广播员：所有的goroutine都处于休眠状态-死锁

huangapple
go评论118阅读模式
英文:

Broadcaster: all goroutines are asleep - deadlock

问题

下面的代码（http://play.golang.org/p/ikUtdoKOo5）应该向多个客户端广播消息。但是它不起作用，我无法弄清楚原因。

  1. package main
  3. import "fmt"
  5. type Broadcaster struct {
  6.     Clients []Client
  7. }
  9. func (b *Broadcaster) Broadcast(msg string) {
  10.     for _, c := range b.Clients {
  11.         go func() {
  12.             c.Inbox() <- msg
  13.         }()
  14.     }
  15. }
  17. type Client interface {
  18.     Inbox() chan string
  19. }
  21. type TestClient struct {
  22.     Messages chan string
  23. }
  25. func (tc TestClient) Inbox() chan string {
  26.     return tc.Messages
  27. }
  29. func main() {
  30.     client1 := TestClient{Messages: make(chan string)}
  31.     client2 := TestClient{Messages: make(chan string)}
  32.     broadcaster := Broadcaster{Clients: []Client{client1, client2}}
  34.     broadcaster.Broadcast("sos")
  36.     fmt.Printf("client1: '%s'\n", <-client1.Messages)
  37.     fmt.Printf("client2: '%s'\n", <-client2.Messages)
  38. }

错误信息：

  1. go run main.go
  2. fatal error: all goroutines are asleep - deadlock!
  4. goroutine 1 [chan receive]:
  5. main.main()
  6.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:36 +0x1f3
  8. goroutine 3 [chan send]:
  9. main.func·001()
  10.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:12 +0x5f
  11. created by main.(*Broadcaster).Broadcast
  12.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:13 +0xcd
  14. goroutine 4 [chan send]:
  15. main.func·001()
  16.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:12 +0x5f
  17. created by main.(*Broadcaster).Broadcast
  18.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:13 +0xcd

更新：

go vet 工具揭示了问题：

  1. % go vet
  2. main.go:12: range variable c enclosed by function
英文:

The code below (http://play.golang.org/p/ikUtdoKOo5) is supposed to broadcast a message to a number of clients. But it does not work and I can't figure out why.

<!-- golang -->

  1. package main
  3. import &quot;fmt&quot;
  5. type Broadcaster struct {
  6. 	Clients []Client
  7. }
  9. func (b *Broadcaster) Broadcast(msg string) {
  10. 	for _, c := range b.Clients {
  11. 		go func() {
  12. 			c.Inbox() &lt;- msg
  13. 		}()
  14. 	}
  15. }
  17. type Client interface {
  18. 	Inbox() chan string
  19. }
  21. type TestClient struct {
  22. 	Messages chan string
  23. }
  25. func (tc TestClient) Inbox() chan string {
  26. 	return tc.Messages
  27. }
  29. func main() {
  30. 	client1 := TestClient{Messages: make(chan string)}
  31. 	client2 := TestClient{Messages: make(chan string)}
  32. 	broadcaster := Broadcaster{Clients: []Client{client1, client2}}
  34. 	broadcaster.Broadcast(&quot;sos&quot;)
  36. 	fmt.Printf(&quot;client1: &#39;%s&#39;\n&quot;, &lt;-client1.Messages)
  37. 	fmt.Printf(&quot;client2: &#39;%s&#39;\n&quot;, &lt;-client2.Messages)
  38. }

<br>
Error:

  1. go run main.go
  2. fatal error: all goroutines are asleep - deadlock!
  4. goroutine 1 [chan receive]:
  5. main.main()
  6.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:36 +0x1f3
  8. goroutine 3 [chan send]:
  9. main.func&#183;001()
  10.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:12 +0x5f
  11. created by main.(*Broadcaster).Broadcast
  12.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:13 +0xcd
  14. goroutine 4 [chan send]:
  15. main.func&#183;001()
  16.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:12 +0x5f
  17. created by main.(*Broadcaster).Broadcast
  18.         /Users/artem/projects/gocode/src/github.com/artemave/broadcaster/main.go:13 +0xcd

<br>
UPDATE:

The go vet tool reveals the problem:

  1. % go vet
  2. main.go:12: range variable c enclosed by function

答案1

得分: 4

你的问题是在闭包中使用变量c的用法。
由于c是一个单一的变量，在执行go协程时，c很可能被设置为client2，并且两条消息都将被广播到该客户端，而不是client1。

修复这个问题的一种方法是使用函数参数：

  1. func (b *Broadcaster) Broadcast(msg string) {
  2.     for _, c := range b.Clients {        
  3.         go func(c Client) {
  4.             c.Inbox() <- msg            
  5.         }(c)
  6.     }
  7. }
英文:

Your problem is the usage of the variable c in the closure.
Since c is a single variable, at the time when the go routines are executed, c will most likely be set to client2, and both messages will be broadcasted to that client, and none to client1.

One way to fix it using function parameter:

  1. func (b *Broadcaster) Broadcast(msg string) {
  2. 	for _, c := range b.Clients {		
  3. 		go func(c Client) {
  4. 			c.Inbox() &lt;- msg    			
  5. 		}(c)
  6. 	}
  7. }

答案2

得分: 4

这是由于在for-range循环中重新分配c而引起的一个微妙的错误。看起来有点奇怪，但在标准库的几个地方可以看到这种模式：

  1. func (b *Broadcaster) Broadcast(msg string) {
  2.     for _, c := range b.Clients {
  3.         c := c  // 为闭包重新声明c
  4.         go func() {
  5.             c.Inbox() <- msg
  6.         }()
  7.     }
  8. }

http://golang.org/doc/faq#closures_and_goroutines

英文:

This is a subtle bug brought on by the reassignment of c in the for-range loop. It looks a little strange, but you see this pattern in the std library in a few places:

  1. func (b *Broadcaster) Broadcast(msg string) {
  2. 	for _, c := range b.Clients {
  3. 		c := c  // redeclare c for the closure
  4. 		go func() {
  5. 			c.Inbox() &lt;- msg
  6. 		}()
  7. 	}
  8. }

http://golang.org/doc/faq#closures_and_goroutines

huangapple
  • 本文由 发表于 2014年5月29日 06:06:44
  • 转载请务必保留本文链接：https://go.coder-hub.com/23922682.html
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