All wake_futex() callers already verify that the we are not dealing with
a pi futex_q, so we can remove the redundant WARN() check, as this is never
triggered anyway.
Cc: Ingo Molnar <mi...@kernel.org>
Cc: Darren Hart <dvh...@linux.intel.com>
Cc: Peter Zijlstra <pet...@infradead.org>
Cc: Thomas Gleixner <t...@linutronix.de>
Cc: Mike Galbraith <efa...@gmx.de>
Cc: Jeff Mahoney <je...@suse.com>
Cc: Linus Torvalds <torva...@linux-foundation.org>
Cc: Scott Norton <scott.nor...@hp.com>
Cc: Tom Vaden <tom.va...@hp.com>
Cc: Aswin Chandramouleeswaran <as...@hp.com>
Cc: Waiman Long <waiman.l...@hp.com>
Cc: Jason Low <jason.l...@hp.com>
Signed-off-by: Davidlohr Bueso <davidl...@hp.com>
---
kernel/futex.c | 3 ---
1 file changed, 3 deletions(-)
diff --git a/kernel/futex.c b/kernel/futex.c
index e6ffe73..0768c68 100644
--- a/kernel/futex.c
+++ b/kernel/futex.c
@@ -844,9 +844,6 @@ static void wake_futex(struct futex_q *q)
{
struct task_struct *p = q->task;
- if (WARN(q->pi_state || q->rt_waiter, "refusing to wake PI futex\n"))
- return;
-
/*
* We set q->lock_ptr = NULL _before_ we wake up the task. If
* a non-futex wake up happens on another CPU then the task
--
1.8.1.4
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