[The other responses have been excellent, but I wanted to point one thing
out.]
On Fri, Feb 22, 2019 at 3:17 AM Jingguo Yao <yaojing...@gmail.com> wrote:
> Consider the following code which launches two goroutines. Each
> goroutine uses a for loop to print messages.
>
> package main
>
> import (
> "fmt"
> "runtime"
> "time"
> )
>
> func main() {
> count := runtime.GOMAXPROCS(1)
> fmt.Printf("GOMAXPROCS: %d\n", count)
>
> // goroutine 1
> go func() {
> for {
> fmt.Printf("1")
> }
> }()
>
> // goroutine 2
> go func() {
> for {
> fmt.Printf("2")
> }
> }()
>
> time.Sleep(20 * time.Second)
> }
>
> The code uses runtime.GOMAXPROCS(1) to make the Go runtime has only
> one processor (P). So there is only one OS thread (M) to run
> goroutines. Without preemption, the M will run goroutine 1 or
> goroutine 2 continuously. The messages sent to the terminal should be
> all 1s or all 2s. But running the code prints interleaved 1s and 2s to
> the terminal, which means that some preemption happens. But
>
> https://www.quora.com/How-does-the-golang-scheduler-work/answer/Ian-Lance-Taylor
> says that:
>
> > A G can only be pre-empted at a safe point, which in the current
> > implementation can only happen when the code makes a function call.
>
> Since both goroutines do not make any function calls, there should be
> some other way to preempt a G. Can anyone give me an explanation of
> such kind of preemption?
>
Not only is fmt.Printf a function call, but it makes a blocking syscall
(write specifically),
which gives the go runtime license to spin up a new OS thread and
immediately schedule the other goroutine.
runtime.GOMAXPROCS(1) only tries to guarantee that one OS thread executing
Go code will be executing at a time.
(this has no bearing on the revised version without a call to fmt.Printf().)
>
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