following is a code sample you can run in a playground
import Foundation
class TempNotifier {
var onChange: (Int) -> Void = {t in }
var currentTemp = 72
init() {
// 1.
onChange = { [unowned self] temp in
self.currentTemp = temp
}
// 2.
onChange = {[unowned self] temp in
self.tempHandler(temp)
}
// 3.
unowned let s = self
onChange = s.tempHandler
}
deinit {
println("deinit")
}
private func tempHandler(temp: Int) {
self.currentTemp = temp
}
}
var tN: TempNotifier? = TempNotifier()
tN = nil
It illustrates 3 ways of assigning a value to a block with potential retain-cycle. Case 1. and 2. create no retain cycle due to unowned self however in case 3. it seems like there is no way to break the retain cycle (deinit is never printed). As you can see, I even tried to create a local unowned reference.
Is this the desired behaviour, is it "by design"? Or am I missing something?
Thanks!
Cross-posted from https://devforums.apple.com/message/1122247
Yes, this is the designed behavior.
Accessing a method without calling it, like s.tempHandler, is equivalent to a closure expression like { x in s.tempHandler(x) }. Here s is not marked unowned or weak, and hence is retained by the closure. If you want it to be captured as unowned or weak, you must explicitly write out the closure.
Related
class Increment {
var number = 0
init(){
print(#function)
}
deinit {
print(#function)
}
// let incrementNumber would give an error — 1
lazy var incrementNumber: (Int) -> () = { [weak self] value in
self?.number += value
print(self?.number)
}
}
do {
let increment = Increment().incrementNumber(3)
}
// output
init()
deinit
nil
if I separate the line, self is still alive when the closure is executed.
let increment = Increment()
increment.incrementNumber(3)
// output
init()
Optional(3)
deinit
Can anyone please explain why the Increment get deinit before the incrementNumber() call?
First of all you need to know definition of lazy variable.
According to apple docs, A lazy stored property is a property whose initial value isn’t calculated until the first time it’s used. You indicate a lazy stored property by writing the lazy modifier before its declaration.
Means that when the class is initial, they don't know they have variable lazy inside which is incrementNumber.
That the reason when you call
do {
let increment = Increment().incrementNumber(3)
}
Increment don't recognized incrementNumber in the class when you access directly. So Swift only see that you do nothing with Increment class in the rest of code then it automatically deinit the unused class.
Update: As Mr. DuncanC's mentioned, because of deinit class first so the compiler create an instance of Increment as AutoReleased, and keep it in memory in order to evaluate the second part of the expression
At the second, you call
let increment = Increment()
increment.incrementNumber(3)
Means that you make a class at first then you make lazy variable ( Swift sees that you do something with that class in second line so it waits until everything in class is called). Then in the rest of code class Increment is unused then it automatically deinit. That's the reason you see lazy is call before deinit.
For more further knowledge, you can do like making a not lazy function to see the difference.
class Increment {
var number = 0
init(){
print(#function)
}
deinit {
print(#function)
}
public func increaseNumber(_ value: Int) {
self.number += value
print(#function)
}
}
do {
let increment = Increment().increaseNumber(3)
}
//init()
//increaseNumber(_:)
//deinit
As you can see that increaseNumber is called before deinit because the class know that it owns func increaseNumber so it was called. Then in the rest of code class Increment is unused then it automatically deinit.
Don't understand why compiler make the error on this code snippet
class Addr {
var num: Int = 0
lazy var increment: (Int) -> () = {[unowned self] value in
self.num += value
print(self.num)
}
deinit {
print("deinit")
}
}
do {
let object = Addr().increment(5) // ERROR
}
Of course, i can change in capture list [unowned self] to [weak self] but I try to understand why this code not working. Why is the obeject is deinit before the call of the property. Will be thanked for the advanced explanation of this mechanism.
The issue is that since you are not storing a reference to the Addr object, it gets deallocated immediately, even before increment would be called on it.
