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boot sector with interrupts
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This is very exciting, we're going to create our own boot sector!
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Theory
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------
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When the computer boots, the BIOS doesn't know how to load the OS, so it
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delegates that task to the boot sector. Thus, the boot sector must be
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placed in a known, standard location. That location is the first sector
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of the disk (cylinder 0, head 0, sector 0) and it takes 512 bytes.
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To make sure that the "disk is bootable", the BIOS checks that bytes
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511 and 512 of the alleged boot sector are bytes `0xAA55`.
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This is the simplest boot sector ever:
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```
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e9 fd ff 00 00 00 00 00 00 00 00 00 00 00 00 00
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00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
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[ 29 more lines with sixteen zero-bytes each ]
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00 00 00 00 00 00 00 00 00 00 00 00 00 00 55 aa
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```
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It is basically all zeros, ending with the 16-bit value
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`0xAA55` (beware of indianness, x86 is little-endian).
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The first three bytes perform an infinite jump
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Simplest boot sector ever
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-------------------------
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You can either write the above 512 bytes
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with a binary editor, or just write a very
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simple assembler code:
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```nasm
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; Infinite loop (e9 fd ff)
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loop:
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jmp loop
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; Fill with 510 zeros minus the size of the previous code
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times 510-($-$$) db 0
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; Magic number
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dw 0xaa55
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```
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To compile:
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`nasm -f bin boot_sect_simple.asm -o boot_sect_simple.bin`
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> OSX warning: if this drops an error, read chapter 00 again
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I know you're anxious to try it out (I am!), so let's do it:
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`qemu boot_sect_simple.bin`
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You will see a window open which says "Booting from Hard Disk..." and
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nothing else. When was the last time you were so excited to see an infinite
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loop? ;-)
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@ -1,6 +0,0 @@
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; A simple boot sector program that loops forever
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loop:
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jmp loop
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times 510-($-$$) db 0
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dw 0xaa55
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