Re: [PATCH 1/3] utils: Improve qemu_strtosz() to have 64 bits of precision

["Daniel P. Berrangé" <berrange@redhat.com>]

On Fri, Feb 05, 2021 at 08:06:53AM -0600, Eric Blake wrote:
> On 2/5/21 4:06 AM, Vladimir Sementsov-Ogievskiy wrote:
> 
> >>> -    /*
> >>> -     * Values near UINT64_MAX overflow to 2**64 when converting to
> >>> double
> >>> -     * precision.  Compare against the maximum representable double
> >>> precision
> >>> -     * value below 2**64, computed as "the next value after 2**64
> >>> (0x1p64) in
> >>> -     * the direction of 0".
> >>> -     */
> >>> -    if ((val * mul > nextafter(0x1p64, 0)) || val < 0) {
> >>> +    if (val > UINT64_MAX / mul) {
> >>
> >> Hmm, do we care about:
> >> 15.9999999999999999999999999999E
> >> where the fractional portion becomes large enough to actually bump our
> >> sum below to 16E which indeed overflows?  Then again, we rejected a
> >> fraction of 1.0 above, and 0.9999999999999999999999999999 parses to 1.0
> >> due to rounding.
> >> Maybe it's just worth a good comment why the overflow check here works
> >> without consulting fraction.
> > 
> > worth a good comment, because I don't follow :)
> > 
> > If mul is big enough and fraction=0.5, why val*mul + fraction*mul will
> > not overflow?
> 
> When mul is a power of 2, we know that fraction*mul does not change the
> number of significant bits, but merely moves the exponent, so starting
> with fraction < 1.0, we know fraction*mul < mul.  But when @unit is
> 1000, there is indeed a rare possibility that the multiplication will
> cause an inexact answer that will trigger rounding, so we could end up
> with fraction*mul == mul.  So I'm not yet 100% confident that there is
> no possible combination where we can't cause an overflow to result in
> val*mul + (uint64_t)(fraction*mul) resulting in 0 instead of UINT64_MAX,
> and I think I will have to tighten this code up for v2.
> 
> 
> > 
> > Also, if we find '.' in the number, why not just reparse the whole
> > number with qemu_strtod_finite? And don't care about 1.0?
> 
> Reparsing the whole number loses precision. Since we already have a
> 64-bit precise integer, why throw it away?

Yep, it isn't acceptable to throw away precision of the non-fractional
part of the input IMHO.

Regards,
Daniel
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