When I add two values in 16 bit assembly, what is the best way to print the result to console?
At the moment I have this code:
;;---CODE START---;;
mov ax, 1 ;put 1 into ax
add ax, 2 ; add 2 to ax current value
mov ah,2 ; 2 is the function number of output char in the DOS Services.
mov dl, ax ; DL takes the value.
int 21h ; calls DOS Services
mov ah,4Ch ; 4Ch is the function number for exit program in DOS Services.
int 21h ; function 4Ch doesn't care about anything in the registers.
;;---CODE END---;;
I think that dl value should be in ASCII code, but I'm not sure how to convert ax value after addition into ASCII.
You basically want to divide by 10, print the remainder (one digit), and then repeat with the quotient.
; assume number is in eax
mov ecx, 10
loophere:
mov edx, 0
div ecx
; now eax <-- eax/10
; edx <-- eax % 10
; print edx
; this is one digit, which we have to convert to ASCII
; the print routine uses edx and eax, so let's push eax
; onto the stack. we clear edx at the beginning of the
; loop anyway, so we don't care if we much around with it
push eax
; convert dl to ascii
add dl, '0'
mov ah,2 ; 2 is the function number of output char in the DOS Services.
int 21h ; calls DOS Services
; now restore eax
pop eax
; if eax is zero, we can quit
cmp eax, 0
jnz loophere
As a side note, you have a bug in your code right here:
mov ax, 1 ;put 1 into ax
add ax, 2 ; add 2 to ax current value
mov ah,2 ; 2 is the function number of output char in the DOS Services.
mov dl, ax ; DL takes the value.
You put 2 in ah, and then you put ax in dl. You're basically junking ax before printing it.
You also have a size mismatch since dl is 8 bits wide and ax is 16 bits wide.
What you should do is flip the last two lines and fix the size mismatch:
mov ax, 1 ;put 1 into ax
add ax, 2 ; add 2 to ax current value
mov dl, al ; DL takes the value.
mov ah,2 ; 2 is the function number of output char in the DOS Services.