Re: [KERNEL]: Avoid divide in IS_ALIGN

[linux-os \(Dick Johnson\)]

On Tue, 20 Nov 2007, Herbert Xu wrote:

> Hi:
>
> [KERNEL]: Avoid divide in IS_ALIGN
>
> I was happy to discover the brand new IS_ALIGN macro and quickly
> used it in my code.  To my dismay I found that the generated code
> used division to perform the test.
>
> This patch fixes it by changing the % test to an &.  This avoids
> the division.
>
> Signed-off-by: Herbert Xu <herbert@xxxxxxxxxxxxxxxxxxx>
>
> Cheers,
> -- 
> Visit Openswan at http://www.openswan.org/
> Email: Herbert Xu ~{PmV>HI~} <herbert@xxxxxxxxxxxxxxxxxxx>
> Home Page: http://gondor.apana.org.au/~herbert/
> PGP Key: http://gondor.apana.org.au/~herbert/pubkey.txt
> --
> diff --git a/include/linux/kernel.h b/include/linux/kernel.h
> index 94bc996..39b3fa6 100644
> --- a/include/linux/kernel.h
> +++ b/include/linux/kernel.h
> @@ -35,7 +35,7 @@ extern const char linux_proc_banner[];
> #define ALIGN(x,a)		__ALIGN_MASK(x,(typeof(x))(a)-1)
> #define __ALIGN_MASK(x,mask)	(((x)+(mask))&~(mask))
> #define PTR_ALIGN(p, a)		((typeof(p))ALIGN((unsigned long)(p), (a)))
> -#define IS_ALIGNED(x,a)		(((x) % ((typeof(x))(a))) == 0)
> +#define IS_ALIGNED(x, a)		(((x) & ((typeof(x))(a) - 1)) == 0)
>

Your macro modification is wrong.

Take 0x12345600, which is aligned on a 16-byte boundary.
Now, with your macro, we have for a 0x10 alignment:
      (0x12345600 & 0x0f) = (0x00 == 0) == TRUE (correct)
For an 0x20 alignment:
      (0x12345600 & 0x1f) = (0x00 == 0) == TRUE (incorrect)

In other words, the macro may work for some alignments, but
not all. It is therefore wrong. You need the modulus, the
remainder after division. Anything else is broken.

Cheers,
Dick Johnson
Penguin : Linux version 2.6.22.1 on an i686 machine (5588.30 BogoMips).
My book : http://www.AbominableFirebug.com/
_


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