问题
I was researching boot loaders and I found this interesting piece of assembly:
;Sends us to the end of the memory
;causing reboot
db 0x0ea
dw 0x0000
dw 0xffff
By the comment I know what it does; sends the computer to the end of memory, but what I can't figure out is how those numbers reboot the computer (x86_64 processor on 16-bit mode).
回答1:
Those bytes correspond to jmp word 0xffff:0000 (you can see this by assembling with NASM and then disassembling the resulting binary), which happens to be a jump to the x86 reset vector in real mode.
回答2:
It's a far jump instruction to the old 8086 reset address. When the 8086 was reset it would start executing instructions at FFFF:0000. For compatibility reasons modern BIOS implementation have a jump to their reset code here, though reset address of modern CPUs is different.
来源：https://stackoverflow.com/questions/31296422/why-do-the-bytes-0xea-0000-ffff-in-a-bootloader-cause-the-computer-to-reboot