在go lang中是否在amqp.Dial具有线程安全性时每次创建连接

问题描述：

As it is mentioned in the RabbitMQ docs that tcp connections are expensive to make. So, for that concept of channel was introduced. Now i came across this example. In the main() it creates the connection everytime a message is publised. conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/"). Shouldn't it be declared globally once and there should be failover mechanism in case connection get closed like singleton object. If amqp.Dial is thread-safe, which i suppose it should be

Edited question :

I am handling the connection error in the following manner. In which i listen on a channel and create a new connection on error. But when i kill the existing connection and try to publish message. I get the following error.

error :

2016/03/30 19:20:08 Failed to open a channel: write tcp 172.16.5.48:51085->172.16.0.20:5672: use of closed network connection
exit status 1
7:25 PM

Code :

 func main() {

        Conn, err := amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")
        failOnError(err, "Failed to connect to RabbitMQ")
         context := &appContext{queueName: "QUEUENAME",exchangeName: "ExchangeName",exchangeType: "direct",routingKey: "RoutingKey",conn: Conn}
        c := make(chan *amqp.Error)

        go func() {
            error := <-c
            if(error != nil){                
                Conn, err = amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")            
                failOnError(err, "Failed to connect to RabbitMQ")            
                Conn.NotifyClose(c)                                           
            }            
        }()

        Conn.NotifyClose(c)
        r := web.New()
        // We pass an instance to our context pointer, and our handler.
        r.Get("/", appHandler{context, IndexHandler})
        graceful.ListenAndServe(":8086", r)  

    }

答

Of course, you shouldn't create a connection for each request. Make it a global variable or better part of an application context which you initialize once at startup.

You can handle connection errors by registering a channel using Connection.NotifyClose:

func initialize() {
  c := make(chan *amqp.Error)
  go func() {
    err := <-c
    log.Println("reconnect: " + err.Error())
    initialize()
  }()

  conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
  if err != nil {
    panic("cannot connect")
  }
  conn.NotifyClose(c)

  // create topology
}

在go lang中是否在amqp.Dial具有线程安全性时每次创建连接

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