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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