Concurrency 这是怎么泄露的?
我试图理解这张幻灯片上概述的问题: 在URL死亡的情况下复制代码:Concurrency 这是怎么泄露的?,concurrency,go,garbage,goroutine,Concurrency,Go,Garbage,Goroutine,我试图理解这张幻灯片上概述的问题: 在URL死亡的情况下复制代码: func sendMsg(msg, addr string) error { conn, err := net.Dial("tcp", addr) if err != nil { return err } defer conn.Close() _, err = fmt.Fprint(conn, msg) return err } func broadcastMs
func sendMsg(msg, addr string) error {
conn, err := net.Dial("tcp", addr)
if err != nil {
return err
}
defer conn.Close()
_, err = fmt.Fprint(conn, msg)
return err
}
func broadcastMsg(msg string, addrs []string) error {
errc := make(chan error)
for _, addr := range addrs {
go func(addr string) {
errc <- sendMsg(msg, addr)
fmt.Println("done")
}(addr)
}
for _ = range addrs {
if err := <-errc; err != nil {
return err
}
}
return nil
}
func main() {
addr := []string{"localhost:8080", "http://google.com"}
err := broadcastMsg("hi", addr)
time.Sleep(time.Second)
if err != nil {
fmt.Println(err)
return
}
fmt.Println("everything went fine")
}
func sendMsg(msg,addr string)错误{
连接,错误:=网络拨号(“tcp”,地址)
如果错误!=零{
返回错误
}
延迟连接关闭()
_,err=fmt.Fprint(连接,消息)
返回错误
}
func broadcastMsg(消息字符串,地址[]字符串)错误{
errc:=制造(chan错误)
对于u,addr:=范围addr{
go func(地址字符串){
errc
函数文字表示匿名函数
FunctionLit = "func" Function .
func(a, b int, z float64) bool { return a*b < int(z) }
自len(addrs)==2
以来启动了两个goroutine。由于在通道errc
上第一次接收时err!=nil
过早退出,只有一个goroutine完成。第二个goroutine在发送时被阻止(写入)对于无缓冲通道errc
,它永远不会完成。因此,仍然存在对errc
的引用,因此它永远不会被垃圾收集。当程序退出时,第二个goroutine最终被放弃
f := func(x, y int) int { return x + y }
func(ch chan int) { ch <- ACK }(replyChan)
SendStmt = Channel "<-" Expression .
Channel = Expression .
func broadcastMsg(msg string, addrs []string) error {
errc := make(chan error)
for _, addr := range addrs {
go func(addr string) {
errc <- sendMsg(msg, addr)
fmt.Println("done")
}(addr)
}
for _ = range addrs {
if err := <-errc; err != nil {
return err
}
}
return nil
}