On 12/23/23 05:15, Paolo Bonzini wrote:
Take advantage of the fact that there can be no 1 bits between SF and OF.
If they were adjacent, you could sum SF and get a carry only if SF was
already set. Then the value of OF in the sum is the XOR of OF itself,
the carry (which is SF) and 0 (the value of the OF bit in the addend):
this is OF^SF exactly.
Because OF and SF are not adjacent, just place more 1 bits to the
left so that the carry propagates, which means summing CC_O - CC_S.
Signed-off-by: Paolo Bonzini<pbonz...@redhat.com>
---
target/i386/tcg/translate.c | 10 ++++------
1 file changed, 4 insertions(+), 6 deletions(-)
Reviewed-by: Richard Henderson <richard.hender...@linaro.org>
r~
