Re: generic swap()

[debian developer <debiandev@xxxxxxxxx>]

I don't seem to understand this patch. What is swap(a, b) returning here? and where is it returning from?

What advantage does using a do{}while(0); have on this?

On Fri, Jan 16, 2009 at 3:15 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:

On Fri, 2009-01-16 at 01:38 -0800, Andrew Morton wrote:
> On Fri, 16 Jan 2009 10:26:34 +0100 Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>
> > Hi,
> >
> > Where did this patch come from? -- I can't seem to find it on lkml at all..
>
> That's a damn good question.  I received:

Hehe, one of the mysteries of life then :-)

> > The reason I ask it that I wonder why swap() has a return value?
>
> Accident, I guess.
>
> > +#define swap(a, b) ({ typeof(a) __tmp = (a); (a) = (b); (b) = __tmp; })
>
> You think it should use the do{}while(0) thing?

That was what I was thinking indeed. Non of the current users appear to
make use of the (somewhat arbitrary) return value, so lets change it
before someone gets creative ;-)

Signed-off-by: Peter Zijlstra <a.p.zijlstra@xxxxxxxxx>
---

diff --git a/include/linux/kernel.h b/include/linux/kernel.h

index 343df9e..7fa3718 100644

--- 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

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