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?
Thomas
