Re: [tip:sched/core] sched: Lower chances of cputime scaling overflow

[Stanislaw Gruszka]

On Thu, Apr 11, 2013 at 03:45:46PM +0200, Peter Zijlstra wrote:
> On Tue, 2013-03-26 at 15:01 +0100, Stanislaw Gruszka wrote:
> > Thoughts?
> 
> Would something like the below work?

Not sure, need to validate that?

> (warning: it's never even been near a compiler)

It compile, but probably has some bugs.
> +	/*
> +	 * Since the stime:utime ratio is already an approximation through
> +	 * the sampling, reducing its resolution isn't too big a deal.
> +	 * And since total = stime+utime; the total_fls will be the biggest
> +	 * of the two;
> +	 */
> +	if (total_fls > 32) {
> +		shift = total_fls - 32; /* a = 2^shift */
> +		stime >>= shift;
> +		total >>= shift;
> +		stime_fls -= shift;
> +		total_fls -= shift;
> +	}
> +
> +	/*
> +	 * Since we limited stime to 32bits the multiplication reduced to 96bit.
> +	 *   stime * rtime = stime * (rl + rh * 2^32) = 
> +	 *                   stime * rl + stime * rh * 2^32
> +	 */
> +	lo = stime * rtime_lo;
> +	hi = stime * rtime_hi;
> +	t = hi << 32;
> +	lo += t;
> +	if (lo < t) /* overflow */
> +		hi += 0x100000000L;
> +	hi >>= 32;

I do not understand why we shift hi value here, is that correct?

> +	/*
> +	 * Pick the 64 most significant bits for division into @lo.
> +	 * 
> +	 * NOTE: res_fls is an approximation (upper-bound) do we want to
> +	 *       properly calculate?
> +	 */
> +	shift = 0;
> +	res_fls = stime_fls + rtime_fls;
> +	if (res_fls > 64) {
> +		shift = res_fls - 64; /* b = 2^shift */
> +		lo >>= shift;
> +		hi <<= 64 - shift;
> +		lo |= hi;
>  	}
>  
> -	return (__force cputime_t) scaled;
> +	/*
> +	 * So here we do:
> +	 *
> +	 *    ((stime / a) * rtime / b)
> +	 *    --------------------------- / b
> +	 *           (total / a)
> +	 */
> +	return div_u64(lo, total) >> shift;

I think it should be:

 ((stime / a) * rtime / b)
--------------------------- * b
        (total / a)

return div_u64(lo, total) << shift;

Thanks
Stanislaw

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