Goroutine Puzzle #1

I came across this “puzzle” when I was studying termbox-go source code and I can’t wrap my head around it. So I wrote a very simple package to test it out. Run it in your head first before you scroll further for answer and my naive reasoning.

package main
import "fmt"
func foo(c chan int) {
  a := 0
    for {
      select {
      case c <- a:
        a = <-c
        fmt.Println("foo received", a)
    }
  }
}
func main() {
  c := make(chan int)
  go foo(c)
  loop:
   for {
     select {
     case r := <-c:
       fmt.Println("main received", r)
       c <- (r + 1)
       if r == 3 {
         break loop
       }
    }
  }
  <-c
}

Please forgive this noob if you found this is too easy. I am not yet clock > 3 hours coding in GO yet, bear with me. So the answer is…

main received 0
foo received 1
main received 1
foo received 2
main received 2
foo received 3
main received 3
foo received 4

It was not obvious to me at the start when I was looking at the foo's select. I thought case c <- a will be selected, based on the wrong assumption that c channel is “empty” now and I can send 1 value inside. I pondered, searched, put more printf and finally this answer from StackOverflow enlightened me. In short…

sends do not complete until there is a receiver to accept the value.

Love GO.