英文:
Parallelisation - Why does sleep pause only once?
问题
为什么只等待第一个goroutine，而其他的则直接执行。我认为这是因为通道c chan string，但我不理解两者之间的行为。
英文:
Why do wait only the first goroutine with
func Sleep(d Duration)
http://golang.org/pkg/time
"Sleep pauses the current goroutine for the duration d."
but the rest is execute directly. I think cause the channel c chan string but i dont understand the behaviour between both.
答案1
得分: 4
你的所有go例程都是并发运行的，所以它们都同时休眠4秒钟，因此它们同时结束。
你像这样调用它们。go意味着所有的getHostName函数同时运行。
for i := 0; i < max; i++ {
go getHostName(haveHost, ipadresse_3, i)
}
这意味着所有的休眠同时发生。
func getHostName(h chan string, ipAdresse string, n int) {
//这里！！！
time.Sleep(4 * time.Second)
ip := ipAdresse + strconv.Itoa(n)
//ip, ok := net.LookupAddr(ip)
if false {
h <- ip + " - "
} else {
h <- "error" + strconv.Itoa(n)
}
}
所以它们都同时完成。
英文:
All your go routines are running concurrently so they all sleep for 4 seconds at the same time, hence they all end at the same time.
You called them like this. The go means that all the getHostName functions run at once.
for i := 0; i < max; i++ {
go getHostName(haveHost, ipadresse_3, i)
}
And this means that all the sleeps happen at the same time
func getHostName(h chan string, ipAdresse string, n int) {
//Here!!!
time.Sleep(4 * time.Second)
ip := ipAdresse + strconv.Itoa(n)
//ip, ok := net.LookupAddr(ip)
if false {
h <- ip + " - "
} else {
h <- "error" + strconv.Itoa(n)
}
}
So they all finish at the same time.
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