Messages in this thread Patch in this message | | | Date | Wed, 20 Apr 2011 21:41:54 -0400 | From | Steven Rostedt <> | Subject | [PATCH 1/7] lockdep: Print a nice description of an irq locking issue |
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From: Steven Rostedt <srostedt@redhat.com>
Locking order inversion due to interrupts is a subtle problem. When a locking inversion due to interrupts is discovered by lockdep, it currently reports something like this:
[ INFO: HARDIRQ-safe -> HARDIRQ-unsafe lock order detected ]
And then writes the locks that are involved as well as back traces. But several developers are confused by what a HARDIRQ->safe to unsafe issue is all about, and sometimes even blow it off as a bug in lockdep. As it is not obvious when lockdep describes this about a lock that is never taken in interrupt context.
After explaining the problems that lockdep is reporting, I decided to add a description of the problem in visual form. Now the following is shown:
--- other info that might help us debug this:
Possible interrupt unsafe locking scenario:
CPU0 CPU1 ---- ---- lock(lockA); local_irq_disable(); lock(&rq->lock); lock(lockA); <Interrupt> lock(&rq->lock);
*** DEADLOCK ***
---
The above is the case when the unsafe lock is taken while holding a lock taken in irq context. But when a lock is taken that also grabs a unsafe lock, the call chain is shown:
--- 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 ***
Acked-by: Peter Zijlstra <a.p.zijlstra@chello.nl> Signed-off-by: Steven Rostedt <rostedt@goodmis.org> --- kernel/lockdep.c | 70 ++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 files changed, 70 insertions(+), 0 deletions(-)
diff --git a/kernel/lockdep.c b/kernel/lockdep.c index 0d2058d..bb77c030 100644 --- a/kernel/lockdep.c +++ b/kernel/lockdep.c @@ -490,6 +490,18 @@ void get_usage_chars(struct lock_class *class, char usage[LOCK_USAGE_CHARS]) usage[i] = '\0'; } +static int __print_lock_name(struct lock_class *class) +{ + char str[KSYM_NAME_LEN]; + const char *name; + + name = class->name; + if (!name) + name = __get_key_name(class->key, str); + + return printk("%s", name); +} + static void print_lock_name(struct lock_class *class) { char str[KSYM_NAME_LEN], usage[LOCK_USAGE_CHARS]; @@ -1325,6 +1337,62 @@ print_shortest_lock_dependencies(struct lock_list *leaf, return; } +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 *safe_class = safe_entry->class; + struct lock_class *unsafe_class = unsafe_entry->class; + struct lock_class *middle_class = hlock_class(prev); + + if (middle_class == safe_class) + middle_class = hlock_class(next); + + /* + * A direct locking problem where unsafe_class lock is taken + * directly by safe_class lock, then all we need to show + * is the deadlock scenario, as it is obvious that the + * unsafe lock is taken under the safe lock. + * + * But if there is a chain instead, where the safe lock takes + * an intermediate lock (middle_class) where this lock is + * not the same as the safe lock, then the lock chain is + * used to describe the problem. Otherwise we would need + * to show a different CPU case for each link in the chain + * from the safe_class lock to the unsafe_class lock. + */ + if (middle_class != unsafe_class) { + printk("Chain exists of:\n "); + __print_lock_name(safe_class); + printk(" --> "); + __print_lock_name(middle_class); + printk(" --> "); + __print_lock_name(unsafe_class); + printk("\n\n"); + } + + printk(" Possible interrupt unsafe locking scenario:\n\n"); + printk(" CPU0 CPU1\n"); + printk(" ---- ----\n"); + printk(" lock("); + __print_lock_name(unsafe_class); + printk(");\n"); + printk(" local_irq_disable();\n"); + printk(" lock("); + __print_lock_name(safe_class); + printk(");\n"); + printk(" lock("); + __print_lock_name(middle_class); + printk(");\n"); + printk(" <Interrupt>\n"); + printk(" lock("); + __print_lock_name(safe_class); + printk(");\n"); + printk("\n *** DEADLOCK ***\n\n"); +} + static int print_bad_irq_dependency(struct task_struct *curr, struct lock_list *prev_root, @@ -1376,6 +1444,8 @@ 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); + lockdep_print_held_locks(curr); printk("\nthe dependencies between %s-irq-safe lock", irqclass); -- 1.7.2.3
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