adding two base-10 numbers

[b][red]This message was edited by stevem222 at 2002-9-16 7:42:13[/red][/b][hr]
[b][red]This message was edited by stevem222 at 2002-9-16 7:37:27[/red][/b][hr]
I am very new at this. Just started taking the Assembler class at school. We are tasked with writing a a recursive program, adding two base-10 numbers (from 1-8), and displaying the answer. The way this is written, it must prompt you for the base-10 numbers. Here is what i have put together thus far, and I am sure it is a far cry from what it should be. Could anybody help me? Thanks

P.S., if there is an easier way of going about this (using turbo assembler) please let me know.

;;program add (a sub-procedure)
;;Stephen Madden

.model small
.stack 100h
Message DB 'Enter a base-10 numeric digit: $'
return DB, 13, 10, '$'
Addup proc

mov ax, @data
mov ds, ax; Initialize DS

mov DX,offset Message
mov ah, 9h
int 21h

mov ah, 1
int 21h; puts character in al

sub al, 30h
mov bl, al

mov dx,offset return
mov ah, 9h
int 21h; fist do carriage return/line feed
mov dx,offset message
int 21h; displays message (ah is still 9)
mov ah, 1; get character input
int 21h; puts character in al

;convert to number, store in al
sub al, 30h

;add the two numbers, result in bl
add bl, al
;if using td, observe register bl

cmp bl, 10
jl outdigit
mov ah, 2
mov dl, bl
int 21h; output the leading '1'

;return to DOS (by hand)
mov ah, 4ch
int 21h

Addup endp
end Addup

Sign In or Register to comment.

Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!