go - Golang channel output order

Question

func main() {
  messages := make(chan string)
  go func() { messages <- "hello" }()
  go func() { messages <- "ping" }()
  msg := <-messages
  msg2 := <-messages
  fmt.Println(msg)
  fmt.Println(msg2)

The above code consistently prints "ping" and then "hello" on my terminal. I am confused about the order in which this prints, so I was wondering if I could get some clarification on my thinking.

I understand that unbuffered channels are blocking while waiting for both a sender and a receiver. So in the above case, when these 2 go routines are executed, there isn't, in both cases,a receiver yet. So I am guessing that both routines block until a receiver is available on the channel.

Now... I would assume that first "hello" is tried into the channel, but has to wait... at the same time, "ping" tries, but again has to wait. Then

msg := <- messages

shows up, so I would assume that at that stage, the program will arbitrarily pick one of the waiting goroutines and allow it to send its message over into the channel, since msg is ready to receive.

However, it seems that no matter how many times I run the program, it always is msg that gets assigned "ping" and msg2 that gets assigned "hello", which gives the impression that "ping" always gets priority to send first (to msg). Why is that?

Answer 1

Waitting for answers