On Mon, Apr 22, 2013 at 11:52:03AM +0200, Borislav Petkov wrote:
> On Mon, Apr 22, 2013 at 12:42:46PM +0300, Gleb Natapov wrote:
> > > Btw, I wanted to ask: when kvm commits the results, does it look at
> > > ctxt->op_bytes to know exactly how many bytes to write to the guest?
> > > Because if it does, we can save ourselves the trouble here.
> > >
> > > Or does it simply write both the full sizeof(unsigned long) bytes of
> > > ->src.val and ->dst.val to the guest?
> > >
> > No, it does this in case of register operand:
> >
> > static void write_register_operand(struct operand *op)
> > {
> > /* The 4-byte case *is* correct: in 64-bit mode we zero-extend. */
> > switch (op->bytes) {
> > case 1:
> > *(u8 *)op->addr.reg = (u8)op->val;
> > break;
> > case 2:
> > *(u16 *)op->addr.reg = (u16)op->val;
> > break;
> > case 4:
> > *op->addr.reg = (u32)op->val;
> > break; /* 64b: zero-extend */
> > case 8:
> > *op->addr.reg = op->val;
> > break;
> > }
> > }
>
> Ok, and for OP_MEM it does look at ctxt->dst.bytes in writeback(),
> AFAICT. And I see other emulated instructions like POPF, for example, do
> this:
>
> ctxt->dst.bytes = ctxt->op_bytes;
>
> Which means, we can drop all the bullshit in em_movbe and even destroy
> some of the bytes in dst.val but only write out the correct ones. Which
> means, a simpler code and a lot less jumping through hoops.
>
> Would that be the more accepted practice?
>
For most instructions the decoder already sets op->bytes to correct value,
given that all flags a correctly specified in opcode table. Explicit
op->bytes setting should be done only if it cannot be expressed by
opcode flags.
--
Gleb.
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