Re: [Qemu-devel] How does a system call work on QEMU?

[Glauber Costa]

On Thu, May 14, 2009 at 3:00 PM, grisu46 <address@hidden> wrote:
> Avi Kivity ha scritto:
>>
>> grisu46 wrote:
>>>>
>>>> No.  Look at target-i386/translate.c, case 0x105 for syscall emulation.
>>>>  Eventually it ends up calling helper_syscall().
>>>>
>>>
>>> Simplify?
>>> I am a newbie.
>>
>> Sorry, you're on your own.  I can give you pointers but it's up to you to
>> follow them.  If you're serious about understanding qemu you should be
>> prepared a lot of time reading the source code and processor documentation.
>>
> I would like to know how a virtualization program can perform a sensitive
> but not privileged operation. I learn about Xen and I like to understand (no
> deep details need) differences between Xen paravirtualization and QEMU
> emulator.
>
As the name suggests, qemu (in this case) is a processor _emulator_. You're not
doing any kind of sensitive operations on your host processor here.
You are emulating
a processor, an all state change happens in the emulated processor.

* qemu reads the instruction.
* qemu updates its internal state accordingly.
* qemu reads next instruction.

No big secret here.

-- 
Glauber  Costa.
"Free as in Freedom"
http://glommer.net

"The less confident you are, the more serious you have to act."