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
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
