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authorSteven Rostedt <srostedt@redhat.com>2011-04-20 21:41:57 -0400
committerIngo Molnar <mingo@elte.hu>2011-04-22 11:06:58 +0200
commitdad3d7435e1d8c254d6877dc06852dc00c5da812 (patch)
tree85f4ab38ff63879b117d363ef1fdb3cf83bf870d
parent48702ecf308e53f176c1f6fdc193d622ded54ac0 (diff)
lockdep: Print a nicer description for irq inversion bugs
Irq inversion and irq dependency bugs are only subtly different. The diffenerence lies where the interrupt occurred. For irq dependency: irq_disable lock(A) lock(B) unlock(B) unlock(A) irq_enable lock(B) unlock(B) <interrupt> lock(A) The interrupt comes in after it has been established that lock A can be held when taking an irq unsafe lock. Lockdep detects the problem when taking lock A in interrupt context. With the irq_inversion the irq happens before it is established and lockdep detects the problem with the taking of lock B: <interrupt> lock(A) irq_disable lock(A) lock(B) unlock(B) unlock(A) irq_enable lock(B) unlock(B) Since the problem with the locking logic for both of these issues is in actuality the same, they both should report the same scenario. This patch implements that and prints this: other info that might help us debug this: Chain exists of: &rq->lock --> lockA --> lockC Possible interrupt unsafe locking scenario: CPU0 CPU1 ---- ---- lock(lockC); local_irq_disable(); lock(&rq->lock); lock(lockA); <Interrupt> lock(&rq->lock); *** DEADLOCK *** Signed-off-by: Steven Rostedt <rostedt@goodmis.org> Acked-by: Peter Zijlstra <a.p.zijlstra@chello.nl> Cc: Frederic Weisbecker <fweisbec@gmail.com> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Andrew Morton <akpm@linux-foundation.org> Link: http://lkml.kernel.org/r/20110421014259.910720381@goodmis.org Signed-off-by: Ingo Molnar <mingo@elte.hu>
-rw-r--r--kernel/lockdep.c34
1 files changed, 29 insertions, 5 deletions
diff --git a/kernel/lockdep.c b/kernel/lockdep.c
index c4cc5d1acf4..0b497dda6ae 100644
--- a/kernel/lockdep.c
+++ b/kernel/lockdep.c
@@ -1395,15 +1395,15 @@ print_shortest_lock_dependencies(struct lock_list *leaf,
static void
print_irq_lock_scenario(struct lock_list *safe_entry,
struct lock_list *unsafe_entry,
- struct held_lock *prev,
- struct held_lock *next)
+ struct lock_class *prev_class,
+ struct lock_class *next_class)
{
struct lock_class *safe_class = safe_entry->class;
struct lock_class *unsafe_class = unsafe_entry->class;
- struct lock_class *middle_class = hlock_class(prev);
+ struct lock_class *middle_class = prev_class;
if (middle_class == safe_class)
- middle_class = hlock_class(next);
+ middle_class = next_class;
/*
* A direct locking problem where unsafe_class lock is taken
@@ -1499,7 +1499,8 @@ print_bad_irq_dependency(struct task_struct *curr,
print_stack_trace(forwards_entry->class->usage_traces + bit2, 1);
printk("\nother info that might help us debug this:\n\n");
- print_irq_lock_scenario(backwards_entry, forwards_entry, prev, next);
+ print_irq_lock_scenario(backwards_entry, forwards_entry,
+ hlock_class(prev), hlock_class(next));
lockdep_print_held_locks(curr);
@@ -2219,6 +2220,10 @@ print_irq_inversion_bug(struct task_struct *curr,
struct held_lock *this, int forwards,
const char *irqclass)
{
+ struct lock_list *entry = other;
+ struct lock_list *middle = NULL;
+ int depth;
+
if (!debug_locks_off_graph_unlock() || debug_locks_silent)
return 0;
@@ -2237,6 +2242,25 @@ print_irq_inversion_bug(struct task_struct *curr,
printk("\n\nand interrupts could create inverse lock ordering between them.\n\n");
printk("\nother info that might help us debug this:\n");
+
+ /* Find a middle lock (if one exists) */
+ depth = get_lock_depth(other);
+ do {
+ if (depth == 0 && (entry != root)) {
+ printk("lockdep:%s bad path found in chain graph\n", __func__);
+ break;
+ }
+ middle = entry;
+ entry = get_lock_parent(entry);
+ depth--;
+ } while (entry && entry != root && (depth >= 0));
+ if (forwards)
+ print_irq_lock_scenario(root, other,
+ middle ? middle->class : root->class, other->class);
+ else
+ print_irq_lock_scenario(other, root,
+ middle ? middle->class : other->class, root->class);
+
lockdep_print_held_locks(curr);
printk("\nthe shortest dependencies between 2nd lock and 1st lock:\n");