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?
Thanks.
--
Regards/Gruss,
Boris.
Sent from a fat crate under my desk. Formatting is fine.
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