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coding fun with MASM

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hey I picked up a book on assembly code with masm and am trying to get the knack of this stuff.... i just want to output a binary digit. i can figure out how to do it with a hex number, but the binary's got me all sorts of confused. here i my code for the hex output
	.MODEL	small
	.STACK	100h
cr	EQU	0Dh
lf	EQU	0Ah

digit_table	DB	'0123456789ABCDEF'
hex_str	DB	'0000h', cr, lf, '$'
divisor	DW	16

number	DW	0F8D2h


	; copy the number to AX so it doesn't get clobbered.
	; As we loop AX will contain the high bits yet to be converted to hex   
        ; digits.
	mov	ax,	number			
	; I'm using DI to index the converted hex string as an array
	; starting with index 3 (the 4th hex char) and working my   
	; way back to index 0 (the 1st hex char).  I have to use DI because
	; memory indexing (like the "mov [di], al" instruction) requires
	; an index register as the first operand
	mov	di,	OFFSET hex_str	; Get the memory address of the hex 
                                        ; string I am filling.
	add	di,	3		; Add 3 to shift the address by 3 bytes.

	mov	cx,	4		; Do the following loop 4 times... CX 
                                        ; is like the "i" of

	; while (cx != 0)								;---------------------------------------------------
	cmp	cx,	0			; Compare CX to zero.
	je	end_hex_loop			; JUMP to the "end_hex_loop" 	
	; Division requires a 32-bit dividend and a 16 bit divisor.  Since our 
        ; registers are only 16 bits, the "div" instruction fakes a 32 bit 
        ; number by assuming the dividend's high 16 bits are in DX and low 16 
        ; bits in AX.  The number we're working on is already in AX and is only
	; 16 bits, but since the instruction requires 32 bits we zero out DX so 
        ; the value of the number isn't affected.  After the division 
        ; instruction completes the quotient is in AX and the remainder in DX.  
        ; Dividing our number by 16 gives a remainder that is the lowest order
	; nibble we're tunring to a hex digit and a quotient that represents 
        ; the rest of the number left to convert.
	mov	dx,	0	; Zero out the high 16 bits of the DX:AX 
	div	divisor		; Divide (unsigned) the number by 16
	push	ax		; The "xlat" instruction below requires the use 
                                ; of AL so we must save our quotient
	mov	bx,	OFFSET digit_table	; Get the address of 
                                                ; the "digit_table".  "xlat" 
                                                ; expects this address in BX.
	mov 	al,	dl		        ; AL needs to indicate 
                                                ; to "xlat" which position of 
                                                ; the array to index.
	xlat					; Grab the hex digit indexed by 
                                                ; BX[AL] and store it in AL
	mov	[di],	al			; Store the hex digit in 
                                                ; the "di"th position of our 
						; output string
	pop	ax				; Get our quotient back from 
                                                ; the stack

	sub	di,	1			; "di--" shifts our string 
                                                ; index BACK a byte to prepare
						; for the next loop iteration
	sub	cx,	1			; Decrement our loop index

	jmp	hex_loop			; Jump back to the top of 
                                                ; our "while" loop.

	; Print our hex string to the screen using the PROCEDURE
	; we developed in class
	mov 	ax,	OFFSET hex_str		
	call	disp_str

; disp_str : Display string of bytes using 
;            DOS int 21h, function 09h
; INPUT : AX should contain the address of 
;         the string of bytes
disp_str	PROC
	push	ax
	push	dx

	mov	dx,	ax
	mov	ah,	09h
	int	21h

	pop	dx
	pop	ax
disp_str	ENDP

i think its easier to work with hex numbers than with binary...

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It's far easier to work with binary than with hex. The processor works using a binary logic, so you just need to use the simple way.

The basic instructions you need are:
* shl (left shift by one bit)
* jc (jump if carry)
If you want to avoid any jumps, you can use adc:

mov eax, my_number ; your number is in eax
mov ecx, 32 ; loop count is in ecx

xor ebx, ebx ; set ebx to 0
shl eax, 1 ; left shift eax by one
; if the MSB is 1, the carry
; flag will be set
adc ebx, 0 ; ebx = ebx + 0 + carry flag

; Now ebx contains either 1 or 0, depending on the MSB of eax
; you can print it using a print routine
; note that if you replace the constant with the value of the '0' char,
; you directly get the character you'll want to display

loop the_loop ; ecx--, then loop to "the_loop" only if ecx != 0

That's all.

Of course, this routine is very specialized - it also don't take the output to a string into account (but this can be solved rather easily).

As you see, it is far easier than the hex version [smile]


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