尝试停止创建更多的goroutine时发生了恐慌。

huangapple
go评论118阅读模式
英文:

Panic while trying to stop creating more goroutines

问题

我正在尝试并行调用一个API以加快速度，但我面临一个问题：如果我从其中一个goroutine调用中收到错误，我需要停止启动goroutine来调用API。由于我在错误处理部分和执行完成时都关闭了通道，所以我得到了一个panic: close of closed channel错误。有没有一种优雅的方式来处理这个问题，而不会导致程序崩溃？任何帮助将不胜感激！

以下是伪代码片段。

  1. for i := 0; i < someNumber; i++ {
  2. 	go func(num int, q chan<- bool) {
  3. 		value, err := callAnAPI()
  4. 		if err != nil {
  5. 			close(q)//退出for循环
  6. 		}
  7. 		// 在这里处理value
  8. 		wg.Done()
  9. 	}(i, quit)
  10. }
  11. close(quit)

为了模拟我的场景，我编写了以下程序。有没有办法在满足条件（已注释）后优雅地退出for循环？

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"sync"
  6. )
  8. func receive(q <-chan bool) {
  9. 	for {
  10. 		select {
  11. 		case <-q:
  12. 			return
  13. 		}
  14. 	}
  15. }
  17. func main() {
  18. 	quit := make(chan bool)
  20. 	var result []int
  21. 	wg := &sync.WaitGroup{}
  22. 	wg.Add(10)
  23. 	for i := 0; i < 10; i++ {
  24. 		go func(num int, q chan<- bool) {
  25. 			//if num == 5 {
  26. 			//	close(q)
  27. 			//}
  28. 			result = append(result, num)
  30. 			wg.Done()
  31. 		}(i, quit)
  32. 	}
  33. 	close(quit)
  34. 	receive(quit)
  36. 	wg.Wait()
  38. 	fmt.Printf("Result: %v", result)
  39. }
英文:

I'm trying to parallelize calls to an API to speed things up, but I'm facing a problem where I need to stop spinning up goroutines to call the API if I receive an error from one of the goroutine calls. Since I am closing the channel twice(once in the error handling part and when the execution is done), I'm getting a panic: close of closed channel error. Is there an elegant way to handle this without the program to panic? Any help would be appreciated!

The following is the pseudo-code snippet.

  1. for i := 0; i &lt; someNumber; i++ {
  2. 	go func(num int, q chan&lt;- bool) {
  3. 		value, err := callAnAPI()
  4. 		if err != nil {
  5. 			close(q)//exit from the for-loop
  6. 		}
  7. 		// process the value here
  8. 		wg.Done()
  9. 	}(i, quit)
  10. }
  11. close(quit)

To mock my scenario, I have written the following program. Is there any way to exit the for-loop gracefully once the condition(commented out) is satisfied?

  1. package main
  3. import (
  4. 	&quot;fmt&quot;
  5. 	&quot;sync&quot;
  6. )
  8. func receive(q &lt;-chan bool) {
  9. 	for {
  10. 		select {
  11. 		case &lt;-q:
  12. 			return
  13. 		}
  14. 	}
  15. }
  17. func main() {
  18. 	quit := make(chan bool)
  20. 	var result []int
  21. 	wg := &amp;sync.WaitGroup{}
  22. 	wg.Add(10)
  23. 	for i := 0; i &lt; 10; i++ {
  24. 		go func(num int, q chan&lt;- bool) {
  25. 			//if num == 5 {
  26. 			//	close(q)
  27. 			//}
  28. 			result = append(result, num)
  30. 			wg.Done()
  31. 		}(i, quit)
  32. 	}
  33. 	close(quit)
  34. 	receive(quit)
  36. 	wg.Wait()
  38. 	fmt.Printf(&quot;Result: %v&quot;, result)
  39. }

答案1

得分: 0

你可以使用context包，该包定义了Context类型，用于在API边界和进程之间传递截止时间、取消信号和其他请求范围的值。

  1. package main
  3. import (
  4. 	"context"
  5. 	"fmt"
  6. 	"sync"
  7. )
  9. func main() {
  10. 	ctx, cancel := context.WithCancel(context.Background())
  11. 	defer cancel() // 在完成后取消，即使没有错误
  13. 	wg := &sync.WaitGroup{}
  15. 	for i := 0; i < 10; i++ {
  16. 		wg.Add(1)
  17. 		go func(num int) {
  18. 			defer wg.Done()
  19. 			select {
  20. 			case <-ctx.Done():
  21. 				return // 发生错误，终止
  22. 			default: // 避免阻塞
  23. 			}
  25. 			// 在这里编写你的代码
  26. 			// res, err := callAnAPI()
  27. 			// if err != nil {
  28. 			// 	cancel()
  29. 			//	return
  30. 			//}
  32. 			if num == 5 {
  33. 				cancel()
  34. 				return
  35. 			}
  37. 			fmt.Println(num)
  39. 		}(i)
  40. 	}
  42. 	wg.Wait()
  44. 	fmt.Println(ctx.Err())
  45. }

在Go Playground上尝试。

你还可以查看这个答案以获取更详细的解释。

英文:

You can use context package which defines the Context type, which carries deadlines, cancellation signals, and other request-scoped values across API boundaries and between processes.

  1. package main
  3. import (
  4. 	&quot;context&quot;
  5. 	&quot;fmt&quot;
  6. 	&quot;sync&quot;
  7. )
  9. func main() {
  10. 	ctx, cancel := context.WithCancel(context.Background())
  11. 	defer cancel() // cancel when we are finished, even without error
  13. 	wg := &amp;sync.WaitGroup{}
  15. 	for i := 0; i &lt; 10; i++ {
  16. 		wg.Add(1)
  17. 		go func(num int) {
  18. 			defer wg.Done()
  19. 			select {
  20. 			case &lt;-ctx.Done():
  21. 				return // Error occured somewhere, terminate
  22. 			default: // avoid blocking
  23. 			}
  24. 			
  25. 			// your code here
  26. 			// res, err := callAnAPI()
  27. 			// if err != nil {
  28. 			// 	cancel()
  29. 			//	return
  30. 			//}
  31. 			
  32. 			if num == 5 {
  33. 				cancel()
  34. 				return
  35. 			}
  37. 			fmt.Println(num)
  39. 		}(i)
  40. 	}
  41. 	
  42. 	wg.Wait()
  43. 	
  44. 	fmt.Println(ctx.Err())
  45. }

Try on: Go Playground

You can also take a look to this answer for more detailed explanation.

huangapple
  • 本文由 发表于 2020年2月5日 11:29:19
  • 转载请务必保留本文链接：https://go.coder-hub.com/60068633.html
  • go
  • goroutine
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