Dont even bother do allocate a channel

client.go

@@ -44,26 +44,14 @@ type Client struct {
 }
 
 // Do performs an asynchronous query. The msg *Msg is the question to ask, the 
-// string addr is the address of the nameserver. The methods returns a channel
-// of *Exchange. If ch is nil, a new channel is allocated and returned. If ch
-// is non nil, ch is used and returned.
-func (c *Client) Do(msg *Msg, addr string, ch chan *Exchange) <-chan *Exchange {
-	switch ch {
-	case nil:
-		ex := make(chan *Exchange)
-		go func() {
-			r, rtt, err := c.Exchange(msg, addr)
-			ex <- &Exchange{msg, r, addr, rtt, err}
-		}()
-		return ex
-	default:
-		go func() {
-			r, rtt, err := c.Exchange(msg, addr)
-			ch <- &Exchange{msg, r, addr, rtt, err}
-		}()
-		return ch
-	}
-	panic("dns: not reached")
+// string addr is the address of the nameserver, ch is a channel of *Exchange.
+// This channel is used to send to the reply from the server to.
+func (c *Client) Do(msg *Msg, addr string, ch chan *Exchange) {
+	go func() {
+		r, rtt, err := c.Exchange(msg, addr)
+		ch <- &Exchange{msg, r, addr, rtt, err}
+	}()
+	return
 }
 
 // Exchange performs an synchronous query. It sends the message m to the address