On 09/16/2015 01:57 AM, Paolo Bonzini wrote: > > > On 15/09/2015 20:45, Richard Henderson wrote: >> + /* Fold the global and local enable bits together into the >> + global fields, then xor to show which registers have >> + changed collective enable state. */ >> + int mod = ((old_dr7 | old_dr7 * 2) ^ (new_dr7 | new_dr7 * 2)) & >> 0xff; > > The AND is not needed at all but, if you add it, you might as well use > "& 0xaa" which is clearer. But even better, just do: > > target_ulong old_dr7 = env->dr[7]; > int mod = old_dr7 ^ new_dr7; > ... > if ((mod & ~0xff) == 0) { > > > and test with > > if (mod & (3 << i * 2)) > > inside the loop.
Nope. I wrote that the first time myself. We're interested in two different things: (1) whether or not something changed outside enable bits, and (2) whether the enable state changed. Since (2) is a combination of both global and local enable bits, we must combine them *and then xor* to see if the enable state actually changes. Just using (mod & (3 << n)) will report "change" when local enable turns off, but global enable remains on. Which is not what we want. Perhaps that comment could stand to be expanded... r~