Wasn't this already answered for you
here?
https://stackoverflow.com/questions/43152776/is-it-one-channel-ops-affect-another-channel-ops
On Saturday, April 1, 2017 at 1:35:12 PM UTC-4, Khalid Adisendjaja wrote:
>
> I made this simple script, trying to understand how is channel working,
> somehow if channel "c" is sent after channel "b" is sent the last routine
> is not being proceed,
> please see the runner function
>
> package main
>
> import (
> "fmt"
> "strconv"
> "time"
> )
>
> func runner(idx int, c chan []int, b chan []int) {
> var temp []int
> fmt.Println("runner " + strconv.Itoa(idx))
> bucket := <-b
> for k, v := range bucket {
> if v != 0 {
> temp = append(temp, v)
> bucket[k] = 0
> }
> if len(temp) == 5 {
> break
> }
> }
>
> //Strange condition if channel c is sent after channel b is sent,
> //somehow the last runner is not being proceed
> b <- bucket
> c <- temp
>
> //All runner ara all proceed if c is sent first
> // c <- temp
> // b <- bucket
>
> }
>
> func printer(c chan []int) {
> for {
> select {
> case msg := <-c:
> fmt.Println(msg)
> time.Sleep(time.Second * 1)
> }
> }
> }
>
> func main() {
>
> c := make(chan []int, 5)
> bucket := make(chan []int)
>
> go runner(1, c, bucket)
> go runner(2, c, bucket)
> go runner(3, c, bucket)
> go runner(4, c, bucket)
>
> bucket <- []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
> 16, 17, 18, 19, 20}
>
> go printer(c)
>
> var input string
> fmt.Scanln(&input)
>
> }
>
>
>
>
>
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