fix int_sqrt64() for very large numbers

If an input number x for int_sqrt64() has the highest bit set, then
fls64(x) is 64.  (1UL << 64) is an overflow and breaks the algorithm.

Subtracting 1 is a better guess for the initial value of m anyway and
that's what also done in int_sqrt() implicitly [*].

[*] Note how int_sqrt() uses __fls() with two underscores, which already
    returns the proper raw bit number.

    In contrast, int_sqrt64() used fls64(), and that returns bit numbers
    illogically starting at 1, because of error handling for the "no
    bits set" case. Will points out that he bug probably is due to a
    copy-and-paste error from the regular int_sqrt() case.

Signed-off-by: Florian La Roche <Florian.LaRoche@googlemail.com>
Acked-by: Will Deacon <will.deacon@arm.com>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>

diff --git a/lib/int_sqrt.c b/lib/int_sqrt.c
index 14436f4ca6bd7f340971a4f85ded981a9c0b25dc..30e0f9770f88cef66f8655c0b377366b7d986db8 100644 (file)
--- a/lib/int_sqrt.c
+++ b/lib/int_sqrt.c
@@ -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;