Question

Would any assembly gurus know the argument (register dl) that signifies the first USB drive?

I'm working through a couple of NASM tutorials, and would like to get a physical boot (I can get a clean one with qemu).

This is the section of code that loads the "kernel" data from disk:

loadkernel:
    mov si, LMSG ;; 'Loading kernel',13,10,0
    call prints  ;; ex puts()

    mov dl, 0x00 ;; The disk to load from
    mov ah, 0x02 ;; Read operation
    mov al, 0x01 ;; Sectors to read
    mov ch, 0x00 ;; Track
    mov cl, 0x02 ;; Sector
    mov dh, 0x00 ;; Head
    mov bx, 0x2000 ;; Buffer end
    mov es, bx
    mov bx, 0x0000 ;; Buffer start

    int 0x13
    jc loadkernel

    mov ax, 0x2000
    mov ds, ax

    jmp 0x2000:0x00

If it makes any difference, I'm running a stock Dell Inspiron 15 BIOS.


Apparently, the correct value for me is 0x80.

The BIOS loads the hard drives and labels them starting at 0x80 according to this answer.

My particular BIOS decides to load the USB drive up as the first, for some reason, so I can boot from there.

Was it helpful?

Solution

The simple answer is that the correct value for dl is in dl.

The happy answer is that the dl value with int 13h, ah=8 GET DRIVE PARAMETERS returns the geometry to use and allows the FAT12 floppy disk image code to run from any BIOS/version that can boot from a USB flash drive.

See my post here: USB Booting Secrets

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