Messages in this thread | | | Date | Sun, 20 Jan 2019 00:01:40 +0000 | From | Will Deacon <> | Subject | Re: fix int_sqrt() for very large numbers |
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On Sat, Jan 19, 2019 at 04:14:50PM +0100, Florian La Roche wrote: > If an input number x for int_sqrt() has the highest bit set, then > __ffs(x) is 64. (1UL << 64) is an overflow and breaks the algorithm.
This is confusing, because the patch doesn't go near an __ffs().
> Just subtracting 1 is an even better guess for the initial > value of m and that's what also used to be done in earlier > versions of this code. > > best regards, > > Florian La Roche > > Signed-off-by: Florian La Roche <Florian.LaRoche@googlemail.com> > --- > lib/int_sqrt.c | 4 ++-- > 1 file changed, 2 insertions(+), 2 deletions(-) > > diff --git a/lib/int_sqrt.c b/lib/int_sqrt.c > index 14436f4ca6bd..ea00e84dc272 100644 > --- a/lib/int_sqrt.c > +++ b/lib/int_sqrt.c > @@ -23,7 +23,7 @@ unsigned long int_sqrt(unsigned long x) > if (x <= 1) > return x; > > - m = 1UL << (__fls(x) & ~1UL); > + m = 1UL << ((__fls(x) - 1) & ~1UL);
I think this one is fine, because __fls() gives you back 0-63 (or undefined, but the previous <= 1 check handles that case).
> while (m != 0) { > b = y + m; > y >>= 1; > @@ -52,7 +52,7 @@ u32 int_sqrt64(u64 x) > if (x <= ULONG_MAX) > return int_sqrt((unsigned long) x); > > - m = 1ULL << (fls64(x) & ~1ULL); > + m = 1ULL << ((fls64(x) - 1) & ~1ULL);
This just looks like a copy-paste error because there isn't an __fls64(). But I think your suggestion here is ok, given the previous check against ULONG_MAX.
Will
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