In the new example, goroutine 2 should get pre-empted at time.Sleep.
goroutine 1 may hog a CPU however. There's an open issue about this,
which is being worked on.
https://github.com/golang/go/issues/10958
I think it's always been the intent that from the programmer's
perspective goroutines don't block one another at all; they're just
like threads but cheaper -- and indeed most of the time you can write
code that acts like that's true and never see a problem -- but the
reality of the implementation is that right now it is possible to
tight-loop and block other goroutines.
-Ian
Quoting Jingguo Yao (2019-02-22 03:31:55)
> Yes, my fault. Thanks for pointing it out.�
> What happens if I remove the fmt.Printf function calls to have the
> following code:
> package main
> import (
> "fmt"
> "runtime"
> "time"
> )
> func main() {
> count := runtime.GOMAXPROCS(1)
> fmt.Printf("GOMAXPROCS: %d\n", count)
> � // goroutine 1
> go func() {
> for {
> }
> }()
> � // goroutine 2
> go func() {
> for {
> }
> }()
> time.Sleep(10 * time.Second)
> }
> Can goroutine 1 or goroutine 2 be preempted? If yes, by which
> mechanism?
> On Friday, February 22, 2019 at 4:20:12 PM UTC+8, Ian Denhardt wrote:
>
> Quoting Jingguo Yao (2019-02-22 03:17:33)
> > � � Since both goroutines do not make any function calls
> fmt.Printf is a function.
>
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