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Re: [PATCH 1/2] target/s390x: Fix determination of overflow condition co


From: Thomas Huth
Subject: Re: [PATCH 1/2] target/s390x: Fix determination of overflow condition code after addition
Date: Wed, 30 Mar 2022 11:42:29 +0200
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:91.0) Gecko/20100101 Thunderbird/91.6.0

On 30/03/2022 11.34, David Hildenbrand wrote:
On 30.03.22 11:29, Thomas Huth wrote:
On 30/03/2022 10.52, David Hildenbrand wrote:
On 23.03.22 17:26, Thomas Huth wrote:
This program currently prints different results when run with TCG instead
of running on real s390x hardware:

   #include <stdio.h>

   int overflow_32 (int x, int y)
   {
     int sum;
     return ! __builtin_add_overflow (x, y, &sum);
   }

   int overflow_64 (long long x, long long y)
   {
     long sum;
     return ! __builtin_add_overflow (x, y, &sum);
   }

   int a1 = -2147483648;
   int b1 = -2147483648;
   long long a2 = -9223372036854775808L;
   long long b2 = -9223372036854775808L;

   int main ()
   {
     {
       int a = a1;
       int b = b1;
       printf ("a = 0x%x, b = 0x%x\n", a, b);
       printf ("no_overflow = %d\n", overflow_32 (a, b));
     }
     {
       long long a = a2;
       long long b = b2;
       printf ("a = 0x%llx, b = 0x%llx\n", a, b);
       printf ("no_overflow = %d\n", overflow_64 (a, b));
     }
   }

Resolves: https://gitlab.com/qemu-project/qemu/-/issues/616
Suggested-by: Bruno Haible <bruno@clisp.org>
Signed-off-by: Thomas Huth <thuth@redhat.com>
---
   target/s390x/tcg/cc_helper.c | 4 ++--
   1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/target/s390x/tcg/cc_helper.c b/target/s390x/tcg/cc_helper.c
index 8d04097f78..e11cdb745d 100644
--- a/target/s390x/tcg/cc_helper.c
+++ b/target/s390x/tcg/cc_helper.c
@@ -136,7 +136,7 @@ static uint32_t cc_calc_subu(uint64_t borrow_out, uint64_t 
result)
static uint32_t cc_calc_add_64(int64_t a1, int64_t a2, int64_t ar)
   {
-    if ((a1 > 0 && a2 > 0 && ar < 0) || (a1 < 0 && a2 < 0 && ar > 0)) {
+    if ((a1 > 0 && a2 > 0 && ar < 0) || (a1 < 0 && a2 < 0 && ar >= 0)) {


Intuitively, I'd have checked for any overflow/underflow by comparing
with one of the input variables:

a) Both numbers are positive

Adding to positive numbers has to result in something that's bigger than
the input parameters.

"a1 > 0 && a2 > 0 && ar < a1"

I think it doesn't really matter whether we compare ar with a1 or 0 here. If
an overflow happens, what's the biggest number that we can get? AFAICT it's
with a1 = 0x7fffffffffffffff and a2 = 0x7fffffffffffffff. You then get:

   0x7fffffffffffffff + 0x7fffffffffffffff = 0xFFFFFFFFFFFFFFFE

and that's still < 0 if treated as a signed value. I don't see a way where
ar could be in the range between 0 and a1.

(OTOH, checking for ar < a1 instead of ar < 0 wouldn't hurt either, I guess).

b) Both numbers are negative

Adding to negative numbers has to result in something that's smaller
than the input parameters.

"a1 < 0 && a2 < 0 && ar > a1"

What about if the uppermost bit gets lost in 64-bit mode:

   0x8000000000000000 + 0x8000000000000000 = 0x0000000000000000

ar > a1 does not work here anymore, does it?


0 > -9223372036854775808, no?

current coffe level < correct coffee level

... sorry, never mind, you're right of course.

Anyway, 0 is the lowest number we can get for an underflow, so comparing with >= 0 should be fine (but comparing with a1 wouldn't hurt either).

 Thomas




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