Golang channel output order

Question

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


        

Answer 1

I just went through this same thing. See my post here: Golang channels, order of execution

Like you, I saw a pattern that was counter-intuitive. In a place where there actually shouldn't be a pattern. Once you launch a go process, you have launched a thread of execution, and basically all bets are off at that point regarding the order that the threads will execute their steps. But if there was going to be an order, logic tells us the first one called would be executed first.

In actual fact, if you recompile that program each time, the results will vary. That's what I found when I started compiling/running it on my local computer. In order to make the results random, I had to "dirty" the file, by adding and removing a space for instance. Then the compiler would re-compile the program, and then I would get a random order of execution. But when compiled in the go sandbox, the result was always the same.

When you use the sandbox, the results are apparently cached. I couldn't get the order to change in the sandbox by using insignificant changes. The only way I got it to change was to issue a time.Sleep(1) command between the launching of the go statements. Then, the first one launched would be the first one executed every time. I still don't think I'd bet my life on that continuing to happen though because they are separate threads of execution and there are no guarantees.

The bottom line is that I was seeing a deterministic result where there should be no determinism. That's what stuck me. I was fully cleared up when I found that the results really are random in a normal environment. The sandbox is a great tool to have. But it's not a normal environment. Compile and run your code locally and you will see the varying results you would expect.

Answer 2

In Golang Spec Channels order is described as:-

Channels act as first-in-first-out queues. For example, if one goroutine sends values on a channel and a second goroutine receives them, the values are received in the order sent.

It will prints which value is available first to be received on other end. If you wants to synchronize them use different channels or add wait Groups.

package main

import (
    "fmt"
)

func main() {
  messages1 := make(chan string)
  messages2 := make(chan string)
  go func(<-chan string) {
        messages2 <- "ping" 
    }(messages2)
  go func(<-chan string) {
        messages1 <- "hello" 
    }(messages1)
  fmt.Println(<-messages1)
  fmt.Println(<-messages2)
}

If you see you can easily receive any value you want according to your choice using different channels.

Go playground

Answer 3

It’s not about order of reading a channel but about order of goroutines execution which is not guaranteed.

Try to ‘Println’ from the function where you are writing to the channel (before and after writing) and I think it should be in same order as reading from the channel.

Answer 4

I was confused when I first met this. but now I am clear about this. the reason cause this is not about channel, but for goroutine.

As The Go Memory Model mention, there's no guaranteed to goroutine's running and exit, so when you create two goroutine, you cannot make sure that they are running in order.

So if you want printing follow FIFO rule, you can change your code like this:

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