The way to guarantee that you wait until goroutines are finished is by using sync.WaitGroup.
See https://www.programming-books.io/essential/go/139-wait-for-goroutines-to-finish for example and longer description. Alternatively, use range over a channel and arrange the code so that you close the channel once when all sends are done. See https://www.programming-books.io/essential/go/144-closing-channels On Saturday, March 24, 2018 at 3:41:19 AM UTC-7, T L wrote: > > In the following example, there are 99 goroutines queuing and blocking on > sending a value to c. > When the only buffered value is received, it looks there is a time > interval until the buffer is filled. > Shouldn't it be that the receive from the only buffer and fill next > queuing value to the only buffer in one atomic operation? > > > package main > > import "time" > > func main() { > c := make(chan int, 1) > for i := 0; i < 100; i++ { > go func() { > c <- 1 > }() > } > > time.Sleep(time.Second) > > n := 0 > for len(c) == cap(c) { > <-c > println(n, len(c)) // here, len(c) is always 0 > n++ > // time.Sleep(time.Second/1000) // if this line is not commented > off, there will be 100 lines output. > } > } > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
