dhcp: do not leave the old socket behind when starting again

dhcp_start() never looked at whether DHCP was already running, and "ip dhcp"
calls it without stopping first, so a second start took a fresh slot from
uip_udp_conns and left the previous one bound.

The leak is the smaller half of it. dhcp_callback() decides whether a call
is for it by comparing the local port, not the connection, and every
leftover socket still holds port 68, so each one runs the state machine on
every poll. The timers live in one shared dhcp_state, so ticks count down
once per socket: two starts and the retransmit and renewal timing runs at
twice the intended rate.

Stop the old one first. dhcp_stop() already does exactly that, and the
state test keeps it away from a connection that was never created.

dhcp.c

@@ -341,6 +341,12 @@ void parse_dhcp(void)
 
 void dhcp_start(void) __banked
 {
+	/* Starting twice would leave the first socket bound: uip_udp_new() takes
+	 * the next free slot while the old one keeps port 68, and dhcp_callback()
+	 * matches on that port rather than on the connection, so every leftover
+	 * runs the state machine too and the timers count down once per socket. */
+	if (dhcp_state.state != DHCP_OFF)
+		dhcp_stop();
 	uip_ipaddr(server, 255,255,255,255);
 	dhcp_state.conn = uip_udp_new(&server, HTONS(DHCPC_SERVER_PORT));
 	dhcp_state.current_ip[0] = dhcp_state.current_ip[1] = dhcp_state.current_ip[2] = dhcp_state.current_ip[3] = 0;