weak self

[Rien]

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

  return self?.myparam ?? 42

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

[glessard (Guillaume Lessard)]

Hi Rien,

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

The object never knows whether a weak reference to it is being used; in order to be safe you must bind the reference — then you get a strong reference out of it, and that guarantees the object stays alive as long as the strong reference is in scope.

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

It does not.

From the perspective of the runtime, weak references are a different type than normal/strong references; what’s important to know here is that getting a strong reference from a weak one is thread-safe. It’s interesting to know that weak references to “dead” objects are nilled out on use, lazily. When a WeakReference is used, the object’s strong reference count is checked, and if that is zero then the WeakReference is nilled out.

You could read Mike Ash’s description at <https://www.mikeash.com/pyblog/friday-qa-2015-12-11-swift-weak-references.html&gt;\. That describes Swift 3 weak references quite well. There’s a newer implementation of reference counting for Swift 4, but the outwardly-visible behaviour is the same.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

There is potential for a race on `self`, as the strong reference count could go to zero between the two uses of the weak reference.

The proper way is

if let myref = self { return myref.myparam }

(or the equivalent guard.)
That’s safe.

Cheers,
Guillaume Lessard

···

On May 1, 2017, at 08:46, Rien via swift-users <swift-users@swift.org> wrote:

[DevAndArtist (Adrian Meister)]

Not answering the questions, but sharing a neat trick.

[weak self] in

guard let `self` = self else { return }

self.foo() // strong self :)

···

--
Adrian Zubarev
Sent with Airmail

Am 1. Mai 2017 um 16:46:33, Rien via swift-users (swift-users@swift.org) schrieb:

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

return self?.myparam ?? 42

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

[Dennis_Weissmann]

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

  return self?.myparam ?? 42

Yes, as you said, you never know when self will be nilled out, that's why you need to create a (temporary) strong reference to it, work on it, and when the block returns, your strong reference is released and self either goes away immediately (incase it was released elsewhere after the local binding to a strong variable and no other objects had a strong reference to it) or it will stay as long as no other object holds a strong reference to it.

When the closure is more involved than a view lines, I often do a guard let `self` = self else { return } to conveniently work with a strong self inside the rest of the closure.

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

- Dennis

···

On May 1, 2017, at 4:46 PM, Rien via swift-users <swift-users@swift.org> wrote:

[Rien]

Thanks Guillaume,

Interesting: this is a kind-of symmetry break between optionals and weak references.
I.e. most (inexperienced?) programmers will see a strong similarity between weak references and optionals.
And for a lot of use cases they do indeed behave similar.
But for weak references I think the guideline should be: Never ‘force-unwrap’ it.
Maybe this should be mentioned in the swift guide?
Maybe even include a warning in the compiler for this?

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

···

On 01 May 2017, at 23:26, Guillaume Lessard <glessard@tffenterprises.com> wrote:

Hi Rien,

On May 1, 2017, at 08:46, Rien via swift-users <swift-users@swift.org> wrote:

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

The object never knows whether a weak reference to it is being used; in order to be safe you must bind the reference — then you get a strong reference out of it, and that guarantees the object stays alive as long as the strong reference is in scope.

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

It does not.

From the perspective of the runtime, weak references are a different type than normal/strong references; what’s important to know here is that getting a strong reference from a weak one is thread-safe. It’s interesting to know that weak references to “dead” objects are nilled out on use, lazily. When a WeakReference is used, the object’s strong reference count is checked, and if that is zero then the WeakReference is nilled out.

You could read Mike Ash’s description at <https://www.mikeash.com/pyblog/friday-qa-2015-12-11-swift-weak-references.html&gt;\. That describes Swift 3 weak references quite well. There’s a newer implementation of reference counting for Swift 4, but the outwardly-visible behaviour is the same.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

There is potential for a race on `self`, as the strong reference count could go to zero between the two uses of the weak reference.

The proper way is

if let myref = self { return myref.myparam }

(or the equivalent guard.)
That’s safe.

Cheers,
Guillaume Lessard

[Rien]

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

  return self?.myparam ?? 42

Yes, as you said, you never know when self will be nilled out, that's why you need to create a (temporary) strong reference to it, work on it, and when the block returns, your strong reference is released and self either goes away immediately (incase it was released elsewhere after the local binding to a strong variable and no other objects had a strong reference to it) or it will stay as long as no other object holds a strong reference to it.

When the closure is more involved than a view lines, I often do a guard let `self` = self else { return } to conveniently work with a strong self inside the rest of the closure.

I was aware of that practise (use it myself) but I was not sure if it would always work.
I.e. I have not found it documented that
  let strongSelf = self
will actually retain ‘self’ an not create a strong reference to an intermediate reference to self.

It makes sense though, otherwise there would be massive failures ‘out there’. ;-)

Thanks,
Rien.

