generic swap(): don't return a value from swap()

The swap() macro is accidentally retuning the value of its first argument.
Change it into a doesn't-return-anything macro before someone goes and
relies upon this behaviour.

Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl>
Cc: Wu Fengguang <wfg@linux.intel.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>

diff --git a/include/linux/kernel.h b/include/linux/kernel.h
index 343df9ef2412393020e90c1ef5227135b3b5a07f..7fa371898e3e76d8e1af870f049844ff71bea931 100644 (file)
--- a/include/linux/kernel.h
+++ b/include/linux/kernel.h
@@ -480,7 +480,8 @@ static inline char *pack_hex_byte(char *buf, u8 byte)
 /*
  * swap - swap value of @a and @b
  */
-#define swap(a, b) ({ typeof(a) __tmp = (a); (a) = (b); (b) = __tmp; })
+#define swap(a, b) \
+       do { typeof(a) __tmp = (a); (a) = (b); (b) = __tmp; } while (0)
 
 /**
  * container_of - cast a member of a structure out to the containing structure