Golang频道并发问题

Question

I am experimenting with channel concept in Golang. I wrote the below program playground to implement counter using channels. But I am not getting any output, although I am doing some printing in the goroutine body.

func main() {
    wg := sync.WaitGroup{}
    ch := make(chan int)

    count := func(ch chan int) {
        var last int
        last = <-ch
        last = last + 1
        fmt.Println(last)
        ch <- last
        wg.Done()
    }

    wg.Add(10)
    for i := 1; i <= 10; i++ {
        go count(ch)

    }
}

I expect at least some output but I am getting none at all.

Answer 1

When the main() function (the main goroutine) ends, your program ends as well (it doesn't wait for other non-main goroutines to finish). You must add a wg.Wait() call to the end. See No output from goroutine in Go.

Once you do this, you'll hit a deadlock. This is because all goroutines start with attempting to receive a value from the channel, and only then would they send something.

So you should first send something on the channel to let at least one of the goroutines to proceed.

Once you do that, you'll see numbers printed 10 times, and again deadlock. This is because when the last goroutine tries to send its incremented number, there will be no one to receive that. An easy way to fix that is to give a buffer to the channel.

Final, working example:

wg := sync.WaitGroup{}
ch := make(chan int, 2)

count := func(ch chan int) {
    var last int
    last = <-ch
    last = last + 1
    fmt.Println(last)
    ch <- last
    wg.Done()
}

wg.Add(10)
for i := 1; i <= 10; i++ {
    go count(ch)

}
go func() {
    ch <- 0
}()
wg.Wait()

Outputs (try it on the Go Playground):

1
2
3
4
5
6
7
8
9
10

Also note that since we made the channel buffered, it's not necessary to use another goroutine to send an initial value, we can do that in the main goroutine:

ch <- 0
wg.Wait()

This will output the same. Try it on the Go Playground.