; this is a program in 8086 assembly language that
; accepts a character string from the keyboard and
; stores it in the string array. the program then converts
; all the lower case characters of the string to upper case.
; if the string is empty (null), it doesn't do anything.

name "neyse artik"

org 100h


jmp start


; first byte is buffer size,
; second byte will hold number
; of used bytes for string,
; all other bytes are for characters:
string db 20, 22 dup('?')
string2 db 20, 22 dup('?')
data1 db 20,20 dup(?)

new_line db 0Dh,0Ah, '$' ; new line code.

start:

; int 21h / ah=0ah - input of a string to ds:dx,
; fist byte is buffer size, second byte is number
; of chars actually read. does not add '$' in the
; end of string. to print using int 21h / ah=09h
; you must set dollar sign at the end of it and
; start printing from address ds:dx + 2.

lea dx, string

mov ah, 0ah
int 21h

mov bx, dx
;mov ah, 0
mov ax, ds:[bx+1]
add bx, ax ; point to end of string.
mov data1,bx

;mov byte ptr [bx+2], '$' ; put dollar to the end.

; int 21h / ah=09h - output of a string at ds:dx.
; string must be terminated by '$' sign.
lea dx, new_line
mov ah, 09h
int 21h


lea dx, string2

mov ah, 0ah
int 21h



mov bx, dx
;mov ah, 0
mov ax, ds:[bx+1]
add bx, ax ; point to end of string.

lea dx, new_line
mov ah, 09h
int 21h

;lea bx, string

;mov ch, 0
;mov cl, [bx+1] ; get string size.

;jcxz null ; is string is empty?

;add bx, 2 ; skip control chars.

mov data1,bx
mov dx, data1



lea dx, data1
mov ah, 09h

int 21h


; wait for any key press....
mov ah, 0
int 16h


null:
ret ; return to operating system.