Reputation: 615
I'm curious about the status of goroutine when I execute the time.Sleep() function, for example:
func main() {
fmt.Println("before test")
time.Sleep(time.Second * 2)
fmt.Println("test")
}
if the goroutine would become the waiting state when execute the time.Sleep() function, how could the goroutine know when to change the state into the ready?
I really want to know the underlying mechanism of time.Sleep() here.
Upvotes: 1
Views: 870
Reputation: 63139
The state of the goroutine will be sleep
. There is very short program you can test it with:
package main
import (
"time"
)
func main() {
go func() {
time.Sleep(3 * time.Second)
}()
time.Sleep(1 * time.Second)
panic("foo")
}
Run it like that GOTRACEBACK=1 go run test.go
to get the state of all goroutines.
Output:
panic: foo
goroutine 1 [running]:
panic(0x45afa0, 0xc42006c000)
/usr/local/go/src/runtime/panic.go:500 +0x1a1
main.main()
/home/user/path/test.go:12 +0x96
goroutine 4 [sleep]:
time.Sleep(0xb2d05e00)
/usr/local/go/src/runtime/time.go:59 +0xe1
main.main.func1()
/home/user/path/test.go:9 +0x2b
created by main.main
/home/user/path/test.go:10 +0x39
exit status 2
Upvotes: 3