using ax instead of dx with interrupt 21

Question

In the following code

data segment
    ; add your data here!
    num db 0,0,0,0 
    sum db 0 
    str db "Sum is : $"
ends

stack segment
    dw   128  dup(0)
ends

code segment
start:
; set segment registers:
    mov ax, data
    mov ds, ax
    mov es, ax

    ;;read array from input     
    mov cx,4;set loop counter
    L1:
    mov ah,7;interupt 7 use for reading character without echo
    int 21h  
    mov num,al;mov al to num 
    add sum,al
    inc num;nex element
    LOOP L1 
    sub num,4;go to first position
    ;;;;;;;;;;;;;;;;;;;;;;;;

    ;;show sum  
    lea dx,str;;-----------------I'm changing this line-----------------------
    mov ah,9;interupt 9 for writing string
    int 21h 
    ;;;;;;;;;;;           

    mov ax, 4c00h ; exit to operating system.
    int 21h    
ends

end start ; set entry point and stop the assembler.

if I change dx to ax

lea dx,str --> lea ax,str

The out put become 1 'Sum is But if I use lea ax,str It is correct > Sum is :

I cant figure out the reason !

Why changing dx to ax causes the wrong output ?

Answer 1

You should not randomly exchange AX for DX. The DOS function to print a string uses DX and that's final!
From your program I see that nothing you do with inputting and summing has any effect on the string you display afterwards.
If with your assembler mov num,al changes the content at the address of 'num' then the line inc num;nex element will certainly not advance to the next element but rather augment the content by 1. The same applies to sub num,4;go to first position