On 06/26/2015 07:35 PM, jd1008 wrote:
I have been googling and read wikis.
None of them really explain clearly
If
1. a drive has no bootable partitions and
2. the boot code in the 1st 446 bytes does not exist (all nulls)
then
how does bios decide it is not bootable, move on to the next in the
sequence?
I didn't get a satisfactory answer from wikis, either, so I did an
experiment. I loaded a bootable image on a flash drive and connected
that to a virtual machine as a USB disk. I also added a bootable ISO to
the VM. I configured the VM to boot from the USB drive first, then the ISO.
The VM successfully booted from the flash drive. I backed up the MBR.
# dd if=/dev/sdb bs=512 count=1 of=sdb.mbr
Then I zeroed 446 byes of the flash drive.
# dd if=/dev/zero bs=446 count=1 of=/dev/sdb
The VM halted when trying to boot, so I restored the boot sector and
wiped the boot signature.
# dd if=sdb.mbr of=/dev/sdb
# dd if=/dev/zero bs=2 count=1 seek=255 of=/dev/sdb
With the boot signature wiped, the VM would boot from the ISO.
Based on testing, we can conclude that at least SeaBIOS will treat a
boot sector with all nul bytes as a valid boot sector and run it. It
will skip a boot sector if the boot signature in that MBR is not present.
Note that as I previously mentioned, the BIOS doesn't use the boot flag
in the partition table. A "bootable" partition is ONLY relevant to DOS
type boot loaders, which use it to identify the C: drive from which they
will boot. It does not matter to BIOS whether a disk has any bootable
partitions or not.
For bios to spend an eternity looking for the boot code on a non-bootable
drive tells me it is a bug, even if implemented according to specs
(thus the
specs themselves would be at fault).
It's not looking for boot code. It identifies a valid boot sector,
where validity is determined by the presence of a boot signature, and
runs that code.
I'm not an expert on BIOS, but the extent to which I've read
documentation is fairly clear and consistent. Execution begins at a
specific memory location where BIOS is expected to reside. BIOS locates
a boot device (possibly a hardware ROM, or a disk) and continues
execution of that code. That code loads a kernel into memory and
continues execution of that code. It's not described as a stack.
Nothing indicates that control will return to the previous chunk of code
if it finishes or does nothing.
"bug" is not a word for "something I don't understand" or "something I
don't like."
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