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Re: [Qemu-devel] [PATCH 4/8] target-i386: Re-introduce optimal breakpoin


From: Paolo Bonzini
Subject: Re: [Qemu-devel] [PATCH 4/8] target-i386: Re-introduce optimal breakpoint removal
Date: Wed, 16 Sep 2015 16:59:48 +0200
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:38.0) Gecko/20100101 Thunderbird/38.2.0


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



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