Converting from MASM to NASM

Question

I've been working on converting some code from MASM format to NASM format for assembly. However I am not very educated in either, and I keep running into issues. I barely understand NASM as is so trying to convert the code has been extremely difficult. If someone could convert it and explain how they converted from MASM to NASM that would be very useful. Also, if there's any modern conversion tool, that would also be quite helpful.

My main issues seem to be with the .DATA section and the offset variables. From my research NASM doesnt have 'offset', but I am confused about what to properly put to substitute for it. Im using dosbox for compiling.

.MODEL SMALL
.STACK 100H
.DATA
    msg db 10, 13, "Enter a three-digit number: $"
    buffer db 4 DUP ('$') ; Buffer to store the input
    result db 10, 13, "Binary value: $"
.CODE
main PROC
    mov ax, @data
    mov ds, ax
    lea dx, msg
    mov ah, 09h ; Display prompt
    int 21h
    mov dx, offset buffer ; Load pointer to the buffer
    mov ah, 0Ah ; Read input
    int 21h    ; Convert ASCII input to binary
    xor cx, cx ; Initialize counter
    mov cl, buffer[1] ; Get the length of input
    mov si, offset buffer + 2 ; Point to the actual input
    ; Loop to convert ASCII digits to binary
    convert_loop:
        mov al, [si]
        sub al, '0' ; Convert ASCII to numeric value
        shl al, cl ; Shift left based on position
        or dl, al ; Combine with result
        dec cl ; Decrement counter
        inc si ; Move to next digit
        loop convert_loop    ; Display the binary value
    lea dx, result
    mov ah, 09h ; Display result prompt
    int 21h    ; Print binary value
    mov ah, 02h ; Print character
    mov dl, '0' ; Initialize with '0'
    print_loop:
        shl dl, 1 ; Shift left
        rcl dh, 1 ; Rotate carry into high byte
        int 21h
        loop print_loop
    mov ah, 4Ch ; Exit program
    int 21h
main ENDP
END main

Answer 1

If you convert this broken software from MASM to NASM, it will still be broken!

There are just too many errors in this program:

• The prompt "Enter a three-digit number: " is ambiguous. It does not clearly state that a binary number is expected, and mandatory for correct operation.

• As @ecm wrote in a comment, the definition of buffer (buffer db 4 DUP ('$')) is not correct for use by the DOS.BufferedInput function 0Ah.

• In the convert_loop, the result gets accumulated in a register that was not setup beforehand.

• In the convert_loop, the 'Shift left based on position' shifts 1 bit too far to the left.

• In the convert_loop, the 'Decrement counter' disrupts the CX loop control variable, making the loop run for too long. Remember that CL is the low byte of CX.

• In the print_loop, the 'Shift left' and 'Rotate carry into high byte' are both non-sense operations.

• The print_loop does not initialize its loop control variable. You forgot to move a suitable value in the CX register.

• The message "Binary value:" is ambiguous. It is not clear whether a print_loop is needed or not. An input of 3 binary digits can produce a result in the range [0,7], and that you can display with a single invokation of the DOS.PrintCharacter function 02h.

What follows is an improved code. It is still written for MASM, but with the link to the NASM manual that @Jester provided in a comment, wouldn't you be able to convert this to NASM easily?

ORG 256

  mov  dx, offset msg
  mov  ah, 09h
  int  21h
  mov  dx, offset buffer
  mov  ah, 0Ah
  int  21h

  mov  bl, 0                  ; Result
  mov  cl, buffer[1]          ; Expected to contain a value in [1,3]
  mov  si, offset buffer + 2
convert_loop:
  mov  al, [si]
  inc  si
  sub  al, '0'
  dec  cl
  shl  al, cl                 ; Shift left based on position
  or   bl, al                 ; Combine with result
  test cl, cl
  jnz  convert_loop

  mov  dx, offset result
  mov  ah, 09h
  int  21h
  mov  dl, '0'
  add  dl, bl                 ; BL=[0,7]
  mov  ah, 02h
  int  21h

  mov  ax, 4C00h
  int  21h
; ----------------------------
msg    db 13, 10, "Enter a three-digit binary number: $"
result db 10, "Binary value: $"
buffer db 4, 0, 4 dup (0)

Bonus hint: NASM will not like mov cl,buffer[1]. Every address component belongs to between the square brackets. Use the plus character for concatenation.