dongshang6062 2016-01-20 08:26
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Golang超时未通过渠道执行

I am using goroutines/channels. Here is my code. Why is the timeout case not getting executed?

func main() {
    c1 := make(chan int, 1)

    go func() {
        for {
            time.Sleep(1500 * time.Millisecond)
            c1 <- 10
        }
    }()

    go func() {
        for {
            select {
            case i := <-c1:
                fmt.Println(i)
            case <-time.After(2000 * time.Millisecond):
                fmt.Println("TIMEOUT") // <-- Not Executed
            }
        }
    }()

    fmt.Scanln()
}
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1条回答 默认 最新

  • dongshao9106 2016-01-20 08:35
    关注

    Your timeout doesn't happen, because one of your goroutine sends a value on your c1 channel in every 1.5 seconds (or so) repeatedly, and your timeout would only happen if there is no value to be received from c1 for 2 seconds.

    Once a value is received from c1, in the next iteration executing select again a new time.After() call will be made which which returns a new channel on which a value will only be sent after another 2 seconds. The timeout channel from the previous select execution is discarded and is not used anymore.

    To receive the timeout after 2 seconds, create the timeout channel only once, e.g.:

    timeout := time.After(2000 * time.Millisecond)
    for {
        select {
        case i := <-c1:
            fmt.Println(i)
        case <-timeout:
            fmt.Println("TIMEOUT") // Will get executed after 2 sec
        }
    }
    

    Output:

    10
    TIMEOUT
    10
    10
    10
    ...
    
    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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