)
print("\(me.name)") //--> Ringo
In normalCall p and me are different variables that happen to point to the same object. When you instantiate and assign a new object to p, they no longer refer to the same object. Hence, further changes to this new object will not be reflected back to me.
Stating that p is a var argument just means that its value can change throughout the function, it does not mean the new value will be assigned to what was passed as argument.
On the other hand, in inoutCall you can think of p and me as aliases. As such, assigning a new object to p is the exact same as assigning a new object to me. Any and every change to p is persisted in me after the function ends.

Assigning Closure to variable in Swift causes 'variable used before being initialized'

I have a problem with a closure that is meant to be created and then being executed within another function over the range of the 2D pixel raster of an image where it shall basically called like this:
filter(i,j) and return a value based on its arguments.
I thought this code should work but it complains that the closure variable I have created is not initialized. I guess that means that I did not gave it arguments, but I wont within this function as the data is known to the closure at the time when it interacts with the image. How can I setup a closure which does not care about initialization?
Thank you in advance :)
func processFilter(type:FilterType){
var x = 0
var y = 0
//create cloure
var closure:(i:Int, j:Int)->Int
if(type == FilterType.MyFilter) {
x = 1024
y = 2048
func filter(i:Int, j:Int)->Int {
return i*j*x*y*4096
}
//compiler does not complain here...
closure = filter
}
//other if statements with different closure definitions follow...
//This call throws error: variable used before being initialized
let image = filterImage(closure)
}
You use the variable closure before the compiler is certain that it is initialized. You can solve this in 2 ways, depending on what you need:
Add an else-clause to your if and set closure to a default closure.
Make closure optional by defining it as var closure: ((i: Int, j: Int) -> Int)? and then you can check if it is optional before using it by using closure?(i, j) or if let filter = closure { filter(i, j)}.
Also, try to use better variable names such as filterClosure. closure on its own doesn't really say much.
The problem is that you define your closure as:
var closure:(i:Int, j:Int)->Int
Then you initialize it only if you enter the if
If not, that var is not initialized, hence the compiler warning
Possible solution:
func processFilter(type:FilterType){
var x = 0
var y = 0
//create cloure
var filterClosure:((i:Int, j:Int)->Int)?
if(type == FilterType.MyFilter) {
x = 1024
y = 2048
func filter(i:Int, j:Int)->Int {
return i*j*x*y*4096
}
//compiler does not complain here...
filterClosure = filter
}
//other if statements with different closure definitions follow...
if let closure = filterClosure {
let image = filterImage(closure)
}
}
Your closure is only initialized if the code enters your if block (i.e. if type == FilterType.MyFilter). In the other case it is left uninitialized.

Swift if let evaluates successfully on Optional(nil)

I have a custom object called Field. I basically use it to define a single field in a form.
class Field {
var name: String
var value: Any?
// initializers here...
}
When the user submits the form, I validate each of the Field objects to make sure they contain valid values. Some fields aren't required so I sometimes deliberately set nil to the value property like this:
field.value = nil
This seems to pose a problem when I use an if-let to determine whether a field is nil or not.
if let value = field.value {
// The field has a value, ignore it...
} else {
// Add field.name to the missing fields array. Later, show the
// missing fields in a dialog.
}
I set breakpoints in the above if-else and when field.value has been deliberately set to nil, it goes through the if-let block, not the else. However, for the fields whose field.value I left uninitialized and unassigned, the program goes to the else block.
I tried printing out field.value and value inside the if-let block:
if let value = field.value {
NSLog("field.value: \(field.value), value: \(value)")
}
And this is what I get:
field.value: Optional(nil), value: nil
So I thought that maybe with optionals, it's one thing to be uninitialized and another to have the value of nil. But even adding another if inside the if-let won't make the compiler happy:
if let value = field.value {
if value == nil { // Cannot invoke '==' with an argument list of type '(Any, NilLiteralConvertible)'
}
}
How do I get around this? I just want to check if the field.value is nil.
I believe this is because Any? allows any value and Optional.None is being interpreted as just another value, since Optional is an enum!
AnyObject? should be unable to do this since it only can contain Optional.Some([any class object]), which does not allow for the case Optional.Some(Optional) with the value Optional.None.
This is deeply confusing to even talk about. The point is: try AnyObject? instead of Any? and see if that works.
More to the point, one of Matt's comment mentions that the reason he wants to use Any is for a selection that could be either a field for text input or a field intended to select a Core Data object.
The Swifty thing to do in this case is to use an enum with associated values, basically the same thing as a tagged/discriminated union. Here's how to declare, assign and use such an enum:
enum DataSelection {
case CoreDataObject(NSManagedObject)
case StringField(String)
}
var selection : DataSelection?
selection = .CoreDataObject(someManagedObject)
if let sel = selection { // if there's a selection at all
switch sel {
case .CoreDataObject(let coreDataObj):
// do what you will with coreDataObj
case .StringField(let string):
// do what you will with string
}
}
Using an enum like this, there's no need to worry about which things could be hiding inside that Any?. There are two cases and they are documented. And of course, the selection variable can be an optional without any worries.
There's a tip to replace my Any? type with an enum but I couldn't get this error out of my head. Changing my approach doesn't change the fact that something is wrong with my current one and I had to figure out how I arrived at an Optional(nil) output.
I was able to reproduce the error by writing the following view controller in a new single-view project. Notice the init signature.
import UIKit
class Field {
var name: String = "Name"
var value: Any?
init(_ name: String, _ value: Any) {
self.name = name
self.value = value
}
}
class AppState {
var currentValue: Field?
}
class ViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
let f = Field("Amount", AppState().currentValue)
NSLog("\(f.value)")
}
}
In short, I was passing a nil value (AppState().currentValue) to an initializer that accepts Any, and assigns it to a property whose type is Any?. The funny thing here is if I directly passed nil instead, the compiler will complain:
let f = Field("Amount", nil) // Type 'Any' does not conform to protocol 'NilLiteralConvertible'
It seems that somewhere along the way, Swift wraps the nil value of AppState().currentValue in an optional, hence Optional(nil).

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