Il 17/06/2014 09:04, Jan Kiszka ha scritto:
+ default1 = vmx_ctl_msr[n].default1;
+ ok = (ctrl.set & default1) == default1 &&
+ ((ctrl.set ^ ctrl.clr) & ~ctrl.clr) == 0;
Thanks, now I can understand what's going on. :) It can still be
simplified though.
This is just (ctrl.set & ~ctrl.clr) == 0:
(a ^ b) & ~b == 0
-> (a & ~b) ^ (b & ~b) == 0
-> a & ~b == 0
This expresses just as well the concept that set=1, clr=0 is an
impossible combination.
Also, it's a bit easier to read if the second line is written as a
separate statement:
ok = ok && (ctrl.set & ~ctrl.clr) == 0;
+ if (ok && basic.ctrl) {
+ true_ctrl.val = rdmsr(vmx_ctl_msr[n].true_index);
+ ok = ctrl.clr == true_ctrl.clr &&
+ ((ctrl.set ^ true_ctrl.set) & ~default1) == 0;
This is
((ctrl.set ^ true_ctrl.set) & ~default1 == 0
-> ((ctrl.set & ~default1) ^ (true_ctrl.set & ~default1) == 0
-> ((ctrl.set & ~default1) == (true_ctrl.set & ~default1)
A bit longer, but clearer: the difference between ctrl.set and
true_ctrl.set is only that true_ctrl.set can clear some default-1 bits.
Or you can simplify it further:
-> (ctrl.set | default1) == (true_ctrl.set | default1)
-> ctrl.set == (true_ctrl.set | default1)
Also clearer: the difference between ctrl.set and true_ctrl.set is only
that default-1 bits must be 1 in ctrl.set. Pick the one you prefer.
Again, using a separate statement makes it easier to read in my opinion;
in fact I would also write both statements as
ok = ok && ...
even though it's redundant for the clr test.
Can you submit v3 of this patch only?
Paolo
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