Weak reference to closure in Swift - swift

I have the following code to create an observable property for data binding. It's in the works so I'm not sure what the final implementation is going to be and I'm still pretty new to Swift.
class Observable<T> {
typealias Observer = T -> Void
var value: T {
didSet {
for observer in self.observers {
observer?(self.value)
}
}
}
var observers: [Observer?] = []
init(_ val: T) {
self.value = val
}
}
I would like to keep weak references to the Observer closures. I don't want to rely on the client to ensure that the closure is weak/unowned before passing it in, via the capture list. Especially because there can be many observable properties on a given class.
Is it possible to make the closure references weak in my Observable class?
UPDATE:
I found a couple of resources that I think will help me accomplish what I want:
Make self weak in methods in Swift
specifically,
func methodPointer<T: AnyObject>(obj: T, method: (T) -> () -> Void) -> (() -> Void) {
return { [unowned obj] in method(obj)() }
}
The following link refers to the above stackoverflow answer and goes into some more detail:
http://blog.xebia.com/2014/10/09/function-references-in-swift-and-retain-cycles/
and this is a two-way binding example:
http://five.agency/solving-the-binding-problem-with-swift/
specifically,
class BondBox<T> {
weak var bond: Bond<T>?
init(_ b: Bond<T>) { bond = b }
}
where the listener is wrapped in a class called Bond, which is weakly referenced in the BondBox.

Is it possible to make the closure references weak in my Observable class
No. Only class instances can be referred to via weak references in Swift, and a function is not a class instance. (And not only must they be class instances, they must be an Optional wrapping a class instance.)
There are some pretty obvious ways around this, or course - the simplest being a wrapper class. But I do not actually recommend that in this situation, because you have not convinced me that weak references to functions are needed here in the first place. Remember, a weak reference to an object to which there is no strong reference will instantly lose the reference and will be pointing at nil. I can't believe that is what you want. I think you're barking up a wrong tree here.

Weak/strong are for memory management of objects, so only apply for variables of reference types (i.e. types which point to objects). Function types in Swift are not reference types, and therefore it does not make sense to talk about weak/strong for their variables.
Plus, you do not actually have variables of function type in your code (except in the middle of the iteration). You simply have a variable of array type. Even in Objective-C, you can only mark variables as weak or strong, not values that are stored inside other things.
And if you were to write the thing you are writing in Objective-C, you would want the "Observable" to have strong references to the closures. Otherwise, who else would have a strong reference to the closure?

Related

swift how to update viewModel parameter while init

I have encountered the problem "Cannot assign to property: 'self' is an immutable capture". How can i call API to check status and update "getUpdateSuccess" parameter
struct HomepageViewModel {
var getUpdateSuccess: Bool = false
init() {
getStatusUpdated.execute().done {
[self] isUpdated in
// Cannot assign to property: 'self' is an immutable capture
self.getUpdateSuccess = isUpdated
}
}
}
If you want to “reference” self later (e.g., update the same instance later), self should be a reference type, not a value type.
So you can solve this by making the view model a class (a reference type) rather than a struct (a value type).
“Model” objects are good candidates for value types, but “view models” are generally reference types. See Value and Reference Types for a discussion of when you might use a value type and when you might use a reference type. For a general discussion, see The Swift Programming Language: Structures and Classes.
By the way, the closure’s capture list syntax (the stuff in between the [ and the ]) has different practical implications if the captured variable is a reference type vs when it is a value type.
With a value type, the capture list makes sure you have a copy of the value:
struct Person {…}
var person = Person(…)
foo { [person] in
// this code will be dealing with a safe copy of the original `Person`
}
// maybe you’ll mutate the original `Person` instance here,
// but the above closure will have a copy of the original one
Contrast that with a reference type, where the capture list makes sure you have a strong reference to the reference type:
class ViewModel {
func bar() {
baz { [self] in
// this has a strong reference to this view model instance, not a copy of the instance
}
}
}
let viewModel = ViewModel(…)
viewModel.bar() // any updates that bar does asynchronously, will update this instance of the view model, not a copy of it
As an aside, with reference types, you will often see a capture list with the weak keyword, e.g., [weak self], if you want a “weak” reference to the class rather than a strong reference. This is common if you need to prevent a “strong reference cycle” (a topic beyond the scope of this conversation). This is discussed in The Swift Programming Language: Strong Reference Cycles for Closures.
But, whether a strong reference or weak reference, you are not dealing with a copy of the object, but rather a “reference” to the same original instance.

