[swift-users] weak self
Guillaume Lessard
glessard at tffenterprises.com
Mon May 1 16:26:52 CDT 2017
Hi Rien,
> On May 1, 2017, at 08:46, Rien via swift-users <swift-users at 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>. 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
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