Bismillahir Rahmanir Rahim
Here the first array is ar=[1,2,3,4,5]
2nd is br=[6,7,8,9,10]
3rd is cr=[11,12,13,14,15]
4th is dr=[16,17,18,19,20]
the equation is sr=ar+br+cr-dr.
I added a extra function to print string.
mov ah,09
mov dx,offset msg ;msg='The result are= '
int 21h
mov ah,09
mov dx,offset coma ;coma=' , '
int 21h
I think, you understand it why i use this string. I am not use this two string the output will be show as
2d4d6d8d10d
instead of this The result is = 2d, 4d, 6d, 8d, 10d
solution:
if 1 include mylib1.lib endif mydata segment aR db 1,2,3,4,5 bR db 6,7,8,9,10 cR db 11,12,13,14,15 dr db 16,17,18,19,20 sR db 0,0,0,0,0 msg db "The result are= $" coma db " , $" COUNT EQU 05 mydata ends mystack segment db 100 dup(0) mystack ends mycode segment assume cs:mycode, ds:mydata, ss:mystack start: mov ax, seg mydata mov ds, ax mov ax, seg mystack mov ss, ax MOV CX,COUNT MOV SI,0000 mov ah,09 mov dx,offset msg ;solved by adhoa int 21h BACK: mov al,[si+ar] add al,[si+br] add al,[si+cr] sub al,[si+dr] mov [si+sr], al inc si push cx mov bh,0 mov bl,al printd pop cx mov ah,09 mov dx,offset coma ;solved by adhoa int 21h loop back mov ax, 4c00h int 21h printfd mycode ends end start
if you have to face any problem to solve or understand it ,comment below or contact admain in facebook .
Best of luck.
