Why Does Golang's Unbuffered Channel Output Order Seem Counterintuitive?

Understanding Channel Output Order in Golang

In Golang, unbuffered channels provide a reliable communication mechanism between goroutines. However, their output order can be confusing at times. Consider the following code:

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

When running this code, it consistently prints "ping" followed by "hello." This may seem counterintuitive, as one might expect the output to match the order in which the goroutines were created.

To understand why this occurs, it's crucial to remember that channels are communication pipelines, while goroutines represent parallel tasks. The execution order of goroutines is not guaranteed, even if they were created in a specific sequence.

In this case, the scheduler may have scheduled the goroutine that sends "ping" before the one that sends "hello." When the first receiver (i.e., msg := <-messages) becomes available, it will accept the message from the already-scheduled "ping" goroutine.

Upon receiving the first message, the program will continue executing until the second receiver (msg2 := <-messages) becomes available. At this point, the "hello" message is ready to be sent, and it will be accepted as the second message.

Therefore, the output order is determined by the execution order of the goroutines that write to the channel, not the order in which they are created or the order in which the messages are read from the channel. To verify this, one can add a Println statement to the goroutines, as suggested in the answer provided:

...
func() { messages <- "hello"; fmt.Println("Hello sent") }()
func() { messages <- "ping"; fmt.Println("Ping sent") }()
...

This will output:

Hello sent
Ping sent
ping
hello

This confirms that the messages were sent and printed in the same order.

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