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


From: Heli
Subject: Re: [Qemu-devel] How does a system call work on QEMU?
Date: Sun, 17 May 2009 10:46:13 +0200
User-agent: Thunderbird 2.0.0.21 (Windows/20090302)

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! http://it.yahoo.com/mail_it/foot/*http://it.messenger.yahoo.com




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