并发执行但序列化输出

huangapple
go评论115阅读模式
英文:

Concurrent execution but serialized output

问题

我需要使用名为g1、g2、g3的3个goroutine，并以轮询的方式在这3个goroutine之间分配1-10的数字。它们将根据提供的数字执行一些假设的工作。程序应以以下方式打印输出。

g1-1<br/>
g2-2<br/>
g3-3<br/>
g1-4<br/>
g2-5<br/>
g3-6<br/>
...

任务必须并发执行，但输出必须按顺序进行。

我已经实现了下面的代码来分配数字并打印输出，但输出的打印顺序不如上述所述。

我需要一些帮助来修复下面的代码，或者对实现上述所需输出的其他方法提出建议。

方法1：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"sync"
  6. )
  8. func main() {
  10. 	chang1 := make(chan int)
  11. 	chang2 := make(chan int)
  12. 	chang3 := make(chan int)
  14. 	var wg sync.WaitGroup
  15. 	wg.Add(3)
  17. 	go func() {
  18. 		for num := range chang1 {
  19. 			fmt.Println("g1", num)
  20. 		}
  21. 		wg.Done()
  22. 	}()
  24. 	go func() {
  25. 		for num := range chang2 {
  26. 			fmt.Println("g2", num)
  27. 		}
  28. 		wg.Done()
  30. 	}()
  32. 	go func() {
  33. 		for num := range chang3 {
  34. 			fmt.Println("g3", num)
  35. 		}
  36. 		wg.Done()
  38. 	}()
  40. 	channels := []chan int{chang1, chang2, chang3}
  42. 	for i := 1; i <= 10; i++ {
  44. 		currentCh := (i - 1) % 3
  45. 		channels[currentCh] <- i
  47. 	}
  49. 	close(chang1)
  50. 	close(chang2)
  51. 	close(chang3)
  52. 	wg.Wait()
  54. }

输出（顺序不正确）

g1- 1<br>
g1- 4<br>
g2- 2<br>
g3- 3<br>
g1- 7<br>
g2- 5<br>
g2- 8<br>
g3- 6<br>
g3- 9<br>
g1- 10<br>
...

英文:

I need to use 3 goroutines named g1, g2, g3. and distribute numbers from 1-10 among the above 3 goroutines in a round-robin fashion. They will do some hypothetical work based on the provided number. And program should print output in the following manner.

g1-1<br/>
g2-2<br/>
g3-3<br/>
g1-4<br/>
g2-5<br/>
g3-6<br/>
...

Tasks must be performed concurrently but the output must be in sequential order.

I have implemented the below code which distributes numbers and prints but output print order is not guaranteed as mentioned above.

I need some help to fix the below code or suggestions on another approach to get the above-desired output.

Approach 1:

  1. package main
  3. import (
  4. 	&quot;fmt&quot;
  5. 	&quot;sync&quot;
  6. )
  8. func main() {
  10. 	chang1 := make(chan int)
  11. 	chang2 := make(chan int)
  12. 	chang3 := make(chan int)
  14. 	var wg sync.WaitGroup
  15. 	wg.Add(3)
  17. 	go func() {
  18. 		for num := range chang1 {
  19. 			fmt.Println(&quot;g1&quot;, num)
  20. 		}
  21. 		wg.Done()
  22. 	}()
  24. 	go func() {
  25. 		for num := range chang2 {
  26. 			fmt.Println(&quot;g2&quot;, num)
  27. 		}
  28. 		wg.Done()
  30. 	}()
  32. 	go func() {
  33. 		for num := range chang3 {
  34. 			fmt.Println(&quot;g3&quot;, num)
  35. 		}
  36. 		wg.Done()
  38. 	}()
  40. 	channels := []chan int{chang1, chang2, chang3}
  42. 	for i := 1; i &lt;= 10; i++ {
  44. 		currentCh := (i - 1) % 3
  45. 		channels[currentCh] &lt;- i
  47. 	}
  49. 	close(chang1)
  50. 	close(chang2)
  51. 	close(chang3)
  52. 	wg.Wait()
  54. }

output (with incorrect sequence)

g1- 1<br>
g1- 4<br>
g2- 2<br>
g3- 3<br>
g1- 7<br>
g2- 5<br>
g2- 8<br>
g3- 6<br>
g3- 9<br>
g1- 10<br>
...

答案1

得分: 2

如果你不想使用切片，我认为像这样的代码可以工作：- (playground)

  1. package main
  3. import (
  4. 	"fmt"
  5. )
  7. func main() {
  9. 	chang1 := make(chan string)
  10. 	chang2 := make(chan string)
  11. 	chang3 := make(chan string)
  13. 	channels := []chan string{chang1, chang2, chang3}
  15. 	for i := 1; i < 10; i += 3 {
  17. 		go g1(i, channels[i%3])
  18. 		go g2(i+1, channels[(i+1)%3])
  19. 		go g3(i+2, channels[(i+2)%3])
  21. 		fmt.Print(<-channels[i%3])
  22. 		fmt.Print(<-channels[(i+1)%3])
  23. 		fmt.Print(<-channels[(i+2)%3])
  24. 	}
  26. }
  28. func g1(i int, chanel chan string) {
  30. 	chanel <- fmt.Sprintln("g1", i)
  32. }
  34. func g2(i int, chanel chan string) {
  36. 	chanel <- fmt.Sprintln("g2", i)
  38. }
  40. func g3(i int, chanel chan string) {
  42. 	chanel <- fmt.Sprintln("g3", i)
  44. }

输出结果：
g1 1
g2 2
g3 3
g1 4
g2 5
g3 6
g1 7
g2 8
g3 9

但请记住，在这个解决方案中，你需要运行3个 goroutine，然后等待它们全部给出结果，然后再继续执行。

如果这对你来说不可行，你需要使用带缓冲的通道。

英文:

if you don't want to use slice then I think something like this will work:- (playground)

  1. package main
  3. import (
  4. 	&quot;fmt&quot;
  5. )
  7. func main() {
  9. 	chang1 := make(chan string)
  10. 	chang2 := make(chan string)
  11. 	chang3 := make(chan string)
  13. 	channels := []chan string{chang1, chang2, chang3}
  15. 	for i := 1; i &lt; 10; i += 3 {
  17. 		go g1(i, channels[i%3])
  18. 		go g2(i+1, channels[(i+1)%3])
  19. 		go g3(i+2, channels[(i+2)%3])
  21. 		fmt.Print(&lt;-channels[i%3])
  22. 		fmt.Print(&lt;-channels[(i+1)%3])
  23. 		fmt.Print(&lt;-channels[(i+2)%3])
  24. 	}
  26. }
  28. func g1(i int, chanel chan string) {
  30. 	chanel &lt;- fmt.Sprintln(&quot;g1&quot;, i)
  32. }
  34. func g2(i int, chanel chan string) {
  36. 	chanel &lt;- fmt.Sprintln(&quot;g2&quot;, i)
  38. }
  40. func g3(i int, chanel chan string) {
  42. 	chanel &lt;- fmt.Sprintln(&quot;g3&quot;, i)
  44. }
  1. output 
  2. g1 1
  3. g2 2
  4. g3 3
  5. g1 4
  6. g2 5
  7. g3 6
  8. g1 7
  9. g2 8
  10. g3 9

but keep in mind in this solution you have to run 3 goroutines then wait for all of them to give the result then go back

if this is not ok for you you need to use bufferd channels.

huangapple
  • 本文由 发表于 2022年2月17日 20:22:07
  • 转载请务必保留本文链接：https://go.coder-hub.com/71158069.html
  • go
  • goroutine
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