sched/debug: Fix SCHED_WARN_ON() to return a value on !CONFIG_SCHED_DEBUG as well

This definition of SCHED_WARN_ON():

 #define SCHED_WARN_ON(x)        ((void)(x))

is not fully compatible with the 'real' WARN_ON_ONCE() primitive, as it
has no return value, so it cannot be used in conditionals.

Fix it.

Cc: Daniel Axtens <dja@axtens.net>
Cc: Konstantin Khlebnikov <khlebnikov@yandex-team.ru>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Thomas Gleixner <tglx@linutronix.de>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: linux-kernel@vger.kernel.org
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Ingo Molnar 2017-06-20 12:24:42 +02:00
parent ebfa4c02fa
commit 6d3aed3d8a
1 changed files with 2 additions and 2 deletions

View File

@ -39,9 +39,9 @@
#include "cpuacct.h" #include "cpuacct.h"
#ifdef CONFIG_SCHED_DEBUG #ifdef CONFIG_SCHED_DEBUG
#define SCHED_WARN_ON(x) WARN_ONCE(x, #x) # define SCHED_WARN_ON(x) WARN_ONCE(x, #x)
#else #else
#define SCHED_WARN_ON(x) ((void)(x)) # define SCHED_WARN_ON(x) ({ (void)(x), 0; })
#endif #endif
struct rq; struct rq;