Also what makes you think that ordering is mandated? Any of the goroutines
can be scheduled to run. The odds of 1 running first seems intuitively
higher but the should be nothing stopping the last to run first.
On Fri, Nov 11, 2016, 07:11 Henrik Johansson <dahankz...@gmail.com> wrote:
> I think there is a race on the WaitGroup. Add all of them before the loop.
> It might not matter for what you are seeing but anyway.
>
> What happens with an unbuffered channel?
>
> On Fri, Nov 11, 2016, 07:08 <mspauldin...@gmail.com> wrote:
>
> Hello,
>
> I've written a small program to demonstrate what I am seeing. If I use a
> channel as a semaphore and set the size of the channel to 1, I would expect
> that the executions of my handle function below would all be executed in
> order, but it's not. For some reason the last item in the list is always
> handled first, and then the remaining items are handled in order. Does
> anyone know why I see this behavior?
>
> Program:
>
> package main
>
> import (
> "fmt"
> "sync"
> "time"
> )
>
> var wg sync.WaitGroup
> var sem chan int
>
> func handle(i int) {
> defer wg.Done()
> fmt.Println("waiting: ", i)
> sem <- 1
> fmt.Println("processing: ", i)
> time.Sleep(1 * time.Second)
> <-sem
> }
>
> func main() {
> sem = make(chan int, 1)
> nums := []int{1, 2, 3, 4}
>
> for _, i := range nums {
> wg.Add(1)
> go handle(i)
> }
> wg.Wait()
> }
>
> Output:
>
> waiting: 4
> processing: 4
> waiting: 1
> waiting: 2
> waiting: 3
> processing: 1
> processing: 2
> processing: 3
>
>
>
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