I think I got the reason.
It is because the compiler is not aware of the println will not change the
value of y.
And it is hard or expensive for the compiler to understand the channel
synchronization.
If the println function changes y, then y must be allocated on heap.
On Tuesday, June 1, 2021 at 9:51:50 AM UTC-4 tapi...@gmail.com wrote:
>
> package main
>
> func newIntPtr(n int) *int {
> return &n
> }
>
> func main() {
> x := newIntPtr(3)
> y := newIntPtr(5)
> c := make(chan bool)
> go func() {
> *y++
> close(c)
> }()
> <-c
> println(*x, *y)
> println(&x)
> //println(&y) // This line makes y escape.
> }
>
>
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to golang-nuts+unsubscr...@googlegroups.com.
To view this discussion on the web visit
https://groups.google.com/d/msgid/golang-nuts/6829f596-8801-4494-a160-de379f33aaf5n%40googlegroups.com.
