section .data
msg db 10,13,"Enter Hex number (max. FFFFH)", 10	
msglen equ $-msg

msg2 db 10,13,"The converted BCD number is:", 10
msglen2 equ $-msg2

nwline db 10
nwlen equ $-nwline

msg3 db 10,13,"Enter BCD number: (Note: Enter 5 digit. E.g. 00018)", 10	
msglen3 equ $-msg3

msg4 db 10,"Choose one of the following:", 10, "(1)Hex to BCD", 10, "(2)BCD to Hex", 10, "(3)Exit", 10
msglen4 equ $-msg4

msg5 db 10,"The converted Hex number is:", 10
msglen5 equ $-msg5

section .bss
cnt resb 1
num resb 5
num1 resb 1
num2 resb 6
numbuff resb 5
num3 resb 4
opt resb 1

%macro disp 2
mov eax,4
mov ebx,1
mov ecx,%1
mov edx,%2
int 80h	
%endmacro
	
%macro accept 2	
mov eax,3
mov ebx,0
mov ecx,%1
mov edx,%2
int 80h		
%endmacro

	
section .text

global _start
_start:

;---------------------------------------------MENU----------------------------------------------------------

	disp msg4, msglen4
	accept opt, 2

	sub byte[opt], 30h

	;mov eax, [opt]

	cmp byte[opt], 01h
	jz hextobcd
	cmp byte[opt], 02h
	jz bcdtohex
	cmp byte[opt], 03h
	jz exit

;---------------------------------------------Hex to BCD----------------------------------------------------

	hextobcd:

		disp msg, msglen

		accept num, 5

		call ascii_to_original

		mov eax, 0
		mov eax, ebx
		mov cx, 10
		mov byte[cnt], 00h

		l1:
			mov edx, 0
			div cx
			push dx		;Pushing Remainder
			inc byte[cnt]
			cmp ax, 0
			jnz l1

		disp msg2, msglen2

		l6:
			pop dx
			add dl, 30h
			mov [num1], dl
			disp num1, 1
			dec byte[cnt]
			jnz l6

		disp nwline, nwlen

	jmp _start

;---------------------------------------------BCD to Hex----------------------------------------------------

	bcdtohex:
	 
		disp msg3, msglen3

		accept num2, 6

		mov esi, num2
		mov eax, 0		;clear since it is used by default by mul instruction
		mov ebx, 10		;since we multiply 10 according to places. E.g. unit place -> *1 ten's place -> *10 etc. and we add
		mov ecx, 05		;5 digit number max. (65,535)

		l5:
			mov edx, 0	;before multiplication, initialize
			mul ebx		
			mov edx, 0	;To clear higher bytes since we are doing operaiton on dl
			mov dl, [esi]	
			sub dl, 30h
			add eax, edx
			inc esi
			dec ecx
		jnz l5


		mov [numbuff], eax

		disp msg5, msglen5

		call original_to_ascii


		disp nwline, nwlen

	jmp _start
		
;Example: 18 BCD to 12h Hex
;8*1 = 08 = 0000 1000
;	  +
;1*10= 0A = 0000 1010
; 	   -----------
;	    0001 0010	

;For 18, it will be stored as: [30][30][30][31][38]
;Step by Step execution:

; (1) eax = 0 dl = 0
;     eax = 0

; (2), (3) same as above since 0 (30)

; (4) eax = 0 dl = 01
;     eax = 01

; (5) eax = 10 (i.e. 0A) dl = 08
;     eax = 12 (i.e. 00C0)     


;---------------------------------------------Exit----------------------------------------------------------

	exit:
		mov eax, 1
		mov ebx, 0
		int 80h

;---------------------------------------------ASCII to Original--------------------------------------------

ascii_to_original:
			
	mov edi,num
	mov ecx,2
	mov bx,0
			
	l2:
		rol bx,4
		mov al,[edi]
		cmp al,39h
	jbe l3
				
		sub al,07h
				
	l3:
		sub al,30h
		mov ah,0h
		add bx,ax
		inc edi
	loop l2
				
	ret

;---------------------------------------------Original to ASCII--------------------------------------------

original_to_ascii:

	;push ecx
	disp nwline, nwlen
	;pop ecx
	mov edi, numbuff
	mov bx, [edi]
	mov ecx, 5
	mov esi, num3

	l7:
		rol bx, 4
		mov dl, bl
		and dl, 0fh
		cmp dl, 09h
	jbe l8
		add dl, 07h

		l8:
		add dl, 30h
		mov [esi], dl
		inc esi
	loop l7

		;inc edi
		;add edi, 4

	disp num3, 4

	ret