Yes, this example is correct. It is important that the wg.Add() happens before the go statement to prevent race conditions. The following would also be correct: func main() { var wg sync.WaitGroup wg.Add(1) go dosomething(200, &wg) wg.Add(1) go dosomething(400, &wg) wg.Add(1) go dosomething(150, &wg) wg.Add(1) go dosomething(600, &wg) wg.Wait() fmt.Println(“Done”) } However, it is rather … Read more
I would spawn 4 worker goroutines that read the tasks from a common channel. Goroutines that are faster than others (because they are scheduled differently or happen to get simple tasks) will receive more task from this channel than others. In addition to that, I would use a sync.WaitGroup to wait for all workers to … Read more
The Go select statement is not biased toward any (ready) cases. Quoting from the spec: If one or more of the communications can proceed, a single one that can proceed is chosen via a uniform pseudo-random selection. Otherwise, if there is a default case, that case is chosen. If there is no default case, the … Read more
You can use sync.WaitGroup. Quoting the linked example: package main import ( “net/http” “sync” ) func main() { var wg sync.WaitGroup var urls = []string{ “http://www.golang.org/”, “http://www.google.com/”, “http://www.somestupidname.com/”, } for _, url := range urls { // Increment the WaitGroup counter. wg.Add(1) // Launch a goroutine to fetch the URL. go func(url string) { // … Read more
Simplest, cleanest and “scalable” way to do it is to use a sync.WaitGroup: var wg = &sync.WaitGroup{} func printElo() { defer wg.Done() fmt.Printf(“Elo\n”) } func printHello() { defer wg.Done() fmt.Printf(“Hello\n”) } func main() { fmt.Printf(“This will print.”) i := 0 for i < 10 { wg.Add(1) go printElo() wg.Add(1) go printHello() i++ } wg.Wait() } … Read more