With time.Sleep, it always prints:
> 31
> 37
> 47
> 47
> 37
> 31
>
Why is each of the finalizers called twice?
On Saturday, April 25, 2020 at 1:08:43 AM UTC-4, Ian Lance Taylor wrote:
>
> On Fri, Apr 24, 2020 at 9:30 PM T L <tapi...@gmail.com <javascript:>>
> wrote:
> >
> > The following program might print:
> >
> >> 31
> >> 37
> >> 47
> >> 47
> >> 37
> >> 31
> >
> >
> > or
> >
> >> 31
> >> 37
> >> 47
> >> 47
> >
> >
> > but with the highest possibility to print
> >
> >> 31
> >> 37
> >> 47
> >
> >
> > Is not normal?
> >
> >
> > package main
> >
> > import (
> > "fmt"
> > "math/rand"
> > "runtime"
> > "time"
> > "unsafe"
> > )
> >
> > type Foo struct {
> > x int
> > a int
> > }
> >
> > func main() {
> > for i := 0; i < 3; i++ {
> > f := NewFoo(i)
> > println(f.a)
> > }
> >
> > runtime.GC()
> >
> > time.After(time.Second)
> > }
> >
> > func do(i *uint) {
> > runtime.SetFinalizer(i, func(f *uint) {
> > fmt.Println(*f)
> > })
> > }
> >
> > //go:noinline
> > func NewFoo(i int) *Foo {
> > f := &Foo{a: rand.Intn(50)}
> >
> > do((*uint)(unsafe.Pointer(&f.a)))
> >
> > return f
> > }
>
> Finalizers are not guaranteed to run.
>
> That said, I'm not sure why you are calling time.After. That doesn't
> make finalizers any more likely to run. You'll probably see fairly
> reliable results if you change that to call time.Sleep instead.
>
> Ian
>
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