Building Windows / DOS - Simple Math

Question

I am currently studying the build of Windows / DOS. I am just doing a small program that adds two base 10 integers and prints the solution to standard output. Here is my current code:

org 100h


MOV al,5
ADD al,3

mov dx,al

mov ah,9
int 21h

ret

I am confused why, when this compiles, I get an error:

error: invalid combination of opcode and operands

Because theoretically, all I do is put 5 in the AL register, adding three to it, taking the contents of the AL register and putting it in the DX register for output, and then displaying it.

Any help would be appreciated, thanks!

Answer 1

DX - 16- , AL - 8-.

AL DL DH 0.

, ; function 9 [ ]. , , 9 , , , .

, 9.

. , Bitdog.

; ------ WDDD = Write a Decimal Digit Dword at the cursor & a CRLF ------
;
; Call with,    DX:AX = Dword value to print
; Returns,  CF set on error, if DX:AX > 655359 max#
;       (see WDECEAX.inc for larger number prints)
align 16
WDDD:   CMP DX,10
    JB  WDDDok
    STC     ;CF=set
    RET     ;error DX:AX exceeded max value
WDDDok: PUSH    AX
    PUSH    BX
    PUSH    CX
    PUSH    DX
    XOR CX,CX   ; clear count register for push count
    MOV BX,10
WDDDnz: DIV BX  ; divide DX:AX by BX=10
    PUSH    DX  ; put least siginificant number (remainder) on stack
    XOR DX,DX   ; clear remainder reciever for next divide
    OR  AX,AX   ; check to see if AX=number is divided to 0 yet
    LOOPNE  WDDDnz  ; get another digit? count the pushes
    MOV AH,2    ; function 2  for interupt 21h write digit
    NEG CX  ; two compliment, reverse CX
    MOV BL,48   ; '0'
WDDDwr: POP DX  ; get digit to print, last pushed=most significant
    ADD DL,BL   ; convert remainder to ASCII character
    INT 21h ; print ascii interupt 21h ( function 2 in AH )
    LOOP    WDDDwr  ; deincrement CX, write another, if CX=0 we done
    MOV DL,13   ; CR carriage return (AH=2 still)
    INT 21h
    MOV DL,10   ; LF line feed
    INT 21h
    POP DX
    POP CX
    POP BX
    POP AX
    CLC     ;CF=clear, sucess
    RET

; A divide error occurs if DX has any value
; when DIV trys to put the remainder into it
; after the DIVide is completed.
; So, DX:AX can be a larger number if the divisor is a larger number.