From: Jong Chun Park <joumon@gmail.com>
To: qemu-devel@nongnu.org
Subject: [Qemu-devel] [Please discard the previous mail] System call from a Guest Linux
Date: Wed, 3 Feb 2010 11:08:58 -0700 [thread overview]
Message-ID: <2e320c2d1002031008y7a01880ay7d17d419b38fb089@mail.gmail.com> (raw)
In-Reply-To: <2e320c2d1002031003i36bc3848hbb4b030a6dfd4156@mail.gmail.com>
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Hi, all?
First of all, I apologize you for the previous incomplete mail and this
spam. I'm working with QEMU v0.12.1 with i386 Linux guest on x86-64 Linux
host. I'm trying to find a point in the source code where a system call from
the guest OS is handled. If I'm not mistaken, QEMU disassembles an
instruction of INT # in a switch statement of target-i386/translate.c and
then calls helper_sysenter of target-i386/op_helper.c. The problem is how to
tell difference between a system call of the guest OS and a system call of
QEMU (I'm not sure of this, though). Assume the following code is executed
after compilation in the guest OS:
int main() {
int fd;
if (access("hello.txt", R_OK) != 0) {
exit(0);
}
fd=open("hello.txt", O_RDONLY);
close(fd);
return 0;
}
For this simple sequence of system calls, acess() -> open() -> close(), QEMU
goes through disas_insn() and helper_sysenter() more than 3 times. This
makes me really confused because I believed those should be called 3 times,
though. It'd be greatly appreciated in advance if someone helps me out this.
Thanks a lot,
Jong
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next prev parent reply other threads:[~2010-02-03 18:09 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
2010-02-03 18:03 [Qemu-devel] System call from a Guest Linux Jong Chun Park
2010-02-03 18:08 ` Jong Chun Park [this message]
2010-02-03 19:25 ` [Qemu-devel] [Please discard the previous mail] " Mulyadi Santosa
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