在两个 goroutine 之间同步

huangapple
go评论100阅读模式
英文:

Sync between 2 goroutines

问题

我的任务是同步两个goroutine，使输出看起来像这样：

foobarfoobarfoobarfoobar

问题在于当我调用它们时，它们的输出完全是随机的。这是我的代码：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"sync"
  6. 	"time"
  7. )
  9. type ConcurrentPrinter struct {
  10. 	sync.WaitGroup
  11. 	sync.Mutex
  12. }
  15. func (cp *ConcurrentPrinter) printFoo(times int) {
  16. 	cp.WaitGroup.Add(times)
  17. 	go func() {
  18. 		cp.Lock()
  19. 		fmt.Print("foo")
  20. 		cp.Unlock()
  21. 	}()
  22. }
  23. func (cp *ConcurrentPrinter) printBar(times int) {
  24. 	cp.WaitGroup.Add(times)
  25. 	go func() {
  26. 		cp.Lock()
  27. 		fmt.Print("bar")
  28. 		cp.Unlock()
  29. 	}()
  30. }
  32. func main() {
  33. 	times := 10
  34. 	cp := &ConcurrentPrinter{}
  36. 	for i := 0; i <= times; i++ {
  37. 		cp.printFoo(i)
  38. 		cp.printBar(i)
  39. 	}
  40. 	time.Sleep(10 * time.Millisecond)
  41. }
英文:

My task is to sync 2 goroutines so the output should look like that:

> foobarfoobarfoobarfoobar

.The issue is that when I call them they come out completely randomized. This is my code:

  1. package main
  3. import (
  4. 	&quot;fmt&quot;
  5. 	&quot;sync&quot;
  6. 	&quot;time&quot;
  7. )
  9. type ConcurrentPrinter struct {
  10. 	sync.WaitGroup
  11. 	sync.Mutex
  12. }
  15. func (cp *ConcurrentPrinter) printFoo(times int) {
  16. 	cp.WaitGroup.Add(times)
  17. 	go func() {
  18. 		cp.Lock()
  19. 		fmt.Print(&quot;foo&quot;)
  20. 		cp.Unlock()
  21. 	}()
  22. }
  23. func (cp *ConcurrentPrinter) printBar(times int) {
  24. 	cp.WaitGroup.Add(times)
  25. 	go func() {
  26. 		cp.Lock()
  27. 		fmt.Print(&quot;bar&quot;)
  28. 		cp.Unlock()
  29. 	}()
  30. }
  32. func main() {
  33. 	times := 10
  34. 	cp := &amp;ConcurrentPrinter{}
  36. 	for i := 0; i &lt;= times; i++ {
  37. 		cp.printFoo(i)
  38. 		cp.printBar(i)
  39. 	}
  40. 	time.Sleep(10 * time.Millisecond)
  41. }

答案1

得分: 2

如评论中所述，使用goroutine可能不是你试图实现的最佳方案，因此这可能是一个XY问题。

话虽如此，如果你想确保两个独立的goroutine以交替的顺序进行工作，你可以实现一组“乒乓”互斥锁：

  1. var ping, pong sync.Mutex
  3. pong.Lock() // 确保第二个goroutine等待，第一个先执行
  5. go func() {
  6.     for {
  7.         ping.Lock()
  8.         foo()
  9.         pong.Unlock()
  10.     }
  11. }()
  13. go func() {
  14.     for {
  15.         pong.Lock()
  16.         bar()
  17.         ping.Unlock()
  18.     }
  19. }()

https://go.dev/play/p/VO2LoMJ8fek

英文:

As outlined in the comments, using goroutines may not be the best use case for what you are trying to achieve - and thus this may be an XY problem.

Having said that, if you want to ensure two independent goroutines interleave their work in an alternating sequence, you can implement a set of "ping-pong" mutexs:

  1. var ping, pong sync.Mutex
  3. pong.Lock() // ensure the 2nd goroutine waits &amp; the 1st goes first
  5. go func() {
  6. 	for {
  7. 		ping.Lock()
  8. 		foo()
  9. 		pong.Unlock()
  10. 	}
  11. }()
  13. go func() {
  14. 	for {
  15. 		pong.Lock()
  16. 		bar()
  17. 		ping.Unlock()
  18. 	}
  19. }()

https://go.dev/play/p/VO2LoMJ8fek

答案2

得分: 0

使用通道：

  1. func printFoo(i int, ch chan<- bool, wg *sync.WaitGroup) {
  2. 	wg.Add(1)
  3. 	go func() {
  4. 		defer wg.Done()
  5. 		fmt.Print("foo")
  6. 		ch <- true
  7. 	}()
  8. }
  10. func printBar(i int, ch chan<- bool, wg *sync.WaitGroup) {
  11. 	wg.Add(1)
  12. 	go func() {
  13. 		defer wg.Done()
  14. 		fmt.Print("bar")
  15. 		ch <- true
  16. 	}()
  18. }
  20. func main() {
  21. 	times := 4
  22. 	firstchan := make(chan bool)
  23. 	secondchan := make(chan bool)
  24. 	var wg sync.WaitGroup
  25. 	for i := 0; i <= times; i++ {
  26. 		printFoo(i, firstchan, &wg)
  27. 		<-firstchan
  28. 		printBar(i, secondchan, &wg)
  29. 		<-secondchan
  30. 	}
  31. 	wg.Wait()
  32. }

https://go.dev/play/p/MlZ9dHkUXGb

英文:

Using channel:

  1. func printFoo(i int, ch chan&lt;- bool, wg *sync.WaitGroup) {
  2. 	wg.Add(1)
  3. 	go func() {
  4. 		defer wg.Done()
  5. 		fmt.Print(&quot;foo&quot;)
  6. 		ch &lt;- true
  7. 	}()
  8. }
  9. func printBar(i int, ch chan&lt;- bool, wg *sync.WaitGroup) {
  10. 	wg.Add(1)
  11. 	go func() {
  12. 		defer wg.Done()
  13. 		fmt.Print(&quot;bar&quot;)
  14. 		ch &lt;- true
  15. 	}()
  17. }
  19. func main() {
  20. 	times := 4
  21. 	firstchan := make(chan bool)
  22. 	secondchan := make(chan bool)
  23. 	var wg sync.WaitGroup
  24. 	for i := 0; i &lt;= times; i++ {
  25. 		printFoo(i, firstchan, &amp;wg)
  26. 		&lt;-firstchan
  27. 		printBar(i, secondchan, &amp;wg)
  28. 		&lt;-secondchan
  29. 	}
  30. 	wg.Wait()
  31. }

https://go.dev/play/p/MlZ9dHkUXGb

huangapple
  • 本文由 发表于 2022年3月14日 23:50:06
  • 转载请务必保留本文链接：https://go.coder-hub.com/71470563.html
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