Re: [swift-users] unowned self in closure in a playground
2017-02-22 12:53 GMT+09:00 Ray Fix :
>
> On Feb 21, 2017, at 7:32 PM, rintaro ishizaki wrote:
>
> lifetime guaranteed to be around until the method call completes
>
>
> AFAIK, this is true.
> However, in this case, it's not a method call. The lifetime is only
> guaranteed until getter:increment completes.
>
> Demo().increment(3)
> |--| lifetime
>
>
> Thanks! This is an interesting explanation.
>
> It is curious that naming the variable cases different behavior.
> IOW, this doesn’t have a problem:
>
> let demo = Demo()
> demo.increment(3)
>
> Lifetime seems to last the entire expression.
>
>
I'm not sure, but I think the compiler is just lazy enough to release the "
demo" value *after* the entire expression or scope finishes.
Again, I'm not sure. :)
Anyway, If I were you, I would define a wrapper method:
class Demo {
var value = 0
lazy var _increment: (Int) -> Void { [unowned self] by in ... }
func increment(by: Int) {
_increment(by)
}
}
Demo().increment(by: 3)
> Ray
>
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Re: [swift-users] unowned self in closure in a playground
>> > The problem with this solution is that this changes the semantics. I want a > value for every Demo instance. My error... I read too fast and missed the () instantiation of Demo in the failing call. I was reading it as Demo.increment(). Never mind. ___ swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
Re: [swift-users] unowned self in closure in a playground
> On Feb 21, 2017, at 7:32 PM, rintaro ishizaki wrote: > > lifetime guaranteed to be around until the method call completes > > AFAIK, this is true. > However, in this case, it's not a method call. The lifetime is only > guaranteed until getter:increment completes. > > Demo().increment(3) > |--| lifetime > Thanks! This is an interesting explanation. It is curious that naming the variable cases different behavior. IOW, this doesn’t have a problem: let demo = Demo() demo.increment(3) Lifetime seems to last the entire expression. Ray___ swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
Re: [swift-users] unowned self in closure in a playground
Thanks. My comment below.
>
> On Feb 21, 2017, at 11:18 AM, Marco S Hyman wrote:
>
>> The following code crashes:
>>
>> class Demo {
>> var value = 0
>> lazy var increment: (Int) -> Void = { [unowned self] by in
>>self.value += by
>>print(self.value)
>> }
>> }
>>
>> Demo().increment(3)
>> error: Playground execution aborted: error: Execution was interrupted,
>> reason: EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0).
>
> value is not a static/class variable. Without an instance of Demo it does
> not exist. This seems to work.
>
> class Demo {
> static var value = 0
> lazy var increment: (Int) -> Void = { [unowned self] by in
>value += by
>print(value)
> }
> }
>
The problem with this solution is that this changes the semantics. I want a
value for every Demo instance.
Ray
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Re: [swift-users] unowned self in closure in a playground
>
> lifetime guaranteed to be around until the method call completes
AFAIK, this is true.
However, in this case, it's not a method call. The lifetime is only
guaranteed until getter:increment completes.
Demo().increment(3)
|--| lifetime
2017-02-22 3:00 GMT+09:00 Ray Fix via swift-users :
> Hi,
>
> The following code crashes:
>
> class Demo {
> var value = 0
> lazy var increment: (Int) -> Void = { [unowned self] by in
> self.value += by
> print(self.value)
> }
> }
>
> Demo().increment(3)
> error: Playground execution aborted: error: Execution was interrupted,
> reason: EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0).
>
> However if I call it this way:
>
> do {
> let demo = Demo()
> demo.increment(3)
> }
>
> All is well. This breaks my mental model of lifetime guaranteed to be
> around until the method call completes.
> Is it me that is wrong or the playground. Is the second way working just
> by luck?
>
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>
>
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Re: [swift-users] unowned self in closure in a playground
I think you are right on ` lifetime guaranteed to be around until the
method call completes`. But`[unowned self]` released the retain manually in
your code. Just removing `[unowned self]` part, you code will work.
Zhaoxin
On Wed, Feb 22, 2017 at 3:18 AM, Marco S Hyman via swift-users <
swift-users@swift.org> wrote:
> > The following code crashes:
> >
> > class Demo {
> > var value = 0
> > lazy var increment: (Int) -> Void = { [unowned self] by in
> > self.value += by
> > print(self.value)
> > }
> > }
> >
> > Demo().increment(3)
> > error: Playground execution aborted: error: Execution was interrupted,
> reason: EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0).
>
> value is not a static/class variable. Without an instance of Demo it
> does not exist. This seems to work.
>
> class Demo {
> static var value = 0
> lazy var increment: (Int) -> Void = { [unowned self] by in
> value += by
> print(value)
> }
> }
>
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Re: [swift-users] unowned self in closure in a playground
> The following code crashes:
>
> class Demo {
> var value = 0
> lazy var increment: (Int) -> Void = { [unowned self] by in
> self.value += by
> print(self.value)
> }
> }
>
> Demo().increment(3)
> error: Playground execution aborted: error: Execution was interrupted,
> reason: EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0).
value is not a static/class variable. Without an instance of Demo it does not
exist. This seems to work.
class Demo {
static var value = 0
lazy var increment: (Int) -> Void = { [unowned self] by in
value += by
print(value)
}
}
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[swift-users] unowned self in closure in a playground
Hi,
The following code crashes:
class Demo {
var value = 0
lazy var increment: (Int) -> Void = { [unowned self] by in
self.value += by
print(self.value)
}
}
Demo().increment(3)
error: Playground execution aborted: error: Execution was interrupted, reason:
EXC_BREAKPOINT (code=EXC_I386_BPT, subcode=0x0).
However if I call it this way:
do {
let demo = Demo()
demo.increment(3)
}
All is well. This breaks my mental model of lifetime guaranteed to be around
until the method call completes.
Is it me that is wrong or the playground. Is the second way working just by
luck?___
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