在这个例子中如何理解goroutines的执行？

Question

i am trying to understand go routines and how it works. in the below example i have two go routines each sending some messege via channels.i was expecting the channel ch will send messege first but why is go readword(ch) execute after go timeout(t). if i change the order of the go routines call inside main function then readword(ch) will execute first.i am getting very confused about go routines? any help?

func readword(ch chan string) {

    fmt.Println("Type a word, then hit Enter.")
    var word string
    fmt.Scanf("%s", &word)
    ch <- word
}

func timeout(t chan bool) {
    t <- false
}

func main() {
    ch := make(chan string)
    go readword(ch)

    t := make(chan bool)
    go timeout(t)

    select {
    case word := <-ch:
        fmt.Println("Received", word)
    case <-t:
        fmt.Println("Timeout.")
    }
}

Answer 1

There is no execution ordering guarantee between goroutines unless they communicate with each other using synchronization mechanisms such as channels. You create two goroutines, then wait until you receive a string from ch, or receive from t. There is no explicit synchronization between the two goroutines, so they can execute in any order in a single thread, or simultaneously on different threads. The select statement will read from the first channel that has something in it.