> 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.
> Regards,
> Rien
>
> Site: http://balancingrock.nl
> Blog: http://swiftrien.blogspot.com
> Github: http://github.com/Balancingrock
> Project: http://swiftfire.nl - A server for websites build in Swift
>
>
>
>
>
>
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- Dennis
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