I am stuck in a strange situation where the write operation to the channel never happens.
package main
import (
"fmt"
"time"
)
func main() {
c := make(chan int)
s := make(chan bool)
k := make(chan bool)
fmt.Println("start")
go func() {
fmt.Println("routine 1")
s <- true
}()
go func() {
fmt.Println("routine 2")
for {
select {
case <-s :
for i := 0; i < 3; i++ {
fmt.Println("before")
c <- i
fmt.Println("after")
}
case <- k :
fmt.Println("k ready")
break
}
}
}()
go func() {
fmt.Println("routine 3")
for {
select {
case x := <- c :
fmt.Println("x=", x)
k <- true
fmt.Println("k done")
}
}
}()
time.Sleep(1000 * time.Millisecond)
}
And here is the output:
start
routine 1
routine 2
before
routine 3
x= 0
after
before
I wonder why the write to the channel k blocks, but the log statement fmt.Println("k ready")
is never printed.
Here is what I think :
- go routine 1 writes true to channel s
- go routine 2 writes 0 to channel c and waits because buffer size is 0, it will not be able to write '1' to it unless someone reads channel c
- go routine 3 comes into the picture, reads channel c (now go routine 2 can write to c once go routine 2 resumes) prints the value of x. NOW IT SHOULD BE ABLE TO WRITE TO CHANNEL K but that is not happening
According to me it should be able to write to channel k then case 2 of goroutine should execute and print "k ready"
Can anyone explain me why write to the channel blocked? As a fix I know I can increase the buffer size of channel c and everything will get printed but I am not interested in fixing this, instead I want to understand this scenario.
A nice blog to understand the above case.