[PATCH v4 08/30] locking/lockdep: Skip checks if direct dependency is already present

[Yuyang Du <duyuyang@gmail.com>]

Given a dependency <prev> -> <next>, two checks are performed:

1. Lock inversion deadlock:

We search whether there is a path from <next> to <prev> in the dependency
graph and if so we have a potential deadlock scenario in check_deadlock_graph().
But if the direct dependency <prev> -> <next> is already in the graph, there
can't be such a path (i.e., <next> to <prev>) because otherwise this path
would have been found when adding the last critical dependency that
completes the circle.

2. IRQ usage violation:

The IRQ usage check searches whether there is a path through <prev> to
<next> that connects an irq-safe lock to an irq-unsafe lock in the
dependency graph in check_irq_usage(). Similarly, if <prev> -> <next> is
already in the graph, there can't be such a path either.

This check skipping should be able to greatly improve performance by
reducing the number of deadlock and IRQ usage checks. This number precisely
equals nr_redundant, which actually is not a small number.

Signed-off-by: Yuyang Du <duyuyang@gmail.com>
---
 kernel/locking/lockdep.c | 34 +++++++++++++++++++---------------
 1 file changed, 19 insertions(+), 15 deletions(-)

diff --git a/kernel/locking/lockdep.c b/kernel/locking/lockdep.c
index 4838c99..de088da 100644
--- a/kernel/locking/lockdep.c
+++ b/kernel/locking/lockdep.c
@@ -2433,6 +2433,25 @@ static inline void inc_chains(void)
 	}
 
 	/*
+	 * Is the <prev> -> <next> dependency already present?
+	 *
+	 * (this may occur even though this is a new chain: consider
+	 *  e.g. the L1 -> L2 -> L3 -> L4 and the L5 -> L1 -> L2 -> L3
+	 *  chains - the second one will be new, but L1 already has
+	 *  L2 added to its dependency list, due to the first chain.)
+	 */
+	list_for_each_entry(entry, &hlock_class(prev)->locks_after, entry) {
+		if (entry->class == hlock_class(next)) {
+			debug_atomic_inc(nr_redundant);
+
+			if (distance == 1)
+				entry->distance = 1;
+
+			return 1;
+		}
+	}
+
+	/*
 	 * Prove that the new <prev> -> <next> dependency would not
 	 * create a deadlock scenario in the graph. (We do this by
 	 * a breadth-first search into the graph starting at <next>,
@@ -2459,21 +2478,6 @@ static inline void inc_chains(void)
 	 */
 	if (next->read == 2 || prev->read == 2)
 		return 1;
-	/*
-	 * Is the <prev> -> <next> dependency already present?
-	 *
-	 * (this may occur even though this is a new chain: consider
-	 *  e.g. the L1 -> L2 -> L3 -> L4 and the L5 -> L1 -> L2 -> L3
-	 *  chains - the second one will be new, but L1 already has
-	 *  L2 added to its dependency list, due to the first chain.)
-	 */
-	list_for_each_entry(entry, &hlock_class(prev)->locks_after, entry) {
-		if (entry->class == hlock_class(next)) {
-			if (distance == 1)
-				entry->distance = 1;
-			return 1;
-		}
-	}
 
 	if (!*trace) {
 		*trace = save_trace();
-- 
1.8.3.1