P2P: Do not use wait_time for SD Response TX without fragmentation

The full SD Response frame is not going to be followed by another Action
frame from the peer, so remove the 200 ms wait time from the offchannel
TX command in that case. This avoids leaving a 200 ms lock on the radio
to remain on the channel unnecessarily.

Signed-off-by: Jouni Malinen <jouni@qca.qualcomm.com>

src/p2p/p2p_sd.c

@@ -426,6 +426,7 @@ void p2p_sd_response(struct p2p_data *p2p, int freq, const u8 *dst,
 {
 	struct wpabuf *resp;
 	size_t max_len;
+	unsigned int wait_time = 200;
 
 	/*
 	 * In the 60 GHz, we have a smaller maximum frame length for management
@@ -460,6 +461,7 @@ void p2p_sd_response(struct p2p_data *p2p, int freq, const u8 *dst,
 					     1, p2p->srv_update_indic, NULL);
 	} else {
 		p2p_dbg(p2p, "SD response fits in initial response");
+		wait_time = 0; /* no more SD frames in the sequence */
 		resp = p2p_build_sd_response(dialog_token,
 					     WLAN_STATUS_SUCCESS, 0,
 					     p2p->srv_update_indic, resp_tlvs);
@@ -470,7 +472,7 @@ void p2p_sd_response(struct p2p_data *p2p, int freq, const u8 *dst,
 	p2p->pending_action_state = P2P_NO_PENDING_ACTION;
 	if (p2p_send_action(p2p, freq, dst, p2p->cfg->dev_addr,
 			    p2p->cfg->dev_addr,
-			    wpabuf_head(resp), wpabuf_len(resp), 200) < 0)
+			    wpabuf_head(resp), wpabuf_len(resp), wait_time) < 0)
 		p2p_dbg(p2p, "Failed to send Action frame");
 
 	wpabuf_free(resp);