How to use int 13 with AH=03h interrupt in assembly to fill a sector with zeroes?

Question

So far i have this:

mov ah,02h
mov cl,11001100001111011101000b ;6,692,584 in dec
mov dl,0
int 13

Answer 1

I can see the following issues with your code:

• AH should be 3 rather than 2 (2 is for reading).
• CL is only eight bits so is unlikely to be holding that big honkin' value.
• You need to set AH, AL, CH, CL, DH, DL and ES:BX as per the following table:

---

AH     03h
AL     Number of sectors to write
CH     Track number
CL     Sector number
DH     Head number
DL     Drive number
ES:BX  Buffer to write

And, if you ever need information on interrupts, you should google for "Ralf Brown". This guy created and maintained the definitive interrupt list back in the days before we were insulated from such things. See, for example, the indexed HTML version.

Your specific needs can be met from the Int 13/AH=03h section.

A sample, though untested and written in real mode, would be something along the following lines:

      mov  ax, 0301h           ; cmd 3 (write sector), 1 sector
      mov  cx, 0001h           ; track 0, sector 1
      mov  dx, 0000h           ; head 0, drive 0
      mov  es, segment buff    ; segment and offset of buffer
      mov  bx, offset buff
      int  13h                 ; make the call
      jc   errp                ; detect error
      ret

errp:                          ; process error here
      ret

buff: db   0(256)              ; buffer to write