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Mar 8, 2018 at 20:33 vote accept sherrellbc
Feb 26, 2018 at 13:37 vote accept sherrellbc
Feb 26, 2018 at 13:37
Feb 24, 2018 at 19:33 comment added sherrellbc @DanielB There are only 512 bytes to work with in the MBR and VBR (if present), so I expected the implementation to be fairly simplified. Also, thanks for the link about ext. I poked around the wiki and could not find it there.
Feb 24, 2018 at 19:23 comment added sherrellbc @DanielB The process is not specific to Linux. DOS was booted this way. The process is well known. The MBR gets execution at 0x0000:0x7c00, it relocates itself, determines the "active" partition and loads its VBR (first sector) it at the same address, 0x0000:0x7c00. This VBR was the first stage of the bootloader specific to the installed OS. Of course, this is how it was originally done, and the process has evolved over the last 40 years. You certainly do not need a VBR, especially if your installed bootloader, like GRUB, overwrites the MBR with its own code specific to its boot scheme
Feb 24, 2018 at 19:09 comment added Kamil Maciorowski Btrfs seems to have 64KiB.
Feb 24, 2018 at 19:05 comment added Daniel B Also, ext2 does have a 1024-byte boot code area.
Feb 24, 2018 at 18:59 answer added Kamil Maciorowski timeline score: 2
Feb 24, 2018 at 18:55 comment added Daniel B You’re mistaken about how the Linux boot process works. There’s no need for a VBR if the MBR code understands enough of the filesystem to load the next stage, which is not subject to space constraints.
Feb 24, 2018 at 18:35 history asked sherrellbc CC BY-SA 3.0