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


From: BALATON Zoltan
Subject: Re: [Qemu-ppc] [PATCH v2 2/2] i6300esb: Fix signed integer overflow
Date: Mon, 23 Mar 2015 10:54:39 +0100 (CET)
User-agent: Alpine 2.01 (GSO 1266 2009-07-14)

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 <address@hidden>
---
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);





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