dtgj8529 2018-01-08 10:21
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开始,将数据传递到通道

I have an issue. Here is example: https://play.golang.org/p/QSWY2INQuSE

func Avg(c chan string, wg *sync.WaitGroup) {
    defer wg.Done()
    c <- "test"
}

func main() {
    var wg sync.WaitGroup
    c := make(chan string)

    timer1 := time.NewTicker(5 * time.Second)

    for {
        select {
        case <-timer1.C:
            wg.Add(1)
            go Avg(c, &wg)
            wg.Wait()
        }
    }
    fmt.Println(<-c)
}

Why data does not reach fmt.Println(<-c)

Thank you!

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1条回答 默认 最新

  • duanfuxing2417 2018-01-08 10:24
    关注

    Because you have an endless for, so the last fmt.Println() statement is never reached.

    You have to break out of the loop if you want the last fmt.Println() statement to ever execute, for example:

    loop:
        for {
            select {
            case <-timer1.C:
                wg.Add(1)
                go Avg(c, &wg)
                wg.Wait()
                break loop
            }
        }
        fmt.Println(<-c)
    

    Note that you have to use a label, else the break would only break out of the select statement (and not from the for loop).

    Also note that this alone won't work, as the channel is unbuffered, and thus Avg() will be blocked forever, trying to send a value on c while noone is ever trying to receive from it.

    This simple example can be made working if you create the channel to be buffered:

    c := make(chan string, 1) // Buffer for 1 value
    

    Now it works and prints (try it on the Go Playground):

    test
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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