4.19-stable review patch. If anyone has any objections, please let me know.
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From: Florian La Roche florian.laroche@googlemail.com
commit fbfaf851902cd9293f392f3a1735e0543016d530 upstream.
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 Signed-off-by: Greg Kroah-Hartman gregkh@linuxfoundation.org
--- lib/int_sqrt.c | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-)
--- 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;