如何在有缓冲的通道上进行循环而不发生死锁?

huangapple go评论149阅读模式
英文:

How to loop on a buffered channel without deadlocking?

问题

我想知道如何排空/关闭缓冲通道,以避免死锁的情况。我正在使用range循环遍历通道,但似乎尽管它们被“读取”,但它们不像非缓冲通道那样被关闭。

package main

func main() {

    cp := 2
    ch := make(chan string, cp)

    for i := 0; i < cp; i++ {
        go send(ch)
    }
    go send(ch)

    for lc := range ch {
        print(lc)

    }

}

func send(ch chan string) {

    ch <- "hello\n"

}

你可以点击这里查看代码示例。

英文:

I'm wondering how can I drain / close the buffered channels so that I don't get into the deadlock? I'm using range to loop through the channels but it seems that although they are "read" they don't get closed like the non-buffered channels do.

package main

func main() {

	cp := 2
	ch := make(chan string, cp)

	for i := 0; i &lt; cp; i++ {
		go send(ch)
	}
	go send(ch)

	for lc := range ch {
		print(lc)

	}

}

func send(ch chan string) {

	ch &lt;- &quot;hello\n&quot;

}

Play

答案1

得分: 1

你可以使用close()内置函数来关闭通道。这个函数必须在所有并发处理完成之后调用。你如何做到这一点取决于你想要做什么。

在你当前的架构中,似乎你需要建立一个全局状态,用于跟踪所有的进程,并确定最后一个进程是否完成。可以通过使用sync.WaitGroup来实现这样的状态。

func send(c chan string, wg *sync.WaitGroup) {
    defer wg.Done()
    // ...
}

wg := &sync.WaitGroup{}

for i := 0; i < cp; i++ {
    wg.Add(1)
    go send(ch, wg)
}
wg.Add(1)
go send(ch, wg)

wg.Wait()
close(ch)

for e := range(ch) {
    // ...
}

请注意,先关闭通道,然后迭代通道将只返回在通道中排队的元素。这意味着任何想要向通道中放入值的goroutine将无法再这样做,因为通道已关闭。

英文:

You can close channels using the close() builtin. This has to be called after all of your concurrent processing is done. How you're doing that depends on what you want to do.

In your current architecture it seems that you have to establish a global state, something that tracks all your processes and determines that the last one finished. Such a state can be achieved by using a sync.WaitGroup for example.

func send(c chan string, wg *sync.WaitGroup) {
    defer wg.Done()
    // ...
}

wg := &amp;sync.WaitGroup{}

for i := 0; i &lt; cp; i++ {
    wg.Add(1)
    go send(ch, wg)
}
wg.Add(1)
go send(ch, wg)

wg.Wait()
close(ch)

for e := range(ch) {
    // ...
}

Note that closing the channel and then iterating over it will give you only the elements that are queued in the channel. This means that any goroutine that wanted to put a value in the channel can't do this anymore as the channel is closed.

huangapple
  • 本文由 发表于 2014年4月12日 04:42:02
  • 转载请务必保留本文链接:https://go.coder-hub.com/23022200.html
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