Alexander,
On Tue, Jul 26, 2011 at 19:05 +0200, Alexander Graf wrote:
> > @@ -2623,6 +2626,7 @@ static void kvm_vcpu_ioctl_x86_get_debugregs(struct kvm_vcpu *vcpu,
> > dbgregs->dr6 = vcpu->arch.dr6;
> > dbgregs->dr7 = vcpu->arch.dr7;
> > dbgregs->flags = 0;
> > + memset(&dbgregs->reserved, 0, sizeof(dbgregs->reserved));
> > }
> >
> > static int kvm_vcpu_ioctl_x86_set_debugregs(struct kvm_vcpu *vcpu,
> > @@ -3106,6 +3110,7 @@ static int kvm_vm_ioctl_get_pit2(struct kvm *kvm, struct kvm_pit_state2 *ps)
> > sizeof(ps->channels));
> > ps->flags = kvm->arch.vpit->pit_state.flags;
> > mutex_unlock(&kvm->arch.vpit->pit_state.lock);
> > + memset(&ps->reserved, 0, sizeof(ps->reserved));
>
> struct kvm_pit_state2 {
> struct kvm_pit_channel_state channels[3];
> __u32 flags;
> __u32 reserved[9];
> };
>
> So memset(&ps->reserved) would give you the a __u32 **, no? Same goes for all the other array sets in here. Or am I understanding some C logic wrong? :)
No, the array name and an address of the array give the same address. I
could use ps->reserved instead of &ps->reserved, but it is a question of
coding style. I got opposite opinions on this question from different
maintainers.
Another thing is that sizeof() of the array name and the pointer to the
first array element differ. But I used sizeof(array) here, so it should
be correct.
Thanks,
--
Vasiliy Kulikov
http://www.openwall.com - bringing security into open computing environments
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