On Fri, Feb 22, 2019 at 9:17 AM Jingguo Yao <yaojing...@gmail.com> wrote:
> Without preemption, the M will run goroutine 1 or goroutine 2 continuously. Not correct, there's also cooperative scheduling/yielding. > The messages sent to the terminal should be all 1s or all 2s. Nothing in the language specs mandates such undesirable behavior. > Since both goroutines do not make any function calls, there should be some other way to preempt a G. Both of the goroutines _do_ call fmt.Printf. > Can anyone give me an explanation of such kind of preemption? Cooperative scheduling happens on function entry*. At least that was the strategy last time I checked. Since that time I think I've heard about plans for doing it preemptively, but I don't know what's the current status. *: Actually, the compiler is free to insert such cooperative check at any point. Another place where it can be inserted is a loop. But again, I don't know if it is happening in this release, if it's still a only plan or an already abandoned plan. -- -j -- 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. For more options, visit https://groups.google.com/d/optout.
