英文:
Deadlock error when trying to sync goroutines
问题
我面临一个烦人的问题。当我尝试使用wg.Add()来同步我的例程时，会出现死锁错误。
package main
import (
"fmt"
"sync"
)
func hello(ch chan int, num int, wg *sync.WaitGroup) {
for {
i := <-ch
if i == num {
fmt.Println("Hello number:", i)
ch <- (num - 1)
defer wg.Done() // Same happens without defer
return
}
ch <- i
}
}
func main() {
fmt.Println("Start")
var wg sync.WaitGroup
ch := make(chan int)
for i := 0; i < 10; i++ {
wg.Add(1)
go hello(ch, i, &wg)
}
ch <- 9
wg.Wait()
fmt.Println("End")
}
输出结果：
Start
Hello number: 9
Hello number: 8
Hello number: 7
Hello number: 6
Hello number: 5
Hello number: 4
Hello number: 3
Hello number: 2
Hello number: 1
Hello number: 0
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [semacquire]:
sync.runtime_Semacquire(0xc04203a20c)
C:/Go/src/runtime/sema.go:47 +0x3b
sync.(*WaitGroup).Wait(0xc04203a200)
C:/Go/src/sync/waitgroup.go:131 +0x81
main.main()
C:/Users/Augusto Dias/Documents/GoLang/MT.go:34 +0x1a0
goroutine 18 [chan send]:
main.hello(0xc0420380c0, 0x0, 0xc04203a200)
C:/Users/Augusto Dias/Documents/GoLang/MT.go:13 +0x197
created by main.main
C:/Users/Augusto Dias/Documents/GoLang/MT.go:29 +0x151
exit status 2
当我在for循环之外使用wg.Add(9)时，就没有错误了。
func main() {
fmt.Println("Start")
var wg sync.WaitGroup
ch := make(chan int)
wg.Add(9) // Use wg.Add(10) will raise deadlock too
for i := 0; i < 10; i++ {
go hello(ch, i, &wg)
}
ch <- 9
wg.Wait()
fmt.Println("End")
}
输出结果：
Start
Hello number: 9
Hello number: 8
Hello number: 7
Hello number: 6
Hello number: 5
Hello number: 4
Hello number: 3
Hello number: 2
Hello number: 1
End
为什么会发生这种情况，我的意思是，为什么例程在我等待它们时会进入睡眠状态？使用相同的通道进行发送和接收可能是问题的根源吗？
英文:
I'm facing an annoying problem. When I try to use wg.Add() to sync my routines, a deadlock error is raised.
package main
import (
"fmt"
"sync"
)
func hello(ch chan int, num int, wg *sync.WaitGroup) {
for {
i := <-ch
if i == num {
fmt.Println("Hello number:", i)
ch <- (num - 1)
defer wg.Done() // Same happens without defer
return
}
ch <- i
}
}
func main() {
fmt.Println("Start")
var wg sync.WaitGroup
ch := make(chan int)
for i := 0; i < 10; i++ {
wg.Add(1)
go hello(ch, i, &wg)
}
ch <- 9
wg.Wait()
fmt.Println("End")
}
Outputs:
Start
Hello number: 9
Hello number: 8
Hello number: 7
Hello number: 6
Hello number: 5
Hello number: 4
Hello number: 3
Hello number: 2
Hello number: 1
Hello number: 0
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [semacquire]:
sync.runtime_Semacquire(0xc04203a20c)
C:/Go/src/runtime/sema.go:47 +0x3b
sync.(*WaitGroup).Wait(0xc04203a200)
C:/Go/src/sync/waitgroup.go:131 +0x81
main.main()
C:/Users/Augusto Dias/Documents/GoLang/MT.go:34 +0x1a0
goroutine 18 [chan send]:
main.hello(0xc0420380c0, 0x0, 0xc04203a200)
C:/Users/Augusto Dias/Documents/GoLang/MT.go:13 +0x197
created by main.main
C:/Users/Augusto Dias/Documents/GoLang/MT.go:29 +0x151
exit status 2
When I use wg.Add(9) outside the for block, I got no error.
func main() {
fmt.Println("Start")
var wg sync.WaitGroup
ch := make(chan int)
wg.Add(9) // Use wg.Add(10) will raise deadlock too
for i := 0; i < 10; i++ {
go hello(ch, i, &wg)
}
ch <- 9
wg.Wait()
fmt.Println("End")
}
Outputs:
Start
Hello number: 9
Hello number: 8
Hello number: 7
Hello number: 6
Hello number: 5
Hello number: 4
Hello number: 3
Hello number: 2
Hello number: 1
End
Why is this happening, I mean, why routines goes asleep when I wait for them all? Use the same channel for send and receive can be the source of this problem?
答案1
得分: 3
通道0（调用go hello(ch, 0, &wg)）因为它是最后一个活动的通道，所以在这一行上被阻塞。
ch <- (num - 1)
它试图发送到一个通道，但没有人接收。因此，该函数将无限期等待，永远不会完成。
解决此问题的一些建议：
- 在主循环中创建一个消费者
- 将通道
ch设置为非阻塞的
英文:
Channel 0 (the call go hello(ch, 0, &wg)) since it is the last channel alive gets stuck on this line
ch <- (num - 1)
It is attempting to send to a channel but no one is there to receive it. Thus the function will wait indefinitely and never be done.
Some suggestions for how to remove this problem
- create a consumer in the main loop
- make the channel
chnon blocking
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