I have a slice of integers, which are manipulated concurrently:
ints := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
I'm using a buffered channel as semaphore in order to have a an upper bound of concurrently running go routines:
sem := make(chan struct{}, 2)
for _, i := range ints {
// acquire semaphore
sem <- struct{}{}
// start long running go routine
go func(id int, sem chan struct{}) {
// do something
// release semaphore
<- sem
}(i, sem)
}
The code above works pretty well until the last or last two integers are reached, because the program ends before those last go routines are finished.
Question: how do I wait for the buffered channel to drain?
See Question&Answers more detail:
os 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…