On Thu, Apr 13, 2006 at 08:49:19PM +0200, Stefan Weil wrote:
> - if ((T0 >> 31) ^ (T1 >> 31) ^ (tmp >> 31)) {
> + if (((tmp ^ T1 ^ (-1)) & (T0 ^ T1)) >> 31) {
> + /* operands of same sign, result different sign */
> CALL_FROM_TB1(do_raise_exception_direct, EXCP_OVERFLOW);
> }
I see this went in, but - huh? The math doesn't make sense.
T0 ^ T1 -> operands of different sign
tmp ^ T1 ^ (-1) -> result has same sign as T1
Which is a "who cares" case. This is addition, it can't overflow if
the operands have the same sign.
> - if (!((T0 >> 31) ^ (T1 >> 31) ^ (tmp >> 31))) {
> + if (((tmp ^ T1) & (tmp ^ T0)) >> 31) {
> + /* operands of different sign, first operand and result
> different sign */
> CALL_FROM_TB1(do_raise_exception_direct, EXCP_OVERFLOW);
> }
tmp ^ T1 -> result and T1 of different sign
tmp ^ T0 -> result and T0 of different sign
Which implies that the operands have the same sign. Again, this case
can't overflow.
I haven't tested the patched qemu, but I did test the expressions
themselves in standalone code, and they definitely do not detect
overflow.
--
Daniel Jacobowitz
CodeSourcery
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