我正在使用golang来实现一个简单的事件驱动的worker。就像这样:
go func() {
for {
select {
case data := <-ch:
time.Sleep(1)
someGlobalMap[data.key] = data.value
}
}
}()
主要功能将创建几个goroutine ,每个人都会做这样的事情:
ch <- data
fmt.Println(someGlobalMap[data.key])
正如你所看到的那样,因为我的工作人员需要一些时间来完成工作,所以我的主要功能无效。我如何正确控制这个工作流程?
答案 0 :(得分:6)
编辑:我可能误解了你的问题,我看到你提到主要会启动许多制作人 goroutines。我认为这是很多消费者 goroutines和一个生产者。在这里留下答案,以防它对寻找该模式的其他人有用,尽管这些要点仍适用于您的案例。
因此,如果我正确理解您的用例,您不能指望在通道上发送并立即读取结果。您不知道工作人员何时处理该发送,您需要在goroutines之间进行通信,这需要通过通道完成。假设只是调用一个带有返回值的函数在你的场景中不起作用,如果你真的需要发送给一个worker,然后阻塞直到你得到结果,你可以发送一个通道作为数据结构的一部分,并阻止 - 发送后收到,即:
resCh := make(chan Result)
ch <- Data{key, value, resCh}
res := <- resCh
但是你应该尝试将工作分解为独立步骤的管道,请参阅我在原始答案中链接到的博客文章。
原始答案我认为这是一个单一的制作人 - 多个消费者/工人模式:
这是Go的goroutines和channel语义非常适合的常见模式。您需要记住以下几点:
主要功能不会自动等待goroutines完成。如果主要内容没有其他任何内容可做,则程序退出并且您没有结果。
您使用的全局映射不是线程安全的。您需要通过互斥锁同步访问,但有一种更好的方法 - 使用输出通道获取已经同步的结果。
您可以在频道上使用for..range,并且可以安全地在多个goroutine之间共享频道。正如我们所看到的,这使得这种模式非常优雅。
游乐场:https://play.golang.org/p/WqyZfwldqp
有关Go管道和并发模式的更多信息,请介绍错误处理,及早取消等。https://blog.golang.org/pipelines
您提到的用例的注释代码:
// could be a command-line flag, a config, etc.
const numGoros = 10
// Data is a similar data structure to the one mentioned in the question.
type Data struct {
key string
value int
}
func main() {
var wg sync.WaitGroup
// create the input channel that sends work to the goroutines
inch := make(chan Data)
// create the output channel that sends results back to the main function
outch := make(chan Data)
// the WaitGroup keeps track of pending goroutines, you can add numGoros
// right away if you know how many will be started, otherwise do .Add(1)
// each time before starting a worker goroutine.
wg.Add(numGoros)
for i := 0; i < numGoros; i++ {
// because it uses a closure, it could've used inch and outch automaticaly,
// but if the func gets bigger you may want to extract it to a named function,
// and I wanted to show the directed channel types: within that function, you
// can only receive from inch, and only send (and close) to outch.
//
// It also receives the index i, just for fun so it can set the goroutines'
// index as key in the results, to show that it was processed by different
// goroutines. Also, big gotcha: do not capture a for-loop iteration variable
// in a closure, pass it as argument, otherwise it very likely won't do what
// you expect.
go func(i int, inch <-chan Data, outch chan<- Data) {
// make sure WaitGroup.Done is called on exit, so Wait unblocks
// eventually.
defer wg.Done()
// range over a channel gets the next value to process, safe to share
// concurrently between all goroutines. It exits the for loop once
// the channel is closed and drained, so wg.Done will be called once
// ch is closed.
for data := range inch {
// process the data...
time.Sleep(10 * time.Millisecond)
outch <- Data{strconv.Itoa(i), data.value}
}
}(i, inch, outch)
}
// start the goroutine that prints the results, use a separate WaitGroup to track
// it (could also have used a "done" channel but the for-loop would be more complex, with a select).
var wgResults sync.WaitGroup
wgResults.Add(1)
go func(ch <-chan Data) {
defer wgResults.Done()
// to prove it processed everything, keep a counter and print it on exit
var n int
for data := range ch {
fmt.Println(data.key, data.value)
n++
}
// for fun, try commenting out the wgResults.Wait() call at the end, the output
// will likely miss this line.
fmt.Println(">>> Processed: ", n)
}(outch)
// send work, wherever that comes from...
for i := 0; i < 1000; i++ {
inch <- Data{"main", i}
}
// when there's no more work to send, close the inch, so the goroutines will begin
// draining it and exit once all values have been processed.
close(inch)
// wait for all goroutines to exit
wg.Wait()
// at this point, no more results will be written to outch, close it to signal
// to the results goroutine that it can terminate.
close(outch)
// and wait for the results goroutine to actually exit, otherwise the program would
// possibly terminate without printing the last few values.
wgResults.Wait()
}
在现实生活场景中,如果事先不知道工作量,那么通道内的关闭可能来自例如:一个SIGINT信号。只需确保在关闭频道后没有代码路径可以发送工作,因为这会引起恐慌。