I am following the examples on tour.golang.org.
I understand the example mostly, the only issue I have is why does it stop when we pass 0 to quit channel? Regardless of whether 0 was passed to quit, there is always a value for x. So shouldn't select always fall on case 'c <- x' ?
func fibonacci(c chan int, quit chan int) {
x, y := 0, 1
for {
select {
case c <- x:
x, y = y, x+y
case <-quit:
return
}
}
close(c)
}
func main() {
c := make(chan int)
quit := make(chan int)
go func() {
for i := 0; i < 10; i++ {
fmt.Println(<-c)
}
quit <- 0
}()
fibonacci(c, quit)
}
there is always a value for x. So shouldn't select always fall on case 'c <- x' ?
No, because this channel is unbuffered, the send will block until someone can receive from it.
Read about channels on Effective Go:
Receivers always block until there is data to receive. If the channel is unbuffered, the sender blocks until the receiver has received the value. If the channel has a buffer, the sender blocks only until the value has been copied to the buffer; if the buffer is full, this means waiting until some receiver has retrieved a value.
Additionally, if 2 cases in a select statement could proceed, one is picked pseudo-randomly.
If one or more of the communications can proceed, a single one that can proceed is chosen via a uniform pseudo-random selection. Otherwise, if there is a default case, that case is chosen. If there is no default case, the "select" statement blocks until at least one of the communications can proceed.