Re: [PATCH] sched: Improve readability in update_cpu_load() code

[Srivatsa Vaddagiri <vatsa@linux.vnet.ibm.com>]

On Thu, May 15, 2008 at 06:34:59PM +0530, Gautham R Shenoy wrote:
> Author: Gautham R Shenoy <ego@in.ibm.com>
> Date:   Thu May 15 17:55:49 2008 +0530
> 
>     sched: Improve readability in update_cpu_load() code
>     
>     Currently the cpu_load[i] is calculated as:
>     	this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
>     
>     However, since scale = 2^i, this can be simplified as:
>     	this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);
>     
>     Makes it easier to read.
>     Signed-off-by: Gautham R Shenoy <ego@in.ibm.com>
> 
> diff --git a/kernel/sched.c b/kernel/sched.c
> index 2d7d8f1..e1a6985 100644
> --- a/kernel/sched.c
> +++ b/kernel/sched.c
> @@ -2921,7 +2921,7 @@ static void update_cpu_load(struct rq *this_rq)
>  		 */
>  		if (new_load > old_load)
>  			new_load += scale-1;
> -		this_rq->cpu_load[i] = (old_load*(scale-1) + new_load) >> i;
> +		this_rq->cpu_load[i] = old_load + ((new_load - old_load) >> i);

This wont work when new_load < old_load ..

For ex: I tried this prog:

#include <stdio.h>

main()
{
	unsigned long old_load = 100, new_load = 90, this_load, this_load1;
	int i = 1, scale = 2 << i;

	this_load = (old_load*(scale-1) + new_load) >> i;
	this_load1 = old_load + ((new_load - old_load) >> i);

	printf ("this_load = %u, this_load1 = %u \n", this_load, this_load1);
}

This is what I get:

$ ./a.out
this_load = 195, this_load1 = 2147483743
$

	

-- 
Regards,
vatsa