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~
