Goroutines频道和"停止短片"

时间:2016-01-04 01:53:09

标签: go concurrency goroutine

我正在阅读/工作Go Concurrency Patterns: Pipelines and cancellation,但我无法理解停止短片部分。我们有以下功能:

func sq(in <-chan int) <-chan int {
    out := make(chan int)
    go func() {
        for n := range in {
            out <- n * n
        }
        close(out)
    }()
    return out
}

func gen(nums ...int) <-chan int {
    out := make(chan int)
    go func() {
        for _, n := range nums {
            out <- n
        }
        close(out)
    }()
    return out
}
func merge(cs ...<-chan int) <-chan int {
    var wg sync.WaitGroup
    out := make(chan int, 1) // enough space for the unread inputs

    // Start an output goroutine for each input channel in cs.  output
    // copies values from c to out until c is closed, then calls wg.Done.
    output := func(c <-chan int) {
        for n := range c {
            out <- n
        }
        wg.Done()
    }
    wg.Add(len(cs))
    for _, c := range cs {
        go output(c)
    }

    // Start a goroutine to close out once all the output goroutines are
    // done.  This must start after the wg.Add call.
    go func() {
        wg.Wait()
        close(out)
    }()
    return out
}

func main() {
    in := gen(2, 3)

    // Distribute the sq work across two goroutines that both read from in.
    c1 := sq(in)
    c2 := sq(in)

    // Consume the first value from output.
    out := merge(c1, c2)
    fmt.Println(<-out) // 4 or 9
    return
    // Apparently if we had not set the merge out buffer size to 1
    // then we would have a hanging go routine. 
}

现在,如果您在2中注意到行merge,则表示我们将chan标记为buffer,因为这是未读输入的足够空间。但是,我几乎肯定我们应该分配chan buffer大小2.根据此代码示例:

c := make(chan int, 2) // buffer size 2
c <- 1  // succeeds immediately
c <- 2  // succeeds immediately
c <- 3  // blocks until another goroutine does <-c and receives 1 

因为此部分意味着chan buffer大小3不会阻止。任何人都可以澄清/协助我的理解吗?

1 个答案:

答案 0 :(得分:1)

程序向通道out发送两个值,并从通道out中读取一个值。其中一个值未收到。

如果通道未缓冲(容量为0),则其中一个发送goroutine将阻塞,直到程序退出。这是泄密。

如果创建的通道容量为1,则两个goroutine都可以发送到通道并退出。发送到频道的第一个值由main接收。第二个值保留在频道中。

如果主函数没有从通道out接收到值,则需要容量为2的通道来防止goroutine无限期地阻塞。