英文:
for loop causing deadlock
问题
我已经从GO Concurrency中为你翻译了一些代码示例：
func gen(numbers ...int) <-chan int {
out := make(chan int)
go func() {
for _, number := range numbers {
out <- number
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for number := range in {
out <- number * number
}
}()
return out
}
所以我尝试在我的主函数中使用上述代码，像这样：
func main() {
result := sq(sq(sq(gen(1, 2, 3, 4))))
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println("-------------------")
for channelValue := range sq(sq(sq(gen(1, 2, 3, 4)))) {
fmt.Println(channelValue)
}
}
当我运行代码时，循环结束后我得到了这个错误信息：
fatal error: all goroutines are asleep - deadlock
请帮助我理解这个问题。根据我的理解，调用fmt.Println(result)四次与for channelValue := range sq(sq(sq(gen(1, 2, 3, 4))))的循环是相同的，这个理解正确吗？
请告诉我为什么在循环结束后我会遇到死锁问题？
英文:
I have written some code example from GO Concurrency :
func gen(numbers ...int) <-chan int {
out := make(chan int)
go func() {
for _, number := range numbers {
out <- number
}
close(out)
}()
return out
}
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for number := range in {
out <- number * number
}
}()
return out
}
so I tried to use above code in my main function like this :
func main() {
result := sq(sq(sq(gen(1, 2, 3, 4))))
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println(<-result)
fmt.Println("-------------------")
for channelValue := range sq(sq(sq(gen(1, 2, 3, 4)))) {
fmt.Println(channelValue)
}
}
I was confused when I run the code I got this message after the loop:
> fatal error: all goroutines are asleep - deadlock
Please help me to understand this. From what I understand is calling the fmt.Prinlnt(result) x 4 times is the same as the for loop on for channelValue := range sq(sq(sq(gen(1, 2, 3, 4)))). is this correct?
Could please tell me why I got deadlock after the loop?
答案1
得分: 2
通道阻塞的原因是通道在sq函数中没有关闭。
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for number := range in {
out <- number * number
}
close(out)
}()
return out
}
解决这种死锁问题的一个好方法是向进程发送SIGQUIT信号。当接收到SIGQUIT信号时，运行时会转储所有goroutine的堆栈信息。堆栈转储通常会指出问题所在。
英文:
The range over the channel blocks because the channel is not closed in sq.
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for number := range in {
out <- number * number
}
close(out)
}()
return out
}
A good way to debug deadlocks like this is to send the process a SIGQUIT. The runtime dumps the stacks of all the goroutines when a SIGQUIT is received. The stack dumps will often point to the issue.
答案2
得分: 1
你没有在sq函数中关闭out通道。
go func() {
for number := range in {
out <- number * number
}
close(out)
}()
https://play.golang.org/p/kk8-08SfwB
英文:
You're not closing the out channel in the sq function
go func() {
for number := range in {
out <- number * number
}
close(out)
}()
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