Storing Addr in a variable and then calling increment on the variable solves the issue.
let object = Addr()
object.increment(5)
I've recently read about capture lists in this article of objc.io. I think it is a great tip and I'm started to use it.
Although it is not totally clarified, I assume there are no retain cycles when capturing this way, so you get a captured strong reference but with no retain cycle to be worried about.
And I've realized that it is possible to even capture methods, not only values:
.subscribe(onNext: { [showErrorAlert, populate] result in
switch result {
case .success(let book):
populate(book)
case .error(_):
showErrorAlert(L10n.errorExecutingOperation.localized)
}
})
I am trying to find some documentation related to this way of capturing but I cannot find any. Is this practice safe? Is this equal to the usual dance of [weak self], strongSelf = self inside the closure?
Is this practice safe? Is this equal to the usual dance of [weak
self], strongSelf = self inside the closure?
Yes and no - capturing methods of objects retains the object as well. The captured method could be accessing anything from the instance so it makes sense that it retains it.
On the other hand capturing a property does not retain the instance.
Here is a short snippet you can paste in playground to see for yourself:
import UIKit
import PlaygroundSupport
PlaygroundPage.current.needsIndefiniteExecution = true
class A {
let name: String
init(name: String) {
self.name = name
}
func a() {
print("Hello from \(name)")
}
func scheduleCaptured() {
print("Scheduling captured method")
DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + 1) { [a] in
a()
}
}
func scheduleCapturedVariable() {
print("Scheduling captured variable")
DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + 1) { [name] in
print("Hello from \(name)")
}
}
deinit {
print("\(name) deinit")
}
}
var a1: A? = A(name: "instance 1")
a1?.scheduleCapturedVariable()
var a2: A? = A(name: "instance 2")
a2?.scheduleCaptured()
a1 = nil
a2 = nil
The output is:
Scheduling captured variable
Scheduling captured method
instance 1 deinit
Hello from instance 1
Hello from instance 2
instance 2 deinit
You can see that instance 2 is not deinitialised until the captured block is fired, while instance 1 is deinitialised immediately after set to nil.
Capturing instance methods isn't safe unless you're sure your closure will be disposed before deinit (f.e. you absolutely sure it will trigger limited amount of times and then sequence always ends). Same for non-reactive use-cases.
The method is captured strongly(and can't be captured weakly), so the closure would keep the reference, prohibiting ARC from destroying it. Hence with strongSelf behavior closure keep only weak ref to object, binds it as strong on execution start, and then releases strong reference at the execution end.
Does accessing a singleton within a closure cause a retain cycle?
Specifically something like this example:
class TheSingleton
{
static let shared = TheSingleton() //THE SINGLETON
enum Temperature //An associated enum
{
case cold, hot
}
var currentTemp: Temperature? //A non-class-type variable
var aPicture: UIImage? //A class-type variable
func giveMeFive() -> Int //A function
{
return 5
}
//Pay attention to this
func doSomething(onDone: #escaping (Int) -> ())
{
OtherSVC.upload("Mr. Server, do async stuff plz") { (inImage) in
TheSingleton.shared.currentTemp = .cold
TheSingleton.shared.aPicture = inImage
onDone(TheSingleton.shared.giveMeFive())
}
}
}
//Fire the thing
TheSingleton.shared.doSomething { _ in}
If so, I don't really know how to write a capture list for this...
[weak TheSingleton.shared] (inImage) in
You can't do that ^
I included three cases because maybe the type of data matters?
I think I'm missing some fundamentals on capture lists and closure retain cycles.
All I know is whenever you access something outside of a closure's curly braces, you have to unown/weak it if it's a class-type object.
That's because closures create strong references by default.
I thought I could be cheeky and get around retain cycles by calling the entire singleton in closures, but I'm probably being dumb by turning a blind eye.