Is it safe to capture properties of `self` without capturing `self`?

You can copy this playground verbatim:
var closures=[() -> Void]()
class Thing {
let name: String
init(_ name: String) { self.name=name }
}
class RefThingWrapper {
let thing: Thing
init(_ t: Thing) {
self.thing=t
closures.append { [thing, weak self] in // Even though `thing` captured strongly, `self` could still get deallocated.
print((self == nil) ? "`self` deallocated" : "`self` not deallocated")
print(thing.name) // Decided to use `thing` within closure to eliminate any chance of optimizations affecting the test results — which I found when I captured `self` strongly without using it.
}
}
}
struct ValThingWrapper {
let thing: Thing
init(_ t: Thing) {
self.thing=t
closures.append { [weak thing] in
print((thing == nil) ? "`thing` deallocated" : "`thing` not deallocated")
}
}
}
var wrapper=Optional(RefThingWrapper(Thing("Thing 1"))) // Test with reference type.
//var wrapper=Optional(ValThingWrapper(Thing("Thing 1"))) // Test with value type.
closures[0]()
wrapper=nil
closures[0]()
It demonstrates how a property of self — whether self is a reference or value type — can be captured within a closure independently of self. Running the program as is demonstrates a captured property existing after self has been deallocated. Testing with the value type wrapper demonstrates that, if weakly captured, the instance will be deallocated once the referencing value instance is deallocated.
I wasn't sure this was possible because when creating the closure at first, I forgot to initialize the property I was capturing. So the compiler complained — in the capture list — 'self' used before all stored properties are initialized. So I figured self was being captured implicitly, and only after digging deeper discovered otherwise.
Is this documented somewhere? I found this post by Joe Groff where he proposes:
For 'let' properties of classes, it'd be reasonable to propose having
closures capture the property directly by default in this way
instead of capturing 'self' (and possibly allowing referencing them
without 'self.', since 'self' wouldn't be involved in any cycle formed
this way).
This was back in 2015, and I didn't find any implemented proposals that arose from the discussion. Is there any authoritative source that communicates this behavior?
If you’re just asking for documentation on capture lists and reference types, see The Swift Programming Language Resolving Strong Reference Cycles for Closures. Also see the Language Reference: Capture Lists
If your capture list includes value type, you’re getting copy of the value.
var foo = 1
let closure = { [foo] in
print(foo)
}
foo = 42
closure() // 1; capturing value of `foo` as it was when the closure was declared
If your capture list includes a reference type, you’re getting a copy of the reference to that current instance.
class Bar {
var value: Int
init(value: Int) { self.value = value }
}
var bar = Bar(value: 1)
let closure = { [bar] in
print(bar.value)
}
bar.value = 2
bar = Bar(value: 3)
closure() // 2; capturing reference to first instance that was subsequently updated
These captured references are strong by default, but can optionally be marked as weak or unowned, as needed, too.
That Capture Lists document outlines the way that you can capture a property without capturing self:
You can also bind an arbitrary expression to a named value in a capture list. The expression is evaluated when the closure is created, and the value is captured with the specified strength. For example:
// Weak capture of "self.parent" as "parent"
myFunction { [weak parent = self.parent] in print(parent!.title) }
I’m not crazy about their code sample, but it illustrates the capture of a property without capturing self, nonetheless.

