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 <br...@clisp.org>
>>> Signed-off-by: Thomas Huth <th...@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?
--
Thanks,
David / dhildenb
