Why is only one channel value displayed?

[Ilya_AGL]

I found this code:

package main
 
import "fmt"
 
func greet(c chan string) {
    fmt.Println(<-c) // for John
    fmt.Println(<-c) // for Mike
}
 
func main() {
 
    c := make(chan string)
 
    go greet(c)
    c <- "John"
 
    c <- "Mike"
 
    //close(c) // closing channel
 
}

In the output there is only “John” without “Mike”. And if I write something like c ← “Adam” with the third fmt.Println(<-c) in the greet function, in the output there will be “John”, “Mike” and “Adam”. Why is it happening?

[Dean_Davidson]

This is happening because main() is exiting before both fmt.Println calls can complete. Just to illustrate this, add a time.Sleep(20 * time.Millisecond) or some arbitrary time after c <- "Mike". This will give greet ample time to print all values.

When you’re dealing with concurrency, you need something to sync/block until your concurrent code has time to execute. Check sync.WaitGroup out for an example of concurrent tasks where the main thread blocks until all have completed. Other useful links:

• Go by Example: Channel Synchronization
• Go by Example: WaitGroups

The channel synchronization pattern above of passing in done chan bool is pretty standard and I’ve seen that plenty of places in the wild.