fix int_sqrt() for very large numbers

[Florian La Roche]

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.

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.laro...@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);
        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);
        while (m != 0) {
                b = y + m;
                y >>= 1;
-- 
2.17.1