[SCSI] libfc: Fix misleading debug statement

The statement reads, "Exchange timed out, notifying the upper layer",
however, this statement is printed whenever the timer is armed. This
is confusing to someone debugging the code because every time an
exchange is initialized, there is an incorrect statement stating that
the timer has already timed out. This patch changes the statement to
read, "Exchange timer armed" which is more accurate.

This patch also adds a debug statement in the timeout handler to
properly indicate that the exchange has timed out.

Signed-off-by: Robert Love <robert.w.love@intel.com>
Signed-off-by: James Bottomley <James.Bottomley@suse.de>

1 files changed, 3 insertions, 1 deletions

diff --git a/drivers/scsi/libfc/fc_exch.c b/drivers/scsi/libfc/fc_exch.c
index 11ddd115efb..40c34274bd8 100644
--- a/drivers/scsi/libfc/fc_exch.c
+++ b/drivers/scsi/libfc/fc_exch.c
@@ -326,7 +326,7 @@ static inline void fc_exch_timer_set_locked(struct fc_exch *ep,
         if (ep->state & (FC_EX_RST_CLEANUP | FC_EX_DONE))
                 return;
 
-        FC_EXCH_DBG(ep, "Exchange timed out, notifying the upper layer\n");
+        FC_EXCH_DBG(ep, "Exchange timer armed\n");
 
         if (schedule_delayed_work(&ep->timeout_work,
                                   msecs_to_jiffies(timer_msec)))
@@ -412,6 +412,8 @@ static void fc_exch_timeout(struct work_struct *work)
         u32 e_stat;
         int rc = 1;
 
+        FC_EXCH_DBG(ep, "Exchange timed out\n");
+
         spin_lock_bh(&ep->ex_lock);
         if (ep->state & (FC_EX_RST_CLEANUP | FC_EX_DONE))
                 goto unlock;