Re: [Qemu-devel] [Qemu-ppc] [PATCH v2 2/2] i6300esb: Fix signed integer overflow

[BALATON Zoltan <balaton@eik.bme.hu>]

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 <david@gibson.dropbear.id.au>
> ---
> 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?

Regards,
BALATON Zoltan

> +    timeout = muldiv64(get_ticks_per_sec(), timeout, 33000000);
>
>     i6300esb_debug("stage %d, timeout %" PRIi64 "\n", d->stage, timeout);