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Date: Sun, 17 May 2009 10:46:13 +0200
From: Heli <helicoterus-elih@yahoo.it>
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Subject: Re: [Qemu-devel] How does a system call work on QEMU?
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To: Avi Kivity <avi@redhat.com>
Cc: qemu-devel@nongnu.org

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.
> 
So it this the architecture of qemu, emulator version?
QEMU is a process that gives to o.s.guest a virtual hw
layer.
Every guest will be run by a QEMU instance.

See picture schema I prepaired: http://yfrog.com/7gqemulayersj

So system calls go from a guest application to its
operating system that communicates with its virtual
hardware. QEMU receives a system call and software emulator
QEMU performs a system call in O.S. host.
It is right?

Thanks
Chiacchiera con i tuoi amici in tempo reale! 
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