On 24/03/2015 01:22, David Gibson wrote: > On Mon, Mar 23, 2015 at 10:54:39AM +0100, BALATON Zoltan wrote: >> On Mon, 23 Mar 2015, David Gibson wrote: >>> If the guest programs a sufficiently large timeout value an >>> integer overflow can occur in i6300esb_restart_timer(). e.g. >>> if the maximum possible timer preload value of 0xfffff is >>> programmed then we end up with the calculation: >>> >>> timeout = get_ticks_per_sec() * (0xfffff << 15) / 33000000; >>> >>> get_ticks_per_sec() returns 1000000000 (10^9) giving: >>> >>> 10^9 * (0xfffff * 2^15) == 0x1dcd632329b000000 (65 bits) >>> >>> Obviously the division by 33MHz brings it back under 64-bits, >>> but the overflow has already occurred. >>> >>> Since signed integer overflow has undefined behaviour in C, in >>> theory this could be arbitrarily bad. In practice, the >>> overflowed value wraps around to something negative, causing >>> the watchdog to immediately expire, killing the guest, which is >>> still fairly bad. >>> >>> The bug can be triggered by running a Linux guest, loading the >>> i6300esb driver with parameter "heartbeat=2046" and opening >>> /dev/watchdog. The watchdog will trigger as soon as the device >>> is opened. >>> >>> This patch corrects the problem by using muldiv64(), which >>> effectively allows a 128-bit intermediate value between the >>> multiplication and division. >>> >>> Signed-off-by: David Gibson <[email protected]> --- >>> hw/watchdog/wdt_i6300esb.c | 10 ++++++++-- 1 file changed, 8 >>> insertions(+), 2 deletions(-) >>> >>> diff --git a/hw/watchdog/wdt_i6300esb.c >>> b/hw/watchdog/wdt_i6300esb.c index e694fa9..c7316f5 100644 --- >>> a/hw/watchdog/wdt_i6300esb.c +++ b/hw/watchdog/wdt_i6300esb.c >>> @@ -125,8 +125,14 @@ static void >>> i6300esb_restart_timer(I6300State *d, int stage) else timeout >>> <<= 5; >>> >>> - /* Get the timeout in units of ticks_per_sec. */ - >>> timeout = get_ticks_per_sec() * timeout / 33000000; + /* Get >>> the timeout in units of ticks_per_sec. + * + * >>> ticks_per_sec is typically 10^9 == 0x3B9ACA00 (30 bits), with + >>> * 20 bits of user supplied preload, and 15 bits of scale, the + >>> * multiply here can exceed 64-bits, before we divide by 33MHz, >>> so + * we use a 128-bit temporary + */ >> >> Is the comment still correct saying "we use a 128-bit temporary" >> when the code does not do that explicitely any more? > > Bother. I fixed the commit message, but not this comment. It's > still kind of correct, in that muldiv64 does effectively have a > 128-bit temporary internally.
Yes, that's how I interpreted it. Though strictly speaking it's 96-bit. I'll change it to "higher-precision intermediate value". Paolo
