On 16/09/2015 16:57, 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.
Ah, I see now. Perhaps
int old_enable = (old_dr7 | (old_dr7 << 1)) & 0xaa;
int new_enable = (new_dr7 | (new_dr7 << 1)) & 0xaa;
?
> Perhaps that comment could stand to be expanded...
I think at least for me it's just the expression that is too complex.
Paolo