Swift issue when storing Object inside a struct

I am confused of how memory is managed here, lets say there is a scenario:
import Foundation
class SomeObject {
deinit {
print("deinitCalled")
}
}
struct SomeStruct {
let object = SomeObject()
var closure: (() -> Void)?
}
func someFunction() {
var someStruct = SomeStruct()
someStruct.closure = {
print(someStruct)
}
someStruct.closure?()
}
someFunction()
print("the end")
What I would expect here is:
Optional(test.SomeStruct(object: test.SomeObject, closure: Optional((Function))))
deinitCalled
the end
However what I get is this:
SomeStruct(object: test.SomeObject, closure: Optional((Function)))
the end
And if I look at memory map:
retain cycle
How do I manage memory in this case
First, you should be very, very careful about putting reference types inside of value types, and especially a mutable reference type that is visible to the outside world. Structs are always value types, but you also want them to have value semantics, and it's challenging to do that while containing reference types. (It's very possible, lots of stdlib types do it in order to implement copy-on-write; it's just challenging.)
So the short version is "you almost certainly don't want to do what you're doing here."
But if you have maintained value semantics in SomeStruct, then the answer is to just make a copy. It's always fine to make a copy of a value type.
someStruct.closure = { [someStruct] in
print(someStruct)
}
This gives the closure it's own immutable value that is a copy of someStruct. Future changes to someStruct won't impact this closure.
If you mean for future changes to someStruct to impact this closure, then you may be violating value semantics, and you should redesign (likely by making SomeStruct a class, if you mean it to have reference semantics).

memory leak with swift due to reference cycle

Can someone please help me to avoid memory leak below
class Base {
var refer: Referenced?
init() {
self.refer = Referenced()
}
deinit {
print("deinit called")
}
}
class Referenced {
var native: Native
init(){
native = Native.init(baseRef: Base())
}
}
struct Native {
var baseRef: Base
}
func testMe () {
var base:Base? = Base()
base?.refer = nil
base = nil
}
testMe()
Looks like there is a cyclic reference in here. Base -> Referenced and Referenced -> Native -> Base
But I am unsure how to break this cycle .
There are two issues here:
As others have pointed out, in general, when you have two or more reference types referring to each other, you need to make one of those references weak to break the strong reference cycle. See Resolving Strong Reference Cycles Between Class Instances.
The more serious problem here is that you have infinite loop. The init of Base is instantiating a new Referenced instance, but the init of Referenced is creating another Base instance (which it’s passing to the Native instance). That new Base instance will then create another Referenced instance, repeating the process, and it will continue doing this until it runs out of memory/stack.
Add print statements inside your various init methods, and you’ll see this infinite loop in action.
You should revisit this model, identify an “ownership” graph. Parent objects can keep strong references to child objects (and possibly instantiate child objects, if appropriate), but child objects should only keep weak references back to their parent objects (and most likely not instantiate new parent objects).
For example, you might do:
class Parent {
var child: Child?
init() {
self.child = Child(parent: self)
}
deinit {
print("deinit called")
}
}
class Child {
weak var parent: Parent?
init(parent: Parent) {
self.parent = parent
}
}
And
func testMe() {
var parent: Parent? = Parent()
parent = nil // in this case, this is unnecessary because `parent` is a local variable, but this is here for illustrative purposes; but note that we don’t have to `nil` the `child`
}
Here, you instantiate a Parent object, which happens to instantiate a child. But note that:
The Parent supplies itself as a parameter to the Child;
The Child only maintains a weak reference back to the Parent; and
The Child obviously does not instantiate a new Parent, but rather just uses the reference that was supplied to it.
You can do a rendition of this with your Native struct, too, but the idea will be the same: Break the strong reference cycle with a weak reference and avoid the infinite loop.
Frankly, in these cases, abstract type names make examples unnecessarily confusing. If you clarify what these various types actual represent with a practical, real-world example, the ownership model will undoubtedly become more self-evident.
Change one of the reference as weak like this :
class Base {
weak var refer: Referenced?
init() {
self.refer = Referenced()
}
deinit {
print("deinit called")
}
}
Hope this helps.
You need to mark either Base.refer or Native.baseRef as weak to break the cycle, i.e. change var to weak var.
First of all, at the code snippet you passed, you have a problem with infinite circular dependency like this:
Base -> Referenced -> Native -> Base...
An instance of Native struct is not referencing the Base instance at the beginning of this chain, it is always creating a new one and this behavior is infinite. Just copy your code to the playground and try to run it. You will receive an "Execution was interrupted" error.
So, your code won't even run. As I suppose, what you are trying to achieve is for a Native struct to hold the reference to the first Base instance. Here is the right solution where we pass the reference from Base to Referenced and to Native that will hold the reference:
class Base {
var refer: Referenced?
init() {
self.refer = Referenced(base: self)
}
deinit {
print("deinit called")
}
}
class Referenced {
var native: Native
init(base: Base){
native = Native.init(baseRef: base)
}
}
struct Native {
var baseRef: Base
}
And now, as Native is referencing to the Base parent, you will struggle with the memory leak you mention in the original question. And this time, to prevent the memory leak you have to make your baseRef var weak like this:
weak var baseRef: Base?

