What is the correct use of select statement in Go? - select

Question

Good Day everyone
I have been learning the fundaments of go and how to use its channel-based concurrency paradigm.
However, while playing with some code I wrote focusing on the select statement I found a strange behavior:
func main() {
even := make(chan int)
odd := make(chan int)
quit := make(chan bool)
//send
go send(even, odd, quit)
//receive
receive(even, odd, quit)
fmt.Println("Exiting")
}
func send(e, o chan<- int, q chan<- bool) {
for i := 0; i < 100; i++ {
if i%2 == 0 {
e <- i
} else {
o <- i
}
}
close(e)
close(o)
q <- true
close(q)
}
func receive(e, o <-chan int, q <-chan bool) {
for cont, i := true, 0; cont; i++ {
fmt.Println("value of i", i, cont)
select {
case v := <-e:
fmt.Println("From even channel:", v)
case v := <-o:
fmt.Println("from odd channel:", v)
case v := <-q:
fmt.Println("Got exit message", v)
// return // have also tried this instead
cont = false
}
}
}
when I run this simple program sometimes the i accumulator ends up with more than a 100 being printed to the console, and instead of finishing up with a "from odd channel: 99", the for loop continues on outputting one or more zeroes from even/odd channels randomly, as if the quit channel's message was being somewhat being delayed onto its case and instead the odd/even channels were sending more integers thus quitting the for loop not exactly after the odd/even channels have been closed.
value of i 97 true
from odd channel: 97
value of i 98 true
From even channel: 98
value of i 99 true
from odd channel: 99
value of i 100 true
From even channel: 0
value of i 101 true
From even channel: 0
value of i 102 true
from odd channel: 0
value of i 103 true
From even channel: 0
value of i 104 true
Got exit message true
Exiting
I have tried to search for the correct use of the case statement but I haven´t been able to find the problem with my code.
It seems like the same behavior can be reproduced on the go playground: my code
thanks for your attention put on my question.

Answer 1

The program is printing 0 because receive on a closed channel returns the zero value. Here's one way to accomplish your goal.
First, eliminate the q channel. Closing the o and e channels is sufficient to indicate that the sender is done.
func send(e, o chan<- int, q chan<- bool) {
for i := 0; i < 100; i++ {
if i%2 == 0 {
e <- i
} else {
o <- i
}
}
close(e)
close(o)
}
When receiving values, use the two value receive to detect when the zero value is returned because the channel is closed. Set the channel to nil when the channel is closed. Receive on a nil channel does not yield a value. Loop until both channels are nil.
func receive(e, o <-chan int, q <-chan bool) {
for e != nil && o != nil {
select {
case v, ok := <-e:
if !ok {
e = nil
continue
}
fmt.Println("From even channel:", v)
case v, ok := <-o:
if !ok {
o = nil
continue
}
fmt.Println("From odd channel:", v)
}
}
}
playground example