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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