NTB: Correctly handle receive buffers of the minimal size

The ring logic of the NTB receive buffer/transmit memory window requires
there to be at least 2 payload sized allotments.  For the minimal size
case, split the buffer into two and set the transport_mtu to the
appropriate size.
Signed-off-by: NJon Mason <jon.mason@intel.com>

drivers/ntb/ntb_transport.c

@@ -490,11 +490,12 @@ static void ntb_transport_setup_qp_mw(struct ntb_transport *nt,
 	rx_size -= sizeof(struct ntb_rx_info);
 
 	qp->rx_buff = qp->remote_rx_info + 1;
-	qp->rx_max_frame = min(transport_mtu, rx_size);
+	/* Due to housekeeping, there must be atleast 2 buffs */
+	qp->rx_max_frame = min(transport_mtu, rx_size / 2);
 	qp->rx_max_entry = rx_size / qp->rx_max_frame;
 	qp->rx_index = 0;
 
-	qp->remote_rx_info->entry = qp->rx_max_entry;
+	qp->remote_rx_info->entry = qp->rx_max_entry - 1;
 
 	/* setup the hdr offsets with 0's */
 	for (i = 0; i < qp->rx_max_entry; i++) {
@@ -818,7 +819,8 @@ static void ntb_transport_init_queue(struct ntb_transport *nt,
 	tx_size -= sizeof(struct ntb_rx_info);
 
 	qp->tx_mw = qp->rx_info + 1;
-	qp->tx_max_frame = min(transport_mtu, tx_size);
+	/* Due to housekeeping, there must be atleast 2 buffs */
+	qp->tx_max_frame = min(transport_mtu, tx_size / 2);
 	qp->tx_max_entry = tx_size / qp->tx_max_frame;
 
 	if (nt->debugfs_dir) {