Would a solution be to do something like:
var updateDis = TheSingleton.shared.aPicture
OtherSVC.upload("ugh this is lame, and more work") { [unowned updateDis] inPic in
updateDis = inPic
}
?
Since you are writing a singleton, TheSingleton.shared is pretty much always going to be the same thing as self, so capture unowned self or weak self instead. I would prefer weak here because self will pretty much always be retained by the class and will only be deallocated when the application terminates.
OtherSVC.upload("Mr. Server, do async stuff plz") { [unowned self] (inImage) in
self..currentTemp = .cold
self.aPicture = inImage
onDone(self.giveMeFive())
}
I was testing swift closure with Xcode playground.
This is my code:
import UIKit
class A{
var closure: ()->() = {}
var name: String = "A"
init() {
self.closure = {
self.name = self.name + " Plus"
}
}
deinit {
print(name + " is deinit")
}
}
var a: A?
a = A()
a = nil
As what is expected, a is self contained by closure, so a is never released.
But, when I add this line before the last line:
a?.closure = { a?.name = "ttt" }
Then, I found "A is deinit" in the output window, which means a is released.
Why? is a not recycle reference?
To be test, I use a function to set the closure, which the code is version 2:
import UIKit
class A{
var closure: ()->() = {}
func funcToSetClosure(){
self.closure = { self.name = "BBB"}
}
var name: String = "A"
init() {
self.closure = {
self.name = self.name + " Plus"
}
}
deinit {
print(name + " is deinit")
}
}
var a: A?
a = A()
a?.funcToSetClosure()
a = nil
Again, a is never released.
So I got the conclusion, when closure is set by init or a function in the class, it will cause recycle reference, when it is set out side the class, it will not cause recycle reference. Am I right?
There are retain cycles in both cases. The difference is the nature of the reference, not the place where closure is set. This difference is manifested in what it takes to break the cycle:
In the "inside" situation, the reference inside the closure is self. When you release your reference to a, that is insufficient to break the cycle, because the cycle is directly self-referential. To break the cycle, you would have had also to set a.closure to nil before setting a to nil, and you didn't do that.
In the "outside" situation, the reference is a. There is a retain cycle so long as your a reference is not set to nil. But you eventually do set it to nil, which is sufficient to break the cycle.
(Illustrations come from Xcode's memory graph feature. So cool.)
As the SIL documentation says, when you capture a local variable in a closure, it will be stored on the heap with reference counting:
Captured local variables and the payloads of indirect value types are
stored on the heap. The type #box T is a reference-counted type that
references a box containing a mutable value of type T.
Therefore when you say:
var a : A? = A()
a?.closure = { a?.name = "ttt" }
you do have a reference cycle (which you can easily verify). This is because the instance of A has a reference to the closure property, which has a reference to the heap-allocated boxed A? instance (due to the fact that it's being captured by the closure), which in turn has a reference to the instance of A.
However, you then say:
a = nil
Which sets the heap-allocated boxed A? instance's value to .none, thus releasing its reference to the instance of A, therefore meaning that you no longer have a reference cycle, and thus A can be deallocated.
Just letting a fall out of scope without assigning a = nil will not break the reference cycle, as the instance of A? on the heap is still being retained by the closure property of A, which is still being retained by the A? instance.
What causes the retain cycle is that you reference self in the closure.
var a: A?
a = A()
a?.closure = { a?.name = "ttt" }
a = nil
You change the closure to no longer reference self, that's why it is deallocated.
In the final example, you make it reference self again in the closure, that is why it does not deallocate. There are ways around this, this post is a great list of when to use each case in swift: How to Correctly handle Weak Self in Swift Blocks with Arguments
I would imagine you are looking for something like this, where you use a weak reference to self inside the block. Swift has some new ways to do this, most commonly using the [unowned self] notation at the front of the block.
init() {
self.closure = { [unowned self] in
self.name = self.name + " Plus"
}
}
More reading on what is going on here: Shall we always use [unowned self] inside closure in Swift