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Why is the Linux boot sequence is organized the way it is?

Power on + BIOS runs hardware initialisation and self tests, LILO/GRUB etc... but why is it organised the way it is?

Would I be right saying it is primarily for debugging purposes?

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migrated from stackoverflow.com Jan 6 '11 at 16:19

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The steps until the bootloader has nothing to do with the operating system. Its due to how the hardware is designed. –  ismail Jan 6 '11 at 15:20
    
And microsoft windows does it the very same way. –  OneOfOne Jan 6 '11 at 15:28
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serverfault.com? –  ajreal Jan 6 '11 at 15:45
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Since you tagged this as homework I guess you misunderstood your teachers question. I think the question is not about bios and bootloaders, etc. It's more about the boot process of linux. Could that be right? Anyhow. I think this link should answer your question. ibm.com/developerworks/linux/library/l-linuxboot –  tombom Jan 6 '11 at 16:01
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Division of labor and abstraction. –  Dennis Williamson Jan 6 '11 at 19:59

1 Answer 1

That is not specifically the Linux boot sequence. That is how the original IBM PC system worked and the PC you're currently using is just an evolved version of that original system due to people putting too much importance on backwards compatibility.

Lots of computers that does not need to be backwards compatible to the IBM PC architecture boot Linux differently. My handphone boots directly into U-Boot which then boots Linux. No BIOS there. Same as my web/SAN server which is ARM based. It boots directly into Redboot which then boots Linux. No BIOS. Google Chrome OS is rumored to have even gotten rid of bootloaders altogether and boots directly into a Linux kernel burned into the first address of memory.

In short, it's mostly legacy junk. There is no real good reason current hardware work the way they do except that if they work differently they won't be able to boot Windows.

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"they won't be able to boot Windows" - chainloading? –  new123456 Aug 1 '11 at 15:46

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