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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? ThanksChiacchiera con i tuoi amici in tempo reale! http://it.yahoo.com/mail_it/foot/*http://it.messenger.yahoo.com
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