Setting a delegate generates a compile error

I want to use a strategy pattern to register a set of objects that implement a protocol. When I set this up, I get a compile error when trying to set the delegate that is part of the protocol.
For discussion purposes, I have slightly reworked the DiceGame from the Swift eBook's Delegation chapter. The changes of significance are:
protocol DiceGame - declares a delegate
class SnakesAndLadders implements DiceGame (and therefore the protocol and delegate)
class Games holds 3 instances of SnakesAndLadders as
1) a concrete class of SnakesAndLadders
2) a 'let' constant of protocol DiceGame
3) a 'var' variable of protocol DiceGame
We can set the delegate fine if we use the concrete class (snakesAndLadders). However, there is a compile error if we use 'let' to hold it as a protocol (diceGameAsLet) but it compiles if we hold the variable as a 'var' (diceGameAsVar).
It is easy to work around, however, the delegate itself never changes so should be held as a 'let' constant, as it is only the internal property that changes. I must not understand something (possibly subtle but significant) about protocols and how they work and should be used.
class Dice
{
func roll() -> Int
{
return 7 // always win :)
}
}
protocol DiceGame
{
// all DiceGames must work with a DiceGameDelegate
var delegate:DiceGameDelegate? {get set}
var dice: Dice {get}
func play()
}
protocol DiceGameDelegate
{
func gameDidStart( game:DiceGame )
func gameDidEnd( game:DiceGame )
}
class SnakesAndLadders:DiceGame
{
var delegate:DiceGameDelegate?
let dice = Dice()
func play()
{
delegate?.gameDidStart(self)
playGame()
delegate?.gameDidEnd(self)
}
private func playGame()
{
print("Playing the game here...")
}
}
class Games : DiceGameDelegate
{
let snakesAndLadders = SnakesAndLadders()
// hold the protocol, not the class
let diceGameAsLet:DiceGame = SnakesAndLadders()
var diceGameAsVar:DiceGame = SnakesAndLadders()
func setupDelegateAsClass()
{
// can assign the delegate if using the class
snakesAndLadders.delegate = self
}
func setupDelegateAsVar()
{
// if we use 'var' we can assign the delegate
diceGameAsVar.delegate = self
}
func setupDelegateAsLet()
{
// DOES NOT COMPILE - Why?
//
// We are not changing the dice game so want to use 'let', but it won't compile
// we are changing the delegate, which is declared as 'var' inside the protocol
diceGameAsLet.delegate = self
}
// MARK: - DiceGameDelegate
func gameDidStart( game:DiceGame )
{
print("Game Started")
}
func gameDidEnd( game:DiceGame )
{
print("Game Ended")
}
}
DiceGame is a heterogeneous protocol that you're using as a type; Swift will treat this type as a value type, and hence (just as for a structures), changing its mutable properties will mutate also the instance of the protocol type itself.
If you, however, add the : class keyword to the DiceGame protocol, Swift will treat it as a reference type, allowing you to mutate members of instances of it, without mutating the instance itself. Note that this will constraint the protocol as conformable to only by class types.
protocol DiceGame: class { ... }
With the addition of the above, the mutation of immutable diceGameAsLet:s properties will be allowed.
In this context, it's worth mentioning that the : class keyword is usually used to constrain protocols used as delegates (e.g., DiceGameDelegate in your example) as conformable to only by class types. With this additional constraint, the delegates can be used as types to which the delegate owner (e.g. some class) only hold a weak reference, useful in contexts where a strong reference to the delegate could create a retain cycle.
See e.g. the 2nd part of this answer for details.
The issue is that when you store something as a Protocol, even if it is a class, swift considers them to be a value type, instead of the reference type you are expecting them to be. Therefore, no part of it is allowed to be changed. Take a look at this reference for more information.