执行time.Sleep（）函数时goroutine的状态是什么

Question

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.

Answer 1

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