On Sunday, 29 January 2017 01:25:20 UTC+11, T L wrote:
>
>
>
> On Saturday, January 28, 2017 at 9:33:51 PM UTC+8, C Banning wrote:
>>
>> From the doc: "The finalizer for obj is scheduled to run at some
>> arbitrary time after obj becomes unreachable. There is no guarantee that
>> finalizers will run before a program exits, so typically they are useful
>> only for releasing non-memory resources associated with an object during a
>> long-running program."
>>
>
>
> If this is true, then the SetFinalizer function would be much meaningless.
>
Yes. The only reasonable way to interpret the operation of SetFinalizer is
to assume it does not thing and program accordingly.
To repeat, you cannot base the correct operation of your program on a
finaliser running before your program exits.
>
> BTW, it looks both finalizer will get executed if let the program sleep
> more than 10us at the beginning:
>
> package main
>
> import "time"
> import "runtime"
>
> func main() {
> time.Sleep(10 * time.Microsecond)
> // if sleep 10 Microsecond here, 2 and 1 will both be out
> // if sleep 1 Microsecond here, still only 2 will out
>
> t1 := new(int)
> t2 := new(int)
>
> runtime.SetFinalizer(t1, func(*int) {println(1)})
> runtime.SetFinalizer(t2, func(*int) {println(2)})
> runtime.GC()
> time.Sleep(time.Second * 20)
> }
>
>
>>
>>
>> On Saturday, January 28, 2017 at 6:17:37 AM UTC-7, T L wrote:
>>>
>>>
>>>
>>> package main
>>>
>>> import "time"
>>> import "runtime"
>>>
>>> type T1 struct{ i int }
>>> type T2 struct{ i int }
>>>
>>> func main() {
>>> t1 := new(T1)
>>> t2 := new(T2)
>>>
>>> runtime.SetFinalizer(t1, func(*T1) {println(1)})
>>> runtime.SetFinalizer(t2, func(*T2) {println(2)})
>>> runtime.GC()
>>> time.Sleep(time.Second * 2)
>>>
>>> // the program will output: 2
>>> // if I adjust the order of the declarations of t1 and t2,
>>> // the program will output: 1
>>> }
>>>
>>> Why the finalizer for the first declaration will not get called?
>>>
>>
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