Yes you are right! Unbuffered channel is solid!
package main
import (
"fmt"
"time"
)
var c = make(chan int)
func report(){
for{
select{
case n:= <- c:
fmt.Println(n, time.Now().Format("2006-01-02 15:04:05"))
default:
An unbuffered channel is a synchronization mechanism. A send on an
unbuffered channel will block until there is a concurrent receive from
another goroutine.
Your program has only one goroutine. A send to an unbuffered channel will
always deadlock, so will a receive from it. So the problem you are
There are numerous ways to create a “dead lock” in any program (this may
actually be a “live lock” but I didn’t fully understand your statement - this
is just one of them.
> On Apr 11, 2022, at 9:29 PM, Zhaoxun Yan wrote:
>
> Hi guys, I have a great demonstration on why an un-cached channel
Hi guys, I have a great demonstration on why an un-cached channel might
malfunction while receiving:
package main
import(
"fmt"
"time"
)
func main(){
t := time.NewTicker(time.Second * 3)
stopnow := make(chan bool)
//stopnow := make(chan bool, 1) //cached channel
