Re: fix int_sqrt() for very large numbers

[Will Deacon <will.deacon@arm.com>]

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