这个 goroutine 是如何失败的？

Question

func serveApp() {
    mux := http.NewServeMux()
    mux.HandleFunc("/", func(resp http.ResponseWriter, req *http.Request) {
        fmt.Fprintln(resp, "Hello, QCon!")
    })
    http.ListenAndServe("0.0.0.0:8080", mux)
}

func serveDebug() {
    http.ListenAndServe("127.0.0.1:8001", http.DefaultServeMux)
}

func main() {
    go serveDebug()
    serveApp()
}

However, serveDebug is run in a separate goroutine and if it returns just that goroutine will exit while the rest of the program continues on. Your operations staff will not be happy to find that they cannot get the statistics out of your application when they want too because the /debug handler stopped working a long time ago.

I am new to Golang and coding in general. I came across an article online, and found this code. I copy and paste it into my editor and type go run main.go . The program runs forever without any errors. I can curl it with no problems. Why is it bad code? I am a noob, and I am trying to get a better understanding of this, if this could be explained in simple terms that would be great.

Answer 1

The program creates two HTTP servers to respond to traffic received on to different ports. The debug server is run in a separate goroutine, and there is no way to detect if that server failed. The program may continue running serving with the App server.

A better implementation would be to stop both servers if one of them failed:

stop:=make(chan struct{},2)
go func() {
    defer func() {
        stop<-struct{}{}
    }()
    serveDebug()
}()

go func() {
    defer func() {
         stop <-struct{}{}
    }{}
    serveApp()
}()
<-stop

Above, the program will create two goroutines and block at <-stop until someone writes to the channel. If any one of the servers fail, the goroutine will write to the channel, which will unblock <-stop , so the program will exit.