Thank you very much for your meticulous reply. I basically understand the scheduling mechanism, but I still don’t understand the third paragraph.
It is "It is also important to note that the behavior of runtime.GC() is not the exactly same as the normal backgound GC that gets run by the runtime. In particular runtime.GC() will block until GC is completed, whereas the noral gc does most of its work while allowing the program to continue running. " 在 2019年4月11日星期四 UTC+8上午12:12:01，jake...@gmail.com写道： > > As Ian said, your question is not completely clear. I think you are Saying > that "I got scheduled!" prints 3 times, then the program appears to "hangs" > on the runtime.GC() call. > > To start with, it is usually helpful to create a *minimal *program when > dealing with questions like this. Your program can be reduced to > https://play.golang.org/p/-9ekKOxyVoc: > package main > > import ( > "fmt" > "runtime" > "time" > ) > > func deadloop() { > for { > //runtime.Gosched() > } > } > func main() { > > go deadloop() > > i := 3 > for { > time.Sleep(time.Second * 1) > i-- > fmt.Println("I got scheduled!", i) > if i == 0 { > runtime.GC() //will set gcwaiting=1， causes other goroutines > not to be scheduled > } > } > } > This prints: > I got scheduled! 2 > I got scheduled! 1 > I got scheduled! 0 > Then hangs. > > I believe that what you are seeing is a result of the GC needing to > briefly pause all goroutines. Go uses something like cooperative > multithreading for goroutines. Your deadloop() function can not be > preempted under the current implentation of the go compiler. So the call to > runtime.GC() hangs waiting to pause all goroutines. If you un comment the > runtime.Gosched() call on line 11, then your program will work as expected. > > The details of premtion are implementation dependant, and work is being > done to improve the situation. There is a lot of information available, but > if you have a long running loop that makes no function calls, then it is a > good idea to call runtime.Gosched() every so often. > > It is also important to note that the behavior of runtime.GC() is not the > exactly same as the normal backgound GC that gets run by the runtime. In > particular runtime.GC() will block until GC is completed, whereas the noral > gc does most of its work while allowing the program to continue running. > > Hope that helps. > > > On Wednesday, April 10, 2019 at 7:51:52 AM UTC-4, mount...@gmail.com > wrote: >> >> Hi, >> >> My macos has 4 cores, but i start 3 goroutines, 2 sub goroutines, 1 main >> goroutine. >> After main goroutine exec 3 times output, triggering gc causes >> gcwaiting=1. >> Finally all goroutine blocking >> >> package main >> >> import ( >> "fmt" >> "log" >> "net/http" >> _ "net/http/pprof" >> "runtime" >> >> // "runtime" >> "time" >> ) >> >> func deadloop() { >> for { >> } >> } >> func main() { >> go func() { >> log.Println(http.ListenAndServe("localhost:6060", nil)) >> }() >> >> go deadloop() >> >> i := 3 >> for { >> time.Sleep(time.Second * 1) >> i-- >> fmt.Println("I got scheduled!") >> if i == 0 { >> runtime.GC() //will set gcwaiting=1， causes other goroutines not to be >> scheduled >> } >> } >> } >> > -- 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.
