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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>
Peter Zijlstra 16 years ago
parent
commit
ac7b900490
1 changed files with 2 additions and 1 deletions
  1. 2 1
      include/linux/kernel.h

+ 2 - 1
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