我試圖弄清楚為什么我有一個(gè)帶有等待組的死鎖。等待（）package mainimport ( "fmt" "sync")var wg sync.WaitGroupfunc 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) }}當(dāng)我像這樣運(yùn)行它時(shí)，它會(huì)打印fatal error: all goroutines are asleep - deadlock!我試圖更改為緩沖通道，這解決了問題。但我真的很想知道為什么會(huì)有死鎖。ch
1 回答
慕碼人8056858
TA貢獻(xiàn)1803條經(jīng)驗(yàn) 獲得超6個(gè)贊
我已經(jīng)注釋掉了程序邏輯不正確的部分：
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) // 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)
}
}
更正的程序：
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)
}
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