英文:
waitgroup.Wait() causing dead lock
问题
我正在尝试弄清楚为什么在使用waitgroup.Wait()时会出现死锁的情况。
当我像这样运行代码时，它会打印出fatal error: all goroutines are asleep - deadlock!。
我尝试将ch更改为带缓冲的通道，问题得到解决。但我真的想知道为什么会出现死锁。
英文:
I'm trying to figure out why I have a dead lock with waitgroup.Wait()
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int)
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
wg.Wait()
close(ch)
for item := range ch {
fmt.Println(item)
}
}
When I run it like this, it prints fatal error: all goroutines are asleep - deadlock!
I tried to change ch to a buffered channel and that solved the problem. But I really want to know why is there a dead lock.
答案1
得分: 2
我已经注释掉了你程序逻辑不正确的部分：
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int) // 无缓冲通道
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
// wg.Wait 等待所有 goroutine 完成，但只有在通道发送成功的情况下才可能完成。
// 在这种情况下，由于你的接收者 "for item := range ch" 在下面，所以不可能完成，导致死锁。
wg.Wait()
// 理想情况下，关闭通道应该是发送者的责任。
// 并且在接收者之前关闭一个无缓冲通道是不正确的。
close(ch)
for item := range ch {
fmt.Println(item)
}
}
修正后的程序：
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int)
go func() {
for item := range ch {
fmt.Println(item)
}
}()
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
wg.Wait()
close(ch)
}
英文:
I've commented out the parts where your program's logic is not correct:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int) // unbuffered channel
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
// wg.Wait is waiting for all goroutines to finish but that's
// only possible if the send to channel succeeds. In this case,
// it is not possible as your receiver "for item := range ch" is below
// this. Hence, a deadlock.
wg.Wait()
// Ideally, it should be the sender's duty to close the channel.
// And closing a channel before the receiver where the channel
// is unbuffered is not correct.
close(ch)
for item := range ch {
fmt.Println(item)
}
}
Corrected program:
package main
import (
"fmt"
"sync"
)
var wg sync.WaitGroup
func foo(c chan int, i int) {
defer wg.Done()
c <- i
}
func main() {
ch := make(chan int)
go func() {
for item := range ch {
fmt.Println(item)
}
}()
for i := 0; i < 10; i++ {
wg.Add(1)
go foo(ch, i)
}
wg.Wait()
close(ch)
}
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