所有的Go协程在使用sync.Cond时都处于休眠状态。

huangapple
go评论118阅读模式
英文:

all go routines are asleep while using sync.Cond

问题

  1. func main() {
  2. 	type Button struct {
  3. 		Clicked *sync.Cond
  4. 	}
  5. 	button := Button{Clicked: sync.NewCond(&sync.Mutex{})}
  7. 	subscribe := func(c *sync.Cond, fn func()) {
  8. 		//var goroutineRunning sync.WaitGroup
  9. 		//goroutineRunning.Add(1)
  10. 		go func() {
  11. 			//goroutineRunning.Done()
  12. 			c.L.Lock()
  13. 			defer c.L.Unlock()
  14. 			c.Wait()
  15. 			fn()
  16. 		}()
  17. 		//goroutineRunning.Wait()
  18. 	}
  20. 	var clickRegistered sync.WaitGroup
  21. 	clickRegistered.Add(3)
  22. 	subscribe(button.Clicked, func() {
  23. 		fmt.Println("Maximizing window")
  24. 		clickRegistered.Done()
  25. 	})
  26. 	subscribe(button.Clicked, func() {
  27. 		fmt.Println("Displaying annoying dialogue box!")
  28. 		clickRegistered.Done()
  29. 	})
  30. 	subscribe(button.Clicked, func() {
  31. 		fmt.Println("Mouse clicked.")
  32. 		clickRegistered.Done()
  33. 	})
  35. 	button.Clicked.Broadcast()
  36. 	clickRegistered.Wait()
  37. }

我正在遵循《Go并发编程》一书中的一些示例，并且我已经在subscribe方法中注释了一些代码语句。有人可以帮忙解释一下为什么这段代码会发生死锁吗？

英文: 
  1. func main() {
  2. 	type Button struct {
  3. 		Clicked *sync.Cond
  4. 	}
  5. 	button := Button{Clicked: sync.NewCond(&sync.Mutex{})}
  7. 	subscribe := func(c *sync.Cond, fn func()) {
  8. 		//var goroutineRunning sync.WaitGroup
  9. 		//goroutineRunning.Add(1)
  10. 		go func() {
  11. 			//goroutineRunning.Done()
  12. 			c.L.Lock()
  13. 			defer c.L.Unlock()
  14. 			c.Wait()
  15. 			fn()
  16. 		}()
  17. 		//goroutineRunning.Wait()
  18. 	}
  20. 	var clickRegistered sync.WaitGroup
  21. 	clickRegistered.Add(3)
  22. 	subscribe(button.Clicked, func() {
  23. 		fmt.Println("Maximizing window")
  24. 		clickRegistered.Done()
  25. 	})
  26. 	subscribe(button.Clicked, func() {
  27. 		fmt.Println("Displaying annoying dialogue box!")
  28. 		clickRegistered.Done()
  29. 	})
  30. 	subscribe(button.Clicked, func() {
  31. 		fmt.Println("Mouse clicked.")
  32. 		clickRegistered.Done()
  33. 	})
  35. 	button.Clicked.Broadcast()
  36. 	clickRegistered.Wait()
  37. }

I am following some examples from concurrency in go book, and I have commented some code statements inside subscribe method. can someone please help in explaining why this code deadlocks.

答案1

得分: 2

一个注意事项：就命名而言，clickRegistered不符合你的代码当前行为。handlerExecuted可能更准确一些。

修复死锁的一种方法：

从你的代码开始：https://play.golang.org/p/XEduyON9j59

你可以传递一个额外的waitGroup或通道给你的goroutine，这样它们就可以向主goroutine发出信号，告诉它们正在运行。但实际上，你也可以使用条件本身来实现这一点：

  1. subscribe := func(c *sync.Cond, fn func()) {
  2.     c.L.Lock()  // 同步地获取锁，一旦goroutine执行了`c.Wait()`，锁就会再次可用
  3.     go func() {
  4.         defer c.L.Unlock()
  5.         c.Wait()
  6.         fn()
  7.     }()
  8. }
  10. subscribe(...)
  11. subscribe(...)
  12. subscribe(...)
  14. button.Clicked.L.Lock() // 只有当所有订阅者都在`.Wait()`时才会返回

https://play.golang.org/p/fFzRolUnaDZ

这是另一种修复死锁的方法，使用WaitGroup让goroutine发出它们正在运行的信号：
https://play.golang.org/p/LSM72HBmo0M

订阅的运行方式可能更加"并发"，但你会注意到，要知道它们是否真正运行了.Wait()指令的唯一直接方法仍然是获取条件的锁。

英文:

a note : as far as naming goes, clickRegistered doesn't fit the current behavior of your code. handlerExecuted would be more accurate.

One way to fix the deadlock :

starting from your code : https://play.golang.org/p/XEduyON9j59

You may pass an extra waitGroup or a channel to your goroutines, so that they can signal the main goroutine that they are running, but actually you can also use the condition itself to do that :

  1. 	subscribe := func(c *sync.Cond, fn func()) {
  2. 		c.L.Lock()  // take the lock synchronously, the lock will be available
  3.                     // again once the goroutine executes 'c.Wait()'
  4. 		go func() {
  5. 			defer c.L.Unlock()
  6. 			c.Wait()
  7. 			fn()
  8. 		}()
  9. 	}
  11.     subscribe(...)
  12.     subscribe(...)
  13.     subscribe(...)
  15.     button.Clicked.L.Lock() // will return only once all subscribers are '.Wait()'ing

https://play.golang.org/p/fFzRolUnaDZ

Here is another way to fix the deadlock, using a Waitgroup to let the goroutines signal that they are running:
https://play.golang.org/p/LSM72HBmo0M

The way the subscriptions are run may feel more "concurrent"; you will note, however, that the only straightforward way to know that they are actually running their .Wait() instruction is still to acquire the condition's lock.

huangapple
  • 本文由 发表于 2021年10月17日 23:01:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/69605585.html
  • concurrency
  • go
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