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Message-ID: <CALCETrUb1kqEpWNSi=AdSNYASEfnfBR-siMvwKjBSwVVxON3hw@mail.gmail.com>
Date: Thu, 26 Jan 2017 09:05:01 -0800
From: Andy Lutomirski <luto@...capital.net>
To: Ricardo Neri <ricardo.neri-calderon@...ux.intel.com>
Cc: Ingo Molnar <mingo@...hat.com>,
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Subject: Re: [v3 PATCH 05/10] x86/insn-kernel: Add support to resolve 16-bit
addressing encodings
On Wed, Jan 25, 2017 at 9:50 PM, Ricardo Neri
<ricardo.neri-calderon@...ux.intel.com> wrote:
> On Wed, 2017-01-25 at 13:58 -0800, Andy Lutomirski wrote:
>> On Wed, Jan 25, 2017 at 12:23 PM, Ricardo Neri
>> <ricardo.neri-calderon@...ux.intel.com> wrote:
>> > Tasks running in virtual-8086 mode will use 16-bit addressing form
>> > encodings as described in the Intel 64 and IA-32 Architecture Software
>> > Developer's Manual Volume 2A Section 2.1.5. 16-bit addressing encodings
>> > differ in several ways from the 32-bit/64-bit addressing form encodings:
>> > the r/m part of the ModRM byte points to different registers and, in some
>> > cases, addresses can be indicated by the addition of the value of two
>> > registers. Also, there is no support for SiB bytes. Thus, a separate
>> > function is needed to parse this form of addressing.
>> >
>> > Furthermore, virtual-8086 mode tasks will use real-mode addressing. This
>> > implies that the segment selectors do not point to a segment descriptor
>> > but are used to compute logical addresses. Hence, there is a need to
>> > add support to compute addresses using the segment selectors. If segment-
>> > override prefixes are present in the instructions, they take precedence.
>> >
>> > Lastly, it is important to note that when a tasks is running in virtual-
>> > 8086 mode and an interrupt/exception occurs, the CPU pushes to the stack
>> > the segment selectors for ds, es, fs and gs. These are accesible via the
>> > struct kernel_vm86_regs rather than pt_regs.
>> >
>> > Code for 16-bit addressing encodings is likely to be used only by virtual-
>> > 8086 mode tasks. Thus, this code is wrapped to be built only if the
>> > option CONFIG_VM86 is selected.
>>
>> That's not true. It's used in 16-bit protected mode, too. And there
>> are (ugh!) six possibilities:
>
> Thanks for the clarification. I will enable the decoding of addresses
> for 16-bit as well... and test the emulation code.
>>
>> - Normal 32-bit protected mode. This should already work.
>> - Normal 64-bit protected mode. This should also already work. (I
>> forget whether a 16-bit SS is either illegal or has no effect in this
>> case.)
>
> For these two cases I am just taking the effective address that the user
> space application provides, given that the segment selectors were set
> beforehand (and with a base of 0).
What do you mean by the base being zero? User code can set a nonzero
DS base if it wants. In 64-bit mode (user_64bit_mode(regs)), the base
is ignored unless there's an fs or gs prefix, and in 32-bit mode the
base is never ignored.
>
>> - Virtual 8086 mode
>
> In this case I calculate the linear address as:
> (segment_select << 4) + effective address.
>
>> - Normal 16-bit protected mode, used by DOSEMU and Wine. (16-bit CS,
>> 16-bit address segment)
>> - 16-bit CS, 32-bit address segment. IIRC this might be used by some
>> 32-bit DOS programs to call BIOS.
>> - 32-bit CS, 16-bit address segment. I don't know whether anything uses this.
>
> In all these protected modes, are you referring to the size in bits of
> the base address of in the descriptor selected in the CS register? In
> such a case I would need to get the base address and add it to the
> effective address given in the operands of the instructions, right?
No, I'm referring to the D/B bit. I'm a bit fuzzy on exactly how the
instruction encoding works, but I think that 16-bit x86 code is
encoded just like real mode code except that the selectors are used
for real.
>> size, but I suspect you'll need to handle 16-bit CS.
>
> Unless I am missing what is special with the 16-bit base address, I only
> would need to add that base address to whatever effective address (aka,
> offset) is encoded in the ModRM and displacement bytes.
Exactly. (And make sure the instruction decoder can decode 16-bit
instructions correctly.)
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