···

On 01 May 2017, at 16:59, Dennis Weissmann <dennis@dennisweissmann.me> wrote:

On May 1, 2017, at 4:46 PM, Rien via swift-users <swift-users@swift.org> wrote:

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

- Dennis

[Chris_McIntyre]

I don’t think there’s any symmetry break. I would argue that forced-unwrapping is rarely a best-practice. The swifty way would be to always use guard let and if let for optionals. Looking through the Swift Programming Language book, the only examples of “if x != nil” I can find just print some explanatory text (and even then I’d argue it’s better to use if let _ = x).

···

--
Chris McIntyre

On May 2, 2017, at 2:05 AM, Rien via swift-users <swift-users@swift.org> wrote:

Thanks Guillaume,

Interesting: this is a kind-of symmetry break between optionals and weak references.
I.e. most (inexperienced?) programmers will see a strong similarity between weak references and optionals.
And for a lot of use cases they do indeed behave similar.
But for weak references I think the guideline should be: Never ‘force-unwrap’ it.
Maybe this should be mentioned in the swift guide?
Maybe even include a warning in the compiler for this?

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

On 01 May 2017, at 23:26, Guillaume Lessard <glessard@tffenterprises.com> wrote:

Hi Rien,

On May 1, 2017, at 08:46, Rien via swift-users <swift-users@swift.org> wrote:

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

The object never knows whether a weak reference to it is being used; in order to be safe you must bind the reference — then you get a strong reference out of it, and that guarantees the object stays alive as long as the strong reference is in scope.

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

It does not.

From the perspective of the runtime, weak references are a different type than normal/strong references; what’s important to know here is that getting a strong reference from a weak one is thread-safe. It’s interesting to know that weak references to “dead” objects are nilled out on use, lazily. When a WeakReference is used, the object’s strong reference count is checked, and if that is zero then the WeakReference is nilled out.

You could read Mike Ash’s description at <https://www.mikeash.com/pyblog/friday-qa-2015-12-11-swift-weak-references.html&gt;\. That describes Swift 3 weak references quite well. There’s a newer implementation of reference counting for Swift 4, but the outwardly-visible behaviour is the same.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

There is potential for a race on `self`, as the strong reference count could go to zero between the two uses of the weak reference.

The proper way is

if let myref = self { return myref.myparam }

(or the equivalent guard.)
That’s safe.

Cheers,
Guillaume Lessard

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

[glessard (Guillaume Lessard)]

I also don’t think there’s a symmetry break there. A weak reference is effectively a shared variable between threads, so its value can change between any two actions. This would be the same if it were an Optional that happens to be shared between threads, but you’d have to handle locking yourself. An `if let` binding to a strong reference is effectively a lightweight lock against deletion of your object.

The more critical thing is that if you need to perform more than one action with a weak reference, you really should bind it to a strong reference first. The optional chaining approach is a fine shortcut for a single action.

Cheers,
Guillaume Lessard

···

On May 2, 2017, at 00:05, Rien <Rien@balancingrock.nl> wrote:

Interesting: this is a kind-of symmetry break between optionals and weak references.
I.e. most (inexperienced?) programmers will see a strong similarity between weak references and optionals.
And for a lot of use cases they do indeed behave similar.
But for weak references I think the guideline should be: Never ‘force-unwrap’ it.
Maybe this should be mentioned in the swift guide?
Maybe even include a warning in the compiler for this?

[Dennis_Weissmann]

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

  return self?.myparam ?? 42

Yes, as you said, you never know when self will be nilled out, that's why you need to create a (temporary) strong reference to it, work on it, and when the block returns, your strong reference is released and self either goes away immediately (incase it was released elsewhere after the local binding to a strong variable and no other objects had a strong reference to it) or it will stay as long as no other object holds a strong reference to it.

When the closure is more involved than a view lines, I often do a guard let `self` = self else { return } to conveniently work with a strong self inside the rest of the closure.

I was aware of that practise (use it myself) but I was not sure if it would always work.
I.e. I have not found it documented that
  let strongSelf = self
will actually retain ‘self’ an not create a strong reference to an intermediate reference to self.

It makes sense though, otherwise there would be massive failures ‘out there’. ;-)

Yes :) local variables (as others too) are strong by default. If you want them weak (you should very rarely need that though) you have to explicitly mark them weak.

will actually retain ‘self’ an not create a strong reference to an intermediate reference to self.

Isn't that the same? I might misunderstand you but they would both reference the same object, no? So in the end the retain message would be sent to the same object, wouldn't it?

Thanks,
Rien.

Regards,
Rien

Site: http://balancingrock.nl <http://balancingrock.nl/&gt;
Blog: http://swiftrien.blogspot.com <http://swiftrien.blogspot.com/&gt;
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl <Indobet - Tempat Daftar Login Rtp Slot Gacor Indonesia; - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org <mailto:swift-users@swift.org>
https://lists.swift.org/mailman/listinfo/swift-users

- Dennis

- Dennis

···

On May 1, 2017, at 5:32 PM, Rien <Rien@Balancingrock.nl> wrote:

On 01 May 2017, at 16:59, Dennis Weissmann <dennis@dennisweissmann.me> wrote:

On May 1, 2017, at 4:46 PM, Rien via swift-users <swift-users@swift.org> wrote:

[Rien]

In my code I use a lot of queues. And (very often) I will use [weak self] to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to [weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

  if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as nil

Is that correct? i.e. should we never use the above if construct but always:

  return self?.myparam ?? 42

Yes, as you said, you never know when self will be nilled out, that's why you need to create a (temporary) strong reference to it, work on it, and when the block returns, your strong reference is released and self either goes away immediately (incase it was released elsewhere after the local binding to a strong variable and no other objects had a strong reference to it) or it will stay as long as no other object holds a strong reference to it.

When the closure is more involved than a view lines, I often do a guard let `self` = self else { return } to conveniently work with a strong self inside the rest of the closure.

I was aware of that practise (use it myself) but I was not sure if it would always work.
I.e. I have not found it documented that
  let strongSelf = self
will actually retain ‘self’ an not create a strong reference to an intermediate reference to self.

It makes sense though, otherwise there would be massive failures ‘out there’. ;-)

Yes :) local variables (as others too) are strong by default. If you want them weak (you should very rarely need that though) you have to explicitly mark them weak.

will actually retain ‘self’ an not create a strong reference to an intermediate reference to self.

Isn't that the same? I might misunderstand you but they would both reference the same object, no? So in the end the retain message would be sent to the same object, wouldn't it?

Just to belabour this point unnecessarily :)

If you are familiar with the compiler innards, then this might be obvious. However I am not, and I can imagine (crazy) scenario’s where an optional woud be “unwrapped” by simply taking the reference to the self from the optional and storing it in a new non-optional variable… :D … go figure… the mind works in devious ways… lol.

Rien.

···

On 01 May 2017, at 17:42, Dennis Weissmann <dennis@dennisweissmann.me> wrote:

On May 1, 2017, at 5:32 PM, Rien <Rien@Balancingrock.nl> wrote:

On 01 May 2017, at 16:59, Dennis Weissmann <dennis@dennisweissmann.me> wrote:

On May 1, 2017, at 4:46 PM, Rien via swift-users <swift-users@swift.org> wrote:

Thanks,
Rien.

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

- Dennis

- Dennis

[Zhao_Xin]

[weak self] and [unowned self] are used to solve the problem of Strong
Reference Cycles Between Class Instances
<The Swift Programming Language: Redirect;
.

If you can grantee self won't be nil during the life time, you should use
`unowned self` instead of `weak self`. If you can't grantee that, you
should use `weak self` and `return self?.myparam ?? 42`. I don't think
using ` if self != nil { return self!.myparam }` is a good idea. The reason
is said by you already.

Zhaoxin

···

On Mon, May 1, 2017 at 11:42 PM, Dennis Weissmann via swift-users < swift-users@swift.org> wrote:

On May 1, 2017, at 5:32 PM, Rien <Rien@Balancingrock.nl> wrote:

On 01 May 2017, at 16:59, Dennis Weissmann <dennis@dennisweissmann.me> > wrote:

On May 1, 2017, at 4:46 PM, Rien via swift-users <swift-users@swift.org> > wrote:

In my code I use a lot of queues. And (very often) I will use [weak self]
to prevent doing things when ‘self’ is no longer available.

Now I am wondering: how does the compiler know that [weak self] is
referenced?

I am assuming it keeps a reverse reference from self to the [weak self] in
order to ‘nil’ the [weak self] when self is nilled.

But when a job is executing it has no control over the exclusive rights to
[weak self].

I.e. [weak self] may be nilled by an outside event in the middle of say:

if self != nil { return self!.myparam }

The if finding [weak self] non nil, but the return finding [weak self] as
nil

Is that correct? i.e. should we never use the above if construct but
always:

return self?.myparam ?? 42

Yes, as you said, you never know when self will be nilled out, that's why
you need to create a (temporary) strong reference to it, work on it, and
when the block returns, your strong reference is released and self either
goes away immediately (incase it was released elsewhere after the local
binding to a strong variable and no other objects had a strong reference to
it) or it will stay as long as no other object holds a strong reference to
it.

When the closure is more involved than a view lines, I often do a guard
let `self` = self else { return } to conveniently work with a strong self
inside the rest of the closure.

I was aware of that practise (use it myself) but I was not sure if it
would always work.
I.e. I have not found it documented that
let strongSelf = self
will actually retain ‘self’ an not create a strong reference to an
intermediate reference to self.

It makes sense though, otherwise there would be massive failures ‘out
there’. ;-)

Yes :) local variables (as others too) are strong by default. If you want
them weak (you should very rarely need that though) you have to explicitly
mark them weak.

will actually retain ‘self’ an not create a strong reference to an
intermediate reference to self.

Isn't that the same? I might misunderstand you but they would both
reference the same object, no? So in the end the retain message would be
sent to the same object, wouldn't it?

Thanks,
Rien.

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: Balancingrock (Rien) · GitHub
Project: http://swiftfire.nl - A server for websites build in Swift

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users

- Dennis

- Dennis

_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users