Re: [Qemu-devel] [Qemu-trivial] [PATCH] hw/i386/pc: reject to boot a wrong header magic kernel

[Stefan Hajnoczi <stefanha@gmail.com>]

On Fri, Mar 29, 2013 at 1:38 AM, li guang <lig.fnst@cn.fujitsu.com> wrote:
> 在 2013-03-28四的 10:42 +0100，Stefan Hajnoczi写道：
>> On Wed, Mar 27, 2013 at 02:10:31PM +0800, liguang wrote:
>> > if head magic is missing or wrong unexpectedly, we'd
>> > better to reject booting.
>> > e.g.
>> > I make a mistake to boot a vmlinuz for MIPS(which
>> > I think it's for x86) like this:
>> > qemu-system-x86_64 -kernel vmlinuz -initrd demord
>> > then qemu report:
>> > "qemu: linux kernel too old to load a ram disk"
>> > that's misleading.
>> >
>> > Signed-off-by: liguang <lig.fnst@cn.fujitsu.com>
>> > ---
>> >  hw/i386/pc.c |    4 +++-
>> >  1 files changed, 3 insertions(+), 1 deletions(-)
>> >
>> > diff --git a/hw/i386/pc.c b/hw/i386/pc.c
>> > index b1e06fa..2b78dfc 100644
>> > --- a/hw/i386/pc.c
>> > +++ b/hw/i386/pc.c
>> > @@ -683,8 +683,10 @@ static void load_linux(void *fw_cfg,
>> >          if (load_multiboot(fw_cfg, f, kernel_filename, initrd_filename,
>> >                             kernel_cmdline, kernel_size, header)) {
>> >              return;
>> > +        } else {
>> > +            fprintf(stderr, "please assure specicified kernel is for x86!\n");
>> > +            exit(1);
>>
>> load_multiboot() can fail for other reasons so this error messing is
>> misleading.  Giving QEMU a non-x86 kernel is just one scenario where
>> this may fail.
>
> according to my check of load_mutiboot function,
> mostly it will return 0 if it's not multboot,
> or 1 it's a multiboot, so print this message,
> or can I just print "wrong kernel image!" ?

Yes, load_multiboot() fails if the image is not a valid multiboot
image.  An error message like "not a valid multiboot image" is good.

>>
>> >          }
>> > -        protocol = 0;
>> >      }
>>
>> Why did you drop protocol = 0?
>
> I think we only want either normal or multi boot linux kernel,
> I can't see meaning let other case go on.
> so, here, if a normal kernel, OK, go on,
> if multiboot, OK, go on,
> others, NO, end up.
> so, "protocol = 0" is meaningless here.

I looked in more detail now and here is why we cannot drop this line of code:

    uint16_t protocol;

Note that the variable is not initialized.

[...]
    if (ldl_p(header+0x202) == 0x53726448)
    protocol = lduw_p(header+0x206);
    else {
    /* This looks like a multiboot kernel. If it is, let's stop
       treating it like a Linux kernel. */
        if (load_multiboot(fw_cfg, f, kernel_filename, initrd_filename,
                           kernel_cmdline, kernel_size, header))
            return;
    protocol = 0;

Set it to 0 here so it's initialized

    }
    if (protocol < 0x200 || !(header[0x211] & 0x01)) {

Use variable here - so we *must* initialize it before use.

Stefan