I read Twelve Go Best Practices and encounter and interesting example on page 30.
func sendMsg(msg, addr string) error {
conn, err := net.Dial("tcp", addr)
if err != nil {
return err
}
defer conn.Close()
_, err = fmt.Fprint(conn, msg)
return err
}
func broadcastMsg(msg string, addrs []string) error {
errc := make(chan error)
for _, addr := range addrs {
go func(addr string) {
errc <- sendMsg(msg, addr)
fmt.Println("done")
}(addr)
}
for _ = range addrs {
if err := <-errc; err != nil {
return err
}
}
return nil
}
func main() {
addr := []string{"localhost:8080", "http://google.com"}
err := broadcastMsg("hi", addr)
time.Sleep(time.Second)
if err != nil {
fmt.Println(err)
return
}
fmt.Println("everything went fine")
}
The programmer mentioned, that happens to the code above:
the goroutine is blocked on the chan write
the goroutine holds a reference to the chan
the chan will never be garbage collected
Why the goroutine is blocked here? The main thread is blocked, until it receive data from goroutine. After it continues the for loop. Not?
Why the errc chan will be never garbage collected? Because I do not close the channel, after goroutine